Implement Homestuck's Catenative Doomsday Dice CascaderRoll Dungeons and Dragons diceRole-Playing Game Dice-RollingSimulate a game of CrapsIterated Dice RollingRoll an ASCII dieHow many dice can you roll without rolling the most probable numberFrequency Distribution of Mixed Dice RollsAverage Idle-Dice RollingSimulating Exploding Dice

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Implement Homestuck's Catenative Doomsday Dice Cascader


Roll Dungeons and Dragons diceRole-Playing Game Dice-RollingSimulate a game of CrapsIterated Dice RollingRoll an ASCII dieHow many dice can you roll without rolling the most probable numberFrequency Distribution of Mixed Dice RollsAverage Idle-Dice RollingSimulating Exploding Dice






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









29














$begingroup$


Challenge



I'm under attack by the larcenous Midnight Crew and I need to summon the Catenative Doomsday Dice Cascader in order to defend myself. Since I'm low on space, I need the code to be as short as possible.



The algorithm for the Catenative Doomsday Dice Cascader is as follows:



First, the six sided die in the Prime Bubble is rolled, and the result will determine how many iterations of the next step take place.



Start with a six sided die. For as many times as the roll of the Prime Bubble die, multiply the number of sides on the next die by the result of the roll of the current die.
For example, if on your first roll of the six-sided die your roll is 2, then your next die will have 6*2 = 12 sides.



Your goal is to write a function or program that takes no input and outputs the final result of the last die rolled. Since this is code-golf, the lowest byte count in each language wins!



Examples



Example #1 (Taken directly from the link above):



The Prime Bubble rolls a 6, meaning that the Cascader will iterate six times

#1: We always start with a 6 sided die, and it rolls a 2, so the next die has 6x2=12 sides
#2: The 12 sided die rolls an 8, meaning that the third die has 12x8=96 sides
#3: The 96 sided die rolls a 35, meaning that die 4 has 96x35=3360 sides
#4: The 3360 sided die rolls a 2922, so die 5 has 3360x2922 = 9,817,920 sides
#5: The 9.8 million sided die rolls a 5,101,894, so the final die has 50,089,987,140,480 sides
#6: The 50 trillion sided die rolls a one. Hooray.
Since the last die rolled gave a 1, your function or program should output 1.


Example #2



The Prime Bubble rolls a 2, meaning that the Cascader will iterate twice.

#1: We always start with a 6 sided die, and it rolls a 4, so the next die has 6x4 = 24 sides
#2: The 24 sided die rolls a 14

Since the last die rolled gave a 14, your function or program should output 14.









share|improve this question












$endgroup$










  • 4




    $begingroup$
    Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
    $endgroup$
    – Kjetil S.
    Jun 1 at 11:53






  • 6




    $begingroup$
    @KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
    $endgroup$
    – Jonathan Allan
    Jun 1 at 12:50







  • 3




    $begingroup$
    Is this supposed to be random? The question doesn't mention anything about randomness at all?
    $endgroup$
    – Sriotchilism O'Zaic
    Jun 1 at 16:35






  • 10




    $begingroup$
    @SriotchilismO'Zaic Dice-rolling implies randomness.
    $endgroup$
    – mbomb007
    Jun 1 at 21:34






  • 6




    $begingroup$
    @SriotchilismO'Zaic xkcd.com/221
    $endgroup$
    – Neyt
    Jun 3 at 9:52

















29














$begingroup$


Challenge



I'm under attack by the larcenous Midnight Crew and I need to summon the Catenative Doomsday Dice Cascader in order to defend myself. Since I'm low on space, I need the code to be as short as possible.



The algorithm for the Catenative Doomsday Dice Cascader is as follows:



First, the six sided die in the Prime Bubble is rolled, and the result will determine how many iterations of the next step take place.



Start with a six sided die. For as many times as the roll of the Prime Bubble die, multiply the number of sides on the next die by the result of the roll of the current die.
For example, if on your first roll of the six-sided die your roll is 2, then your next die will have 6*2 = 12 sides.



Your goal is to write a function or program that takes no input and outputs the final result of the last die rolled. Since this is code-golf, the lowest byte count in each language wins!



Examples



Example #1 (Taken directly from the link above):



The Prime Bubble rolls a 6, meaning that the Cascader will iterate six times

#1: We always start with a 6 sided die, and it rolls a 2, so the next die has 6x2=12 sides
#2: The 12 sided die rolls an 8, meaning that the third die has 12x8=96 sides
#3: The 96 sided die rolls a 35, meaning that die 4 has 96x35=3360 sides
#4: The 3360 sided die rolls a 2922, so die 5 has 3360x2922 = 9,817,920 sides
#5: The 9.8 million sided die rolls a 5,101,894, so the final die has 50,089,987,140,480 sides
#6: The 50 trillion sided die rolls a one. Hooray.
Since the last die rolled gave a 1, your function or program should output 1.


Example #2



The Prime Bubble rolls a 2, meaning that the Cascader will iterate twice.

#1: We always start with a 6 sided die, and it rolls a 4, so the next die has 6x4 = 24 sides
#2: The 24 sided die rolls a 14

Since the last die rolled gave a 14, your function or program should output 14.









share|improve this question












$endgroup$










  • 4




    $begingroup$
    Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
    $endgroup$
    – Kjetil S.
    Jun 1 at 11:53






  • 6




    $begingroup$
    @KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
    $endgroup$
    – Jonathan Allan
    Jun 1 at 12:50







  • 3




    $begingroup$
    Is this supposed to be random? The question doesn't mention anything about randomness at all?
    $endgroup$
    – Sriotchilism O'Zaic
    Jun 1 at 16:35






  • 10




    $begingroup$
    @SriotchilismO'Zaic Dice-rolling implies randomness.
    $endgroup$
    – mbomb007
    Jun 1 at 21:34






  • 6




    $begingroup$
    @SriotchilismO'Zaic xkcd.com/221
    $endgroup$
    – Neyt
    Jun 3 at 9:52













29












29








29


3



$begingroup$


Challenge



I'm under attack by the larcenous Midnight Crew and I need to summon the Catenative Doomsday Dice Cascader in order to defend myself. Since I'm low on space, I need the code to be as short as possible.



The algorithm for the Catenative Doomsday Dice Cascader is as follows:



First, the six sided die in the Prime Bubble is rolled, and the result will determine how many iterations of the next step take place.



Start with a six sided die. For as many times as the roll of the Prime Bubble die, multiply the number of sides on the next die by the result of the roll of the current die.
For example, if on your first roll of the six-sided die your roll is 2, then your next die will have 6*2 = 12 sides.



Your goal is to write a function or program that takes no input and outputs the final result of the last die rolled. Since this is code-golf, the lowest byte count in each language wins!



Examples



Example #1 (Taken directly from the link above):



The Prime Bubble rolls a 6, meaning that the Cascader will iterate six times

#1: We always start with a 6 sided die, and it rolls a 2, so the next die has 6x2=12 sides
#2: The 12 sided die rolls an 8, meaning that the third die has 12x8=96 sides
#3: The 96 sided die rolls a 35, meaning that die 4 has 96x35=3360 sides
#4: The 3360 sided die rolls a 2922, so die 5 has 3360x2922 = 9,817,920 sides
#5: The 9.8 million sided die rolls a 5,101,894, so the final die has 50,089,987,140,480 sides
#6: The 50 trillion sided die rolls a one. Hooray.
Since the last die rolled gave a 1, your function or program should output 1.


Example #2



The Prime Bubble rolls a 2, meaning that the Cascader will iterate twice.

#1: We always start with a 6 sided die, and it rolls a 4, so the next die has 6x4 = 24 sides
#2: The 24 sided die rolls a 14

Since the last die rolled gave a 14, your function or program should output 14.









share|improve this question












$endgroup$




Challenge



I'm under attack by the larcenous Midnight Crew and I need to summon the Catenative Doomsday Dice Cascader in order to defend myself. Since I'm low on space, I need the code to be as short as possible.



The algorithm for the Catenative Doomsday Dice Cascader is as follows:



First, the six sided die in the Prime Bubble is rolled, and the result will determine how many iterations of the next step take place.



Start with a six sided die. For as many times as the roll of the Prime Bubble die, multiply the number of sides on the next die by the result of the roll of the current die.
For example, if on your first roll of the six-sided die your roll is 2, then your next die will have 6*2 = 12 sides.



Your goal is to write a function or program that takes no input and outputs the final result of the last die rolled. Since this is code-golf, the lowest byte count in each language wins!



Examples



Example #1 (Taken directly from the link above):



The Prime Bubble rolls a 6, meaning that the Cascader will iterate six times

#1: We always start with a 6 sided die, and it rolls a 2, so the next die has 6x2=12 sides
#2: The 12 sided die rolls an 8, meaning that the third die has 12x8=96 sides
#3: The 96 sided die rolls a 35, meaning that die 4 has 96x35=3360 sides
#4: The 3360 sided die rolls a 2922, so die 5 has 3360x2922 = 9,817,920 sides
#5: The 9.8 million sided die rolls a 5,101,894, so the final die has 50,089,987,140,480 sides
#6: The 50 trillion sided die rolls a one. Hooray.
Since the last die rolled gave a 1, your function or program should output 1.


Example #2



The Prime Bubble rolls a 2, meaning that the Cascader will iterate twice.

#1: We always start with a 6 sided die, and it rolls a 4, so the next die has 6x4 = 24 sides
#2: The 24 sided die rolls a 14

Since the last die rolled gave a 14, your function or program should output 14.






code-golf random






share|improve this question
















share|improve this question













share|improve this question




share|improve this question








edited Jun 4 at 11:26









LegionMammal978

15.5k4 gold badges18 silver badges53 bronze badges




15.5k4 gold badges18 silver badges53 bronze badges










asked Jun 1 at 2:48









Bazinga_9000Bazinga_9000

4771 gold badge5 silver badges10 bronze badges




4771 gold badge5 silver badges10 bronze badges










  • 4




    $begingroup$
    Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
    $endgroup$
    – Kjetil S.
    Jun 1 at 11:53






  • 6




    $begingroup$
    @KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
    $endgroup$
    – Jonathan Allan
    Jun 1 at 12:50







  • 3




    $begingroup$
    Is this supposed to be random? The question doesn't mention anything about randomness at all?
    $endgroup$
    – Sriotchilism O'Zaic
    Jun 1 at 16:35






  • 10




    $begingroup$
    @SriotchilismO'Zaic Dice-rolling implies randomness.
    $endgroup$
    – mbomb007
    Jun 1 at 21:34






  • 6




    $begingroup$
    @SriotchilismO'Zaic xkcd.com/221
    $endgroup$
    – Neyt
    Jun 3 at 9:52












  • 4




    $begingroup$
    Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
    $endgroup$
    – Kjetil S.
    Jun 1 at 11:53






  • 6




    $begingroup$
    @KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
    $endgroup$
    – Jonathan Allan
    Jun 1 at 12:50







  • 3




    $begingroup$
    Is this supposed to be random? The question doesn't mention anything about randomness at all?
    $endgroup$
    – Sriotchilism O'Zaic
    Jun 1 at 16:35






  • 10




    $begingroup$
    @SriotchilismO'Zaic Dice-rolling implies randomness.
    $endgroup$
    – mbomb007
    Jun 1 at 21:34






  • 6




    $begingroup$
    @SriotchilismO'Zaic xkcd.com/221
    $endgroup$
    – Neyt
    Jun 3 at 9:52







4




4




$begingroup$
Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
$endgroup$
– Kjetil S.
Jun 1 at 11:53




$begingroup$
Whats the maximum output? If every roll results in the maximum side? I think it's 7958661109946400884391936 = ((((6^2)^2)^2)^2)^2 = 6^(2^5) = 6^32
$endgroup$
– Kjetil S.
Jun 1 at 11:53




6




6




$begingroup$
@KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
$endgroup$
– Jonathan Allan
Jun 1 at 12:50





$begingroup$
@KjetilS. Indeed, and the likelihood of that output should be $frac16*prod_i=0^5(6^2^i)=frac16times 6times 6^2times 6^4times 6^8times 6^16times 6^32=frac12155416739906037495048372267884096782336$
$endgroup$
– Jonathan Allan
Jun 1 at 12:50





3




3




$begingroup$
Is this supposed to be random? The question doesn't mention anything about randomness at all?
$endgroup$
– Sriotchilism O'Zaic
Jun 1 at 16:35




$begingroup$
Is this supposed to be random? The question doesn't mention anything about randomness at all?
$endgroup$
– Sriotchilism O'Zaic
Jun 1 at 16:35




10




10




$begingroup$
@SriotchilismO'Zaic Dice-rolling implies randomness.
$endgroup$
– mbomb007
Jun 1 at 21:34




$begingroup$
@SriotchilismO'Zaic Dice-rolling implies randomness.
$endgroup$
– mbomb007
Jun 1 at 21:34




6




6




$begingroup$
@SriotchilismO'Zaic xkcd.com/221
$endgroup$
– Neyt
Jun 3 at 9:52




$begingroup$
@SriotchilismO'Zaic xkcd.com/221
$endgroup$
– Neyt
Jun 3 at 9:52










20 Answers
20






active

oldest

votes


















8
















$begingroup$


Python 2, 76 69 bytes





from random import*
R=randint
v=6;exec"p=R(1,v);v*=p;"*R(1,6)
print p


Try it online!






share|improve this answer












$endgroup$






















    8
















    $begingroup$


    Perl 6, 43 37 bytes



    -6 bytes thanks to nwellnhof





    (6,roll 1..[*] @_:...*)[1+6.rand]


    Try it online!



    Anonymous code block that returns the doomsday dice result.



    Explanation:



     # Anonymous code block
    ( )[1+6.rand] # Take a random number from
    ...* # The infinite list of
    6,roll 1..[*] @_: # Cascading dice values
    6, # Starting from 6
    roll : # And choosing a random value from
    1.. # One to
    [*] @_ # The product of every value so far





    share|improve this answer












    $endgroup$






















      7
















      $begingroup$


      Wolfram Language (Mathematica), 43 bytes



      (r=RandomInteger)@1,Nest[r@1,##&,6,r@5]


      Try it online!






      share|improve this answer












      $endgroup$






















        5
















        $begingroup$


        J, 21 bytes



        1+[:?(*1+?)^:(?`])@6x


        Try it online!



        +6 bytes thanks to a logic problem spotted by FrownyFrog



        NOTE: J has no niladic verbs. However, this verb will work the same no matter what argument you give it. In the TIO example, I'm calling it with 0, but I could have used 99 or '' just as well.



        how




        • 1+ add one to...


        • [:? a single roll of an n-sided die (sides reading 0 to n-1), where the number n is determined by...


        • (*1+?) take the current argument y and roll ? to produce a random number between 0 and y-1. 1+ makes that 1 to y, inclusive. Finally the * creates a J hook, which will multiply that by y again.


        • ^: do the above this many times...


        • (?`]) ? roll the initial argument, which is 6, to determine how many times to repeat. If we roll 0 (corresponding to a 1 on the Prime Bubble), the argument will pass through unchanged. The ] indicates that 6, unchanged, will be the starting value of repeated (*1+?) verb that determines the die value for the final roll.


        • @6x attaches the constant verb 6, so that we can call it with anything, and the x forces J to use extended integer computation which we need for the possibly huge numbers.





        share|improve this answer












        $endgroup$














        • $begingroup$
          in this case 0 executes the previous verb once, 1 twice, etc why is that?
          $endgroup$
          – FrownyFrog
          Jun 1 at 23:15











        • $begingroup$
          because i made a mistake :(. will fix soon.
          $endgroup$
          – Jonah
          Jun 1 at 23:42










        • $begingroup$
          Fixed now. Thanks.
          $endgroup$
          – Jonah
          Jun 1 at 23:55


















        4
















        $begingroup$


        K (oK), 32 bytes



        Solution:



        *|x,1+1?x:(*).x/[*a;6,a:1+1?6]


        Try it online!



        Start with 6 and "1 choose 6", iterate over "1 choose 6" times:



        *|x,1+1?x:(*).x/[*a;6,a:1+1?6] / the solution
        /[n; c ] / iterate over lambda n times with starting condition c
        1?6 / 1 choose 6, between 0..5 (returns a list of 1 item)
        1+ / add 1 (so between 1..6)
        a: / store as 'a'
        6, / prepend 6, the number of sides of the first dice
        *a / we are iterating between 0 and 5 times, take first (*)
        (*).x / multi-argument apply (.) multiply (*) to x, e.g. 6*2 => 12
        x: / save that as 'x'
        1? / 1 choose x, between 0..x-1
        1+ / add 1 (so between 1..x)
        x, / prepend x
        *| / reverse-first aka 'last'


        You can see the iterations by switching the over for a scan, e.g.



        (6 3 / 1 choose 6 => 3, so perform 3 iterations
        18 15 / 1 choose (6*3=18) => 15
        270 31 / 1 choose (18*15=270) => 31
        8370 5280) / 1 choose (270*31=8730) => 5280





        share|improve this answer












        $endgroup$










        • 1




          $begingroup$
          (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
          $endgroup$
          – ngn
          Jun 7 at 23:16


















        4
















        $begingroup$


        Jelly, 9 bytes



        6X×$5СXX


        A niladic Link which yields a positive integer.



        Try it online!



        Saving a byte over the more obvious 6X×$6X’¤¡X



        How?



        6X×$5СXX - Link: no arguments
        6 - initialise left argument to 6
        5С - repeat five times, collecting up as we go: -> a list of 6 possible dice sizes
        $ - last two links as a monad = f(v): e.g [6,18,288,4032,1382976,216315425088]
        X - random number in [1,v] or [6,6,6,6,6,6]
        × - multiply (by v) or [6,36,1296,1679616,2821109907456,7958661109946400884391936]
        X - random choice (since the input is now a list) -> faces (on final die)
        X - random number in [1,faces]





        share|improve this answer












        $endgroup$














        • $begingroup$
          Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
          $endgroup$
          – Nick Kennedy
          Jun 1 at 13:50










        • $begingroup$
          Heh, I missed that you'd posted almost that exact answer!
          $endgroup$
          – Jonathan Allan
          Jun 1 at 14:01


















        3
















        $begingroup$


        05AB1E, 10 bytes



        X6DLΩF*DLΩ


        The random choice builtin for large lists is pretty slow, so may result in a timeout if the Prime Bubble roll is for example a 6.



        Try it online or try it online with added prints to see the rolls. (TIO uses the legacy version of 05AB1E, since it's slightly faster.)



        Explanation:





        X # Push a 1 to the stack
        6 # Push a 6 to the stack
        D # Push another 6 to the stack
        L # Pop the top 6, and push a list [1,2,3,4,5,6] to the stack
        Ω # Pop and push a random item from this list (this is out Prime Bubble roll)
        F # Loop that many times:
        * # Multiply the top two values on the stack
        # (which is why we had the initial 1 and duplicated 6 before the loop)
        D # Duplicate this result
        LΩ # Pop and push a random value from its ranged list again
        # (after the loop, output the top of the stack implicitly)





        share|improve this answer










        $endgroup$






















          3
















          $begingroup$


          Jelly, 9 bytes



          6×X$X’$¡X


          Try it online!



          Jonathan Allan's answer claims that it's




          Saving a byte over the more obvious 6X×$6X’¤¡X




          . In fact, we don't need to make such a big modification. Therefore, this is an alternative approach to Jonathan Allan's answer, and, also, a resting place for my initial invalid 6-byter. :(






          share|improve this answer












          $endgroup$






















            2
















            $begingroup$


            Perl 5, 54 bytes





            $a=6;map$a*=$r=1+int rand$a,0..rand 6;say 1+int rand$r


            Try it online!






            share|improve this answer










            $endgroup$














            • $begingroup$
              That's one rand too many, you should just say$r
              $endgroup$
              – Grimy
              Jun 1 at 19:25


















            2
















            $begingroup$


            Charcoal, 16 bytes



            ⊞υ⁶F⊕‽⁶⊞υ⊕‽ΠυI⊟υ


            Try it online! Link is to verbose version of code. Explanation:



            ⊞υ⁶


            Push 6 to the predefined list.



            F⊕‽⁶


            Repeat a random number of times from 1 to 6...



            ⊞υ⊕‽Πυ


            ... push a random number between 1 and the product of the list to the list.



            I⊟υ


            Output the last number pushed to the list.



            Alternative approach, also 16 bytes



            ≔⁶θF‽⁶≧×⊕‽θθI⊕‽θ


            Try it online! Link is to verbose version of code. Explanation:



            ≔⁶θ


            Set the number of sides to 6.



            F‽⁶


            Repeat a random number between 0 and 5 times...



            ≧×⊕‽θθ


            ... multiply the number of sides by a random number from 1 to the number of sides.



            I⊕‽θ


            Print a random number from 1 to the number of sides.






            share|improve this answer










            $endgroup$






















              2
















              $begingroup$


              Python 3, 76 bytes





              from random import*
              r=randint
              a=6
              for i in" "*r(0,5):a*=r(1,a)
              print(r(1,a))


              Try it online!



              -2 bytes thanks to TFeld






              share|improve this answer












              $endgroup$










              • 1




                $begingroup$
                -2 bytes by changing the import
                $endgroup$
                – TFeld
                Jun 1 at 7:34










              • $begingroup$
                @TFeld thanks [filler]
                $endgroup$
                – HyperNeutrino
                Jun 1 at 19:23


















              2
















              $begingroup$


              R, 43 bytes





              s=sample
              for(i in 1:s(k<-6))T=s(k<-k*T,1)
              T


              Try it online!



              k keeps track of the current number of faces on the die. Uses the fact that T is initialized as 1.



              I tried a few other things, but couldn't beat this simple, straightforward approach.






              share|improve this answer










              $endgroup$






















                1
















                $begingroup$


                Jelly, 10 bytes



                6×X$6X’¤¡X


                Try it online!



                Explanation



                 ¤ | Following as a nilad:
                6X | - A random number between 1 and 6
                ’ | - Decrease by 1 (call this N)
                6 | Now, start with 6
                $ ¡ | Repeat the following N times, as a monad
                × | - Multiply by:
                X | - A random number between 1 and the current total
                X | Finally, generate a random number between 1 and the output of the above loop





                share|improve this answer










                $endgroup$






















                  1
















                  $begingroup$


                  Ruby, 41 bytes





                  r=6;rand(2..7).timesr*=$s=rand 1..r;p$s


                  Try it online!



                  Explanation



                  r=6 # Set dice number to 6

                  rand(2..7).times # Repeat X times, where X=dice roll+1
                  r*=$s=rand 1..r # Multiply dice number by a dice roll
                  # Save the most recent dice roll

                  p$s # Print last dice roll (this is why
                  # we did the last step one extra time)





                  share|improve this answer










                  $endgroup$






















                    1
















                    $begingroup$

                    Java 10, 214 93 86 bytes





                    v->int r=6,n=0;for(var d=Math.random()*6;d-->0;n*=Math.random(),r*=++n)n=r;return n;


                    Try it online or try it online with additional print-lines to see the steps.



                    -128 bytes by using int instead of java.math.BigInteger. $6^32$, the largest possible result, doesn't fit inside an int nor long, which is why BigInteger was used initially. OP allowed to use int and assuming it's infinitely large, so that saved more than 125 bytes here. :) (Here the previous 214 bytes version using BigIntegers.)



                    Explanation:



                    v-> // Method with empty unused parameter & integer return-type
                    int r=6, // The range in which to roll, starting at 6
                    n=0; // The roll itself (which must be initialized, therefor is 0)
                    for(var d=Math.random()*6;// Roll the Prime Bubble Dice
                    d-->0 // Loop that many times:
                    ; // After every iteration:
                    n*=Math.random(), // Roll a random dice in the range [0, n)
                    r*=++n) // Increase `n` by 1 first with `++n`, so the range is [1,n]
                    // And then multiply `r` by `n` for the new range
                    n=r; // Set `n` to `r`
                    return n; // After the loop, return `n` as result





                    share|improve this answer












                    $endgroup$














                    • $begingroup$
                      Post the solution that doesn't do BigInteger as your competing solution.
                      $endgroup$
                      – Stackstuck
                      Jun 3 at 16:16










                    • $begingroup$
                      @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                      $endgroup$
                      – Kevin Cruijssen
                      Jun 3 at 16:26










                    • $begingroup$
                      I'll see if I can find anything on meta about this.
                      $endgroup$
                      – Stackstuck
                      Jun 3 at 16:29






                    • 1




                      $begingroup$
                      OP says don't worry about integer size limits. Use the int type.
                      $endgroup$
                      – Stackstuck
                      Jun 8 at 1:26






                    • 1




                      $begingroup$
                      @Stackstuck Done, and golfed 7 bytes in the process. :)
                      $endgroup$
                      – Kevin Cruijssen
                      Jun 8 at 12:00


















                    0
















                    $begingroup$


                    PHP, 59 bytes



                    $r=$q=rand(1,$s=6);while($l++<$q)$r=rand(1,$s*=$r);print$r;


                    expanded:



                    $r=$q=rand(1,$s=6);
                    while($l++<$q)$
                    r=rand(1,$s*=$r);
                    print$r;


                    Not sure if I'm supposed to include the open tag.



                    On my machine, it crashes if $s*$r is too large, so it doesn't print on $q>=5 sometimes... because the numbers get so big. Not sure of a fix.






                    share|improve this answer










                    $endgroup$






















                      0
















                      $begingroup$


                      Pyth, 14 bytes



                      uhO=*|Z6GO6hO6


                      Try it online!



                      uhO=*|Z6GO6hO6 
                      O6 Random number in range [0-6)
                      u Perform the following the above number of times...
                      hO6 ... with starting value G a random number in range [1-6]:
                      * G Multiply G with...
                      |Z6 The value of Z, or 6 if it's the first time through (Z is 0 at program start)
                      = Z Assign the above back into Z
                      O Random number in range [0-Z)
                      h Increment
                      Implicit print result of final iteration





                      share|improve this answer












                      $endgroup$






















                        0
















                        $begingroup$


                        C# (.NET Core), 136 bytes





                        class Astatic void Main()var r=new System.Random();int i=r.Next(6),j=6;while(i-->0)j*=1+r.Next(j);System.Console.Write(r.Next(j)+1);


                        Try it online!



                        I'm pretty sure this works, given the assumption of infinite integer length that we're fond of here. If I have to actually handle the overflow, I'd need to bust out an entirely different class.






                        share|improve this answer












                        $endgroup$














                        • $begingroup$
                          It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                          $endgroup$
                          – Kevin Cruijssen
                          Jun 3 at 9:34











                        • $begingroup$
                          @KevinCruijssen yes, that's the whole point of my comment.
                          $endgroup$
                          – Stackstuck
                          Jun 3 at 16:12


















                        0
















                        $begingroup$


                        Julia 1.0, 60 bytes





                        g(b=big(6),r=rand)=(for i in 1:r(0:5) b=b*r(1:b) end;r(1:b))


                        b=big(6) makes it work with arbitrary sized integers
                        Try it online!






                        share|improve this answer










                        $endgroup$






















                          0
















                          $begingroup$


                          Gaia, 13 bytes



                          :(ṛ×
                          6₅ṛ(↑ₓ(ṛ


                          Try it online!






                          share|improve this answer












                          $endgroup$
















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                            8
















                            $begingroup$


                            Python 2, 76 69 bytes





                            from random import*
                            R=randint
                            v=6;exec"p=R(1,v);v*=p;"*R(1,6)
                            print p


                            Try it online!






                            share|improve this answer












                            $endgroup$



















                              8
















                              $begingroup$


                              Python 2, 76 69 bytes





                              from random import*
                              R=randint
                              v=6;exec"p=R(1,v);v*=p;"*R(1,6)
                              print p


                              Try it online!






                              share|improve this answer












                              $endgroup$

















                                8














                                8










                                8







                                $begingroup$


                                Python 2, 76 69 bytes





                                from random import*
                                R=randint
                                v=6;exec"p=R(1,v);v*=p;"*R(1,6)
                                print p


                                Try it online!






                                share|improve this answer












                                $endgroup$




                                Python 2, 76 69 bytes





                                from random import*
                                R=randint
                                v=6;exec"p=R(1,v);v*=p;"*R(1,6)
                                print p


                                Try it online!







                                share|improve this answer















                                share|improve this answer




                                share|improve this answer








                                edited Jun 1 at 3:28

























                                answered Jun 1 at 3:12









                                Chas BrownChas Brown

                                6,8641 gold badge6 silver badges27 bronze badges




                                6,8641 gold badge6 silver badges27 bronze badges


























                                    8
















                                    $begingroup$


                                    Perl 6, 43 37 bytes



                                    -6 bytes thanks to nwellnhof





                                    (6,roll 1..[*] @_:...*)[1+6.rand]


                                    Try it online!



                                    Anonymous code block that returns the doomsday dice result.



                                    Explanation:



                                     # Anonymous code block
                                    ( )[1+6.rand] # Take a random number from
                                    ...* # The infinite list of
                                    6,roll 1..[*] @_: # Cascading dice values
                                    6, # Starting from 6
                                    roll : # And choosing a random value from
                                    1.. # One to
                                    [*] @_ # The product of every value so far





                                    share|improve this answer












                                    $endgroup$



















                                      8
















                                      $begingroup$


                                      Perl 6, 43 37 bytes



                                      -6 bytes thanks to nwellnhof





                                      (6,roll 1..[*] @_:...*)[1+6.rand]


                                      Try it online!



                                      Anonymous code block that returns the doomsday dice result.



                                      Explanation:



                                       # Anonymous code block
                                      ( )[1+6.rand] # Take a random number from
                                      ...* # The infinite list of
                                      6,roll 1..[*] @_: # Cascading dice values
                                      6, # Starting from 6
                                      roll : # And choosing a random value from
                                      1.. # One to
                                      [*] @_ # The product of every value so far





                                      share|improve this answer












                                      $endgroup$

















                                        8














                                        8










                                        8







                                        $begingroup$


                                        Perl 6, 43 37 bytes



                                        -6 bytes thanks to nwellnhof





                                        (6,roll 1..[*] @_:...*)[1+6.rand]


                                        Try it online!



                                        Anonymous code block that returns the doomsday dice result.



                                        Explanation:



                                         # Anonymous code block
                                        ( )[1+6.rand] # Take a random number from
                                        ...* # The infinite list of
                                        6,roll 1..[*] @_: # Cascading dice values
                                        6, # Starting from 6
                                        roll : # And choosing a random value from
                                        1.. # One to
                                        [*] @_ # The product of every value so far





                                        share|improve this answer












                                        $endgroup$




                                        Perl 6, 43 37 bytes



                                        -6 bytes thanks to nwellnhof





                                        (6,roll 1..[*] @_:...*)[1+6.rand]


                                        Try it online!



                                        Anonymous code block that returns the doomsday dice result.



                                        Explanation:



                                         # Anonymous code block
                                        ( )[1+6.rand] # Take a random number from
                                        ...* # The infinite list of
                                        6,roll 1..[*] @_: # Cascading dice values
                                        6, # Starting from 6
                                        roll : # And choosing a random value from
                                        1.. # One to
                                        [*] @_ # The product of every value so far






                                        share|improve this answer















                                        share|improve this answer




                                        share|improve this answer








                                        edited Jun 1 at 9:42

























                                        answered Jun 1 at 6:23









                                        Jo KingJo King

                                        33.1k4 gold badges75 silver badges142 bronze badges




                                        33.1k4 gold badges75 silver badges142 bronze badges
























                                            7
















                                            $begingroup$


                                            Wolfram Language (Mathematica), 43 bytes



                                            (r=RandomInteger)@1,Nest[r@1,##&,6,r@5]


                                            Try it online!






                                            share|improve this answer












                                            $endgroup$



















                                              7
















                                              $begingroup$


                                              Wolfram Language (Mathematica), 43 bytes



                                              (r=RandomInteger)@1,Nest[r@1,##&,6,r@5]


                                              Try it online!






                                              share|improve this answer












                                              $endgroup$

















                                                7














                                                7










                                                7







                                                $begingroup$


                                                Wolfram Language (Mathematica), 43 bytes



                                                (r=RandomInteger)@1,Nest[r@1,##&,6,r@5]


                                                Try it online!






                                                share|improve this answer












                                                $endgroup$




                                                Wolfram Language (Mathematica), 43 bytes



                                                (r=RandomInteger)@1,Nest[r@1,##&,6,r@5]


                                                Try it online!







                                                share|improve this answer















                                                share|improve this answer




                                                share|improve this answer








                                                edited Jun 1 at 11:47

























                                                answered Jun 1 at 11:40









                                                attinatattinat

                                                3,1063 silver badges11 bronze badges




                                                3,1063 silver badges11 bronze badges
























                                                    5
















                                                    $begingroup$


                                                    J, 21 bytes



                                                    1+[:?(*1+?)^:(?`])@6x


                                                    Try it online!



                                                    +6 bytes thanks to a logic problem spotted by FrownyFrog



                                                    NOTE: J has no niladic verbs. However, this verb will work the same no matter what argument you give it. In the TIO example, I'm calling it with 0, but I could have used 99 or '' just as well.



                                                    how




                                                    • 1+ add one to...


                                                    • [:? a single roll of an n-sided die (sides reading 0 to n-1), where the number n is determined by...


                                                    • (*1+?) take the current argument y and roll ? to produce a random number between 0 and y-1. 1+ makes that 1 to y, inclusive. Finally the * creates a J hook, which will multiply that by y again.


                                                    • ^: do the above this many times...


                                                    • (?`]) ? roll the initial argument, which is 6, to determine how many times to repeat. If we roll 0 (corresponding to a 1 on the Prime Bubble), the argument will pass through unchanged. The ] indicates that 6, unchanged, will be the starting value of repeated (*1+?) verb that determines the die value for the final roll.


                                                    • @6x attaches the constant verb 6, so that we can call it with anything, and the x forces J to use extended integer computation which we need for the possibly huge numbers.





                                                    share|improve this answer












                                                    $endgroup$














                                                    • $begingroup$
                                                      in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                      $endgroup$
                                                      – FrownyFrog
                                                      Jun 1 at 23:15











                                                    • $begingroup$
                                                      because i made a mistake :(. will fix soon.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:42










                                                    • $begingroup$
                                                      Fixed now. Thanks.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:55















                                                    5
















                                                    $begingroup$


                                                    J, 21 bytes



                                                    1+[:?(*1+?)^:(?`])@6x


                                                    Try it online!



                                                    +6 bytes thanks to a logic problem spotted by FrownyFrog



                                                    NOTE: J has no niladic verbs. However, this verb will work the same no matter what argument you give it. In the TIO example, I'm calling it with 0, but I could have used 99 or '' just as well.



                                                    how




                                                    • 1+ add one to...


                                                    • [:? a single roll of an n-sided die (sides reading 0 to n-1), where the number n is determined by...


                                                    • (*1+?) take the current argument y and roll ? to produce a random number between 0 and y-1. 1+ makes that 1 to y, inclusive. Finally the * creates a J hook, which will multiply that by y again.


                                                    • ^: do the above this many times...


                                                    • (?`]) ? roll the initial argument, which is 6, to determine how many times to repeat. If we roll 0 (corresponding to a 1 on the Prime Bubble), the argument will pass through unchanged. The ] indicates that 6, unchanged, will be the starting value of repeated (*1+?) verb that determines the die value for the final roll.


                                                    • @6x attaches the constant verb 6, so that we can call it with anything, and the x forces J to use extended integer computation which we need for the possibly huge numbers.





                                                    share|improve this answer












                                                    $endgroup$














                                                    • $begingroup$
                                                      in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                      $endgroup$
                                                      – FrownyFrog
                                                      Jun 1 at 23:15











                                                    • $begingroup$
                                                      because i made a mistake :(. will fix soon.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:42










                                                    • $begingroup$
                                                      Fixed now. Thanks.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:55













                                                    5














                                                    5










                                                    5







                                                    $begingroup$


                                                    J, 21 bytes



                                                    1+[:?(*1+?)^:(?`])@6x


                                                    Try it online!



                                                    +6 bytes thanks to a logic problem spotted by FrownyFrog



                                                    NOTE: J has no niladic verbs. However, this verb will work the same no matter what argument you give it. In the TIO example, I'm calling it with 0, but I could have used 99 or '' just as well.



                                                    how




                                                    • 1+ add one to...


                                                    • [:? a single roll of an n-sided die (sides reading 0 to n-1), where the number n is determined by...


                                                    • (*1+?) take the current argument y and roll ? to produce a random number between 0 and y-1. 1+ makes that 1 to y, inclusive. Finally the * creates a J hook, which will multiply that by y again.


                                                    • ^: do the above this many times...


                                                    • (?`]) ? roll the initial argument, which is 6, to determine how many times to repeat. If we roll 0 (corresponding to a 1 on the Prime Bubble), the argument will pass through unchanged. The ] indicates that 6, unchanged, will be the starting value of repeated (*1+?) verb that determines the die value for the final roll.


                                                    • @6x attaches the constant verb 6, so that we can call it with anything, and the x forces J to use extended integer computation which we need for the possibly huge numbers.





                                                    share|improve this answer












                                                    $endgroup$




                                                    J, 21 bytes



                                                    1+[:?(*1+?)^:(?`])@6x


                                                    Try it online!



                                                    +6 bytes thanks to a logic problem spotted by FrownyFrog



                                                    NOTE: J has no niladic verbs. However, this verb will work the same no matter what argument you give it. In the TIO example, I'm calling it with 0, but I could have used 99 or '' just as well.



                                                    how




                                                    • 1+ add one to...


                                                    • [:? a single roll of an n-sided die (sides reading 0 to n-1), where the number n is determined by...


                                                    • (*1+?) take the current argument y and roll ? to produce a random number between 0 and y-1. 1+ makes that 1 to y, inclusive. Finally the * creates a J hook, which will multiply that by y again.


                                                    • ^: do the above this many times...


                                                    • (?`]) ? roll the initial argument, which is 6, to determine how many times to repeat. If we roll 0 (corresponding to a 1 on the Prime Bubble), the argument will pass through unchanged. The ] indicates that 6, unchanged, will be the starting value of repeated (*1+?) verb that determines the die value for the final roll.


                                                    • @6x attaches the constant verb 6, so that we can call it with anything, and the x forces J to use extended integer computation which we need for the possibly huge numbers.






                                                    share|improve this answer















                                                    share|improve this answer




                                                    share|improve this answer








                                                    edited Jun 2 at 0:25

























                                                    answered Jun 1 at 4:15









                                                    JonahJonah

                                                    5,4372 gold badges13 silver badges25 bronze badges




                                                    5,4372 gold badges13 silver badges25 bronze badges














                                                    • $begingroup$
                                                      in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                      $endgroup$
                                                      – FrownyFrog
                                                      Jun 1 at 23:15











                                                    • $begingroup$
                                                      because i made a mistake :(. will fix soon.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:42










                                                    • $begingroup$
                                                      Fixed now. Thanks.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:55
















                                                    • $begingroup$
                                                      in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                      $endgroup$
                                                      – FrownyFrog
                                                      Jun 1 at 23:15











                                                    • $begingroup$
                                                      because i made a mistake :(. will fix soon.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:42










                                                    • $begingroup$
                                                      Fixed now. Thanks.
                                                      $endgroup$
                                                      – Jonah
                                                      Jun 1 at 23:55















                                                    $begingroup$
                                                    in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                    $endgroup$
                                                    – FrownyFrog
                                                    Jun 1 at 23:15





                                                    $begingroup$
                                                    in this case 0 executes the previous verb once, 1 twice, etc why is that?
                                                    $endgroup$
                                                    – FrownyFrog
                                                    Jun 1 at 23:15













                                                    $begingroup$
                                                    because i made a mistake :(. will fix soon.
                                                    $endgroup$
                                                    – Jonah
                                                    Jun 1 at 23:42




                                                    $begingroup$
                                                    because i made a mistake :(. will fix soon.
                                                    $endgroup$
                                                    – Jonah
                                                    Jun 1 at 23:42












                                                    $begingroup$
                                                    Fixed now. Thanks.
                                                    $endgroup$
                                                    – Jonah
                                                    Jun 1 at 23:55




                                                    $begingroup$
                                                    Fixed now. Thanks.
                                                    $endgroup$
                                                    – Jonah
                                                    Jun 1 at 23:55











                                                    4
















                                                    $begingroup$


                                                    K (oK), 32 bytes



                                                    Solution:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6]


                                                    Try it online!



                                                    Start with 6 and "1 choose 6", iterate over "1 choose 6" times:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6] / the solution
                                                    /[n; c ] / iterate over lambda n times with starting condition c
                                                    1?6 / 1 choose 6, between 0..5 (returns a list of 1 item)
                                                    1+ / add 1 (so between 1..6)
                                                    a: / store as 'a'
                                                    6, / prepend 6, the number of sides of the first dice
                                                    *a / we are iterating between 0 and 5 times, take first (*)
                                                    (*).x / multi-argument apply (.) multiply (*) to x, e.g. 6*2 => 12
                                                    x: / save that as 'x'
                                                    1? / 1 choose x, between 0..x-1
                                                    1+ / add 1 (so between 1..x)
                                                    x, / prepend x
                                                    *| / reverse-first aka 'last'


                                                    You can see the iterations by switching the over for a scan, e.g.



                                                    (6 3 / 1 choose 6 => 3, so perform 3 iterations
                                                    18 15 / 1 choose (6*3=18) => 15
                                                    270 31 / 1 choose (18*15=270) => 31
                                                    8370 5280) / 1 choose (270*31=8730) => 5280





                                                    share|improve this answer












                                                    $endgroup$










                                                    • 1




                                                      $begingroup$
                                                      (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                      $endgroup$
                                                      – ngn
                                                      Jun 7 at 23:16















                                                    4
















                                                    $begingroup$


                                                    K (oK), 32 bytes



                                                    Solution:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6]


                                                    Try it online!



                                                    Start with 6 and "1 choose 6", iterate over "1 choose 6" times:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6] / the solution
                                                    /[n; c ] / iterate over lambda n times with starting condition c
                                                    1?6 / 1 choose 6, between 0..5 (returns a list of 1 item)
                                                    1+ / add 1 (so between 1..6)
                                                    a: / store as 'a'
                                                    6, / prepend 6, the number of sides of the first dice
                                                    *a / we are iterating between 0 and 5 times, take first (*)
                                                    (*).x / multi-argument apply (.) multiply (*) to x, e.g. 6*2 => 12
                                                    x: / save that as 'x'
                                                    1? / 1 choose x, between 0..x-1
                                                    1+ / add 1 (so between 1..x)
                                                    x, / prepend x
                                                    *| / reverse-first aka 'last'


                                                    You can see the iterations by switching the over for a scan, e.g.



                                                    (6 3 / 1 choose 6 => 3, so perform 3 iterations
                                                    18 15 / 1 choose (6*3=18) => 15
                                                    270 31 / 1 choose (18*15=270) => 31
                                                    8370 5280) / 1 choose (270*31=8730) => 5280





                                                    share|improve this answer












                                                    $endgroup$










                                                    • 1




                                                      $begingroup$
                                                      (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                      $endgroup$
                                                      – ngn
                                                      Jun 7 at 23:16













                                                    4














                                                    4










                                                    4







                                                    $begingroup$


                                                    K (oK), 32 bytes



                                                    Solution:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6]


                                                    Try it online!



                                                    Start with 6 and "1 choose 6", iterate over "1 choose 6" times:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6] / the solution
                                                    /[n; c ] / iterate over lambda n times with starting condition c
                                                    1?6 / 1 choose 6, between 0..5 (returns a list of 1 item)
                                                    1+ / add 1 (so between 1..6)
                                                    a: / store as 'a'
                                                    6, / prepend 6, the number of sides of the first dice
                                                    *a / we are iterating between 0 and 5 times, take first (*)
                                                    (*).x / multi-argument apply (.) multiply (*) to x, e.g. 6*2 => 12
                                                    x: / save that as 'x'
                                                    1? / 1 choose x, between 0..x-1
                                                    1+ / add 1 (so between 1..x)
                                                    x, / prepend x
                                                    *| / reverse-first aka 'last'


                                                    You can see the iterations by switching the over for a scan, e.g.



                                                    (6 3 / 1 choose 6 => 3, so perform 3 iterations
                                                    18 15 / 1 choose (6*3=18) => 15
                                                    270 31 / 1 choose (18*15=270) => 31
                                                    8370 5280) / 1 choose (270*31=8730) => 5280





                                                    share|improve this answer












                                                    $endgroup$




                                                    K (oK), 32 bytes



                                                    Solution:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6]


                                                    Try it online!



                                                    Start with 6 and "1 choose 6", iterate over "1 choose 6" times:



                                                    *|x,1+1?x:(*).x/[*a;6,a:1+1?6] / the solution
                                                    /[n; c ] / iterate over lambda n times with starting condition c
                                                    1?6 / 1 choose 6, between 0..5 (returns a list of 1 item)
                                                    1+ / add 1 (so between 1..6)
                                                    a: / store as 'a'
                                                    6, / prepend 6, the number of sides of the first dice
                                                    *a / we are iterating between 0 and 5 times, take first (*)
                                                    (*).x / multi-argument apply (.) multiply (*) to x, e.g. 6*2 => 12
                                                    x: / save that as 'x'
                                                    1? / 1 choose x, between 0..x-1
                                                    1+ / add 1 (so between 1..x)
                                                    x, / prepend x
                                                    *| / reverse-first aka 'last'


                                                    You can see the iterations by switching the over for a scan, e.g.



                                                    (6 3 / 1 choose 6 => 3, so perform 3 iterations
                                                    18 15 / 1 choose (6*3=18) => 15
                                                    270 31 / 1 choose (18*15=270) => 31
                                                    8370 5280) / 1 choose (270*31=8730) => 5280






                                                    share|improve this answer















                                                    share|improve this answer




                                                    share|improve this answer








                                                    edited Jun 1 at 7:24

























                                                    answered Jun 1 at 7:17









                                                    streetsterstreetster

                                                    3,2756 silver badges16 bronze badges




                                                    3,2756 silver badges16 bronze badges










                                                    • 1




                                                      $begingroup$
                                                      (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                      $endgroup$
                                                      – ngn
                                                      Jun 7 at 23:16












                                                    • 1




                                                      $begingroup$
                                                      (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                      $endgroup$
                                                      – ngn
                                                      Jun 7 at 23:16







                                                    1




                                                    1




                                                    $begingroup$
                                                    (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                    $endgroup$
                                                    – ngn
                                                    Jun 7 at 23:16




                                                    $begingroup$
                                                    (*).x -> */x and /[*a;6,a:1+1?6] -> a /6,a:*1+1?6
                                                    $endgroup$
                                                    – ngn
                                                    Jun 7 at 23:16











                                                    4
















                                                    $begingroup$


                                                    Jelly, 9 bytes



                                                    6X×$5СXX


                                                    A niladic Link which yields a positive integer.



                                                    Try it online!



                                                    Saving a byte over the more obvious 6X×$6X’¤¡X



                                                    How?



                                                    6X×$5СXX - Link: no arguments
                                                    6 - initialise left argument to 6
                                                    5С - repeat five times, collecting up as we go: -> a list of 6 possible dice sizes
                                                    $ - last two links as a monad = f(v): e.g [6,18,288,4032,1382976,216315425088]
                                                    X - random number in [1,v] or [6,6,6,6,6,6]
                                                    × - multiply (by v) or [6,36,1296,1679616,2821109907456,7958661109946400884391936]
                                                    X - random choice (since the input is now a list) -> faces (on final die)
                                                    X - random number in [1,faces]





                                                    share|improve this answer












                                                    $endgroup$














                                                    • $begingroup$
                                                      Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                      $endgroup$
                                                      – Nick Kennedy
                                                      Jun 1 at 13:50










                                                    • $begingroup$
                                                      Heh, I missed that you'd posted almost that exact answer!
                                                      $endgroup$
                                                      – Jonathan Allan
                                                      Jun 1 at 14:01















                                                    4
















                                                    $begingroup$


                                                    Jelly, 9 bytes



                                                    6X×$5СXX


                                                    A niladic Link which yields a positive integer.



                                                    Try it online!



                                                    Saving a byte over the more obvious 6X×$6X’¤¡X



                                                    How?



                                                    6X×$5СXX - Link: no arguments
                                                    6 - initialise left argument to 6
                                                    5С - repeat five times, collecting up as we go: -> a list of 6 possible dice sizes
                                                    $ - last two links as a monad = f(v): e.g [6,18,288,4032,1382976,216315425088]
                                                    X - random number in [1,v] or [6,6,6,6,6,6]
                                                    × - multiply (by v) or [6,36,1296,1679616,2821109907456,7958661109946400884391936]
                                                    X - random choice (since the input is now a list) -> faces (on final die)
                                                    X - random number in [1,faces]





                                                    share|improve this answer












                                                    $endgroup$














                                                    • $begingroup$
                                                      Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                      $endgroup$
                                                      – Nick Kennedy
                                                      Jun 1 at 13:50










                                                    • $begingroup$
                                                      Heh, I missed that you'd posted almost that exact answer!
                                                      $endgroup$
                                                      – Jonathan Allan
                                                      Jun 1 at 14:01













                                                    4














                                                    4










                                                    4







                                                    $begingroup$


                                                    Jelly, 9 bytes



                                                    6X×$5СXX


                                                    A niladic Link which yields a positive integer.



                                                    Try it online!



                                                    Saving a byte over the more obvious 6X×$6X’¤¡X



                                                    How?



                                                    6X×$5СXX - Link: no arguments
                                                    6 - initialise left argument to 6
                                                    5С - repeat five times, collecting up as we go: -> a list of 6 possible dice sizes
                                                    $ - last two links as a monad = f(v): e.g [6,18,288,4032,1382976,216315425088]
                                                    X - random number in [1,v] or [6,6,6,6,6,6]
                                                    × - multiply (by v) or [6,36,1296,1679616,2821109907456,7958661109946400884391936]
                                                    X - random choice (since the input is now a list) -> faces (on final die)
                                                    X - random number in [1,faces]





                                                    share|improve this answer












                                                    $endgroup$




                                                    Jelly, 9 bytes



                                                    6X×$5СXX


                                                    A niladic Link which yields a positive integer.



                                                    Try it online!



                                                    Saving a byte over the more obvious 6X×$6X’¤¡X



                                                    How?



                                                    6X×$5СXX - Link: no arguments
                                                    6 - initialise left argument to 6
                                                    5С - repeat five times, collecting up as we go: -> a list of 6 possible dice sizes
                                                    $ - last two links as a monad = f(v): e.g [6,18,288,4032,1382976,216315425088]
                                                    X - random number in [1,v] or [6,6,6,6,6,6]
                                                    × - multiply (by v) or [6,36,1296,1679616,2821109907456,7958661109946400884391936]
                                                    X - random choice (since the input is now a list) -> faces (on final die)
                                                    X - random number in [1,faces]






                                                    share|improve this answer















                                                    share|improve this answer




                                                    share|improve this answer








                                                    edited Jun 1 at 13:58

























                                                    answered Jun 1 at 13:47









                                                    Jonathan AllanJonathan Allan

                                                    61.9k5 gold badges44 silver badges188 bronze badges




                                                    61.9k5 gold badges44 silver badges188 bronze badges














                                                    • $begingroup$
                                                      Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                      $endgroup$
                                                      – Nick Kennedy
                                                      Jun 1 at 13:50










                                                    • $begingroup$
                                                      Heh, I missed that you'd posted almost that exact answer!
                                                      $endgroup$
                                                      – Jonathan Allan
                                                      Jun 1 at 14:01
















                                                    • $begingroup$
                                                      Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                      $endgroup$
                                                      – Nick Kennedy
                                                      Jun 1 at 13:50










                                                    • $begingroup$
                                                      Heh, I missed that you'd posted almost that exact answer!
                                                      $endgroup$
                                                      – Jonathan Allan
                                                      Jun 1 at 14:01















                                                    $begingroup$
                                                    Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                    $endgroup$
                                                    – Nick Kennedy
                                                    Jun 1 at 13:50




                                                    $begingroup$
                                                    Nice. I was trying to think of a way from moving beyond my ‘obvious’ answer, but didn’t think of generating all the dice and then picking one at random.
                                                    $endgroup$
                                                    – Nick Kennedy
                                                    Jun 1 at 13:50












                                                    $begingroup$
                                                    Heh, I missed that you'd posted almost that exact answer!
                                                    $endgroup$
                                                    – Jonathan Allan
                                                    Jun 1 at 14:01




                                                    $begingroup$
                                                    Heh, I missed that you'd posted almost that exact answer!
                                                    $endgroup$
                                                    – Jonathan Allan
                                                    Jun 1 at 14:01











                                                    3
















                                                    $begingroup$


                                                    05AB1E, 10 bytes



                                                    X6DLΩF*DLΩ


                                                    The random choice builtin for large lists is pretty slow, so may result in a timeout if the Prime Bubble roll is for example a 6.



                                                    Try it online or try it online with added prints to see the rolls. (TIO uses the legacy version of 05AB1E, since it's slightly faster.)



                                                    Explanation:





                                                    X # Push a 1 to the stack
                                                    6 # Push a 6 to the stack
                                                    D # Push another 6 to the stack
                                                    L # Pop the top 6, and push a list [1,2,3,4,5,6] to the stack
                                                    Ω # Pop and push a random item from this list (this is out Prime Bubble roll)
                                                    F # Loop that many times:
                                                    * # Multiply the top two values on the stack
                                                    # (which is why we had the initial 1 and duplicated 6 before the loop)
                                                    D # Duplicate this result
                                                    LΩ # Pop and push a random value from its ranged list again
                                                    # (after the loop, output the top of the stack implicitly)





                                                    share|improve this answer










                                                    $endgroup$



















                                                      3
















                                                      $begingroup$


                                                      05AB1E, 10 bytes



                                                      X6DLΩF*DLΩ


                                                      The random choice builtin for large lists is pretty slow, so may result in a timeout if the Prime Bubble roll is for example a 6.



                                                      Try it online or try it online with added prints to see the rolls. (TIO uses the legacy version of 05AB1E, since it's slightly faster.)



                                                      Explanation:





                                                      X # Push a 1 to the stack
                                                      6 # Push a 6 to the stack
                                                      D # Push another 6 to the stack
                                                      L # Pop the top 6, and push a list [1,2,3,4,5,6] to the stack
                                                      Ω # Pop and push a random item from this list (this is out Prime Bubble roll)
                                                      F # Loop that many times:
                                                      * # Multiply the top two values on the stack
                                                      # (which is why we had the initial 1 and duplicated 6 before the loop)
                                                      D # Duplicate this result
                                                      LΩ # Pop and push a random value from its ranged list again
                                                      # (after the loop, output the top of the stack implicitly)





                                                      share|improve this answer










                                                      $endgroup$

















                                                        3














                                                        3










                                                        3







                                                        $begingroup$


                                                        05AB1E, 10 bytes



                                                        X6DLΩF*DLΩ


                                                        The random choice builtin for large lists is pretty slow, so may result in a timeout if the Prime Bubble roll is for example a 6.



                                                        Try it online or try it online with added prints to see the rolls. (TIO uses the legacy version of 05AB1E, since it's slightly faster.)



                                                        Explanation:





                                                        X # Push a 1 to the stack
                                                        6 # Push a 6 to the stack
                                                        D # Push another 6 to the stack
                                                        L # Pop the top 6, and push a list [1,2,3,4,5,6] to the stack
                                                        Ω # Pop and push a random item from this list (this is out Prime Bubble roll)
                                                        F # Loop that many times:
                                                        * # Multiply the top two values on the stack
                                                        # (which is why we had the initial 1 and duplicated 6 before the loop)
                                                        D # Duplicate this result
                                                        LΩ # Pop and push a random value from its ranged list again
                                                        # (after the loop, output the top of the stack implicitly)





                                                        share|improve this answer










                                                        $endgroup$




                                                        05AB1E, 10 bytes



                                                        X6DLΩF*DLΩ


                                                        The random choice builtin for large lists is pretty slow, so may result in a timeout if the Prime Bubble roll is for example a 6.



                                                        Try it online or try it online with added prints to see the rolls. (TIO uses the legacy version of 05AB1E, since it's slightly faster.)



                                                        Explanation:





                                                        X # Push a 1 to the stack
                                                        6 # Push a 6 to the stack
                                                        D # Push another 6 to the stack
                                                        L # Pop the top 6, and push a list [1,2,3,4,5,6] to the stack
                                                        Ω # Pop and push a random item from this list (this is out Prime Bubble roll)
                                                        F # Loop that many times:
                                                        * # Multiply the top two values on the stack
                                                        # (which is why we had the initial 1 and duplicated 6 before the loop)
                                                        D # Duplicate this result
                                                        LΩ # Pop and push a random value from its ranged list again
                                                        # (after the loop, output the top of the stack implicitly)






                                                        share|improve this answer













                                                        share|improve this answer




                                                        share|improve this answer










                                                        answered Jun 1 at 13:50









                                                        Kevin CruijssenKevin Cruijssen

                                                        52.4k7 gold badges85 silver badges252 bronze badges




                                                        52.4k7 gold badges85 silver badges252 bronze badges
























                                                            3
















                                                            $begingroup$


                                                            Jelly, 9 bytes



                                                            6×X$X’$¡X


                                                            Try it online!



                                                            Jonathan Allan's answer claims that it's




                                                            Saving a byte over the more obvious 6X×$6X’¤¡X




                                                            . In fact, we don't need to make such a big modification. Therefore, this is an alternative approach to Jonathan Allan's answer, and, also, a resting place for my initial invalid 6-byter. :(






                                                            share|improve this answer












                                                            $endgroup$



















                                                              3
















                                                              $begingroup$


                                                              Jelly, 9 bytes



                                                              6×X$X’$¡X


                                                              Try it online!



                                                              Jonathan Allan's answer claims that it's




                                                              Saving a byte over the more obvious 6X×$6X’¤¡X




                                                              . In fact, we don't need to make such a big modification. Therefore, this is an alternative approach to Jonathan Allan's answer, and, also, a resting place for my initial invalid 6-byter. :(






                                                              share|improve this answer












                                                              $endgroup$

















                                                                3














                                                                3










                                                                3







                                                                $begingroup$


                                                                Jelly, 9 bytes



                                                                6×X$X’$¡X


                                                                Try it online!



                                                                Jonathan Allan's answer claims that it's




                                                                Saving a byte over the more obvious 6X×$6X’¤¡X




                                                                . In fact, we don't need to make such a big modification. Therefore, this is an alternative approach to Jonathan Allan's answer, and, also, a resting place for my initial invalid 6-byter. :(






                                                                share|improve this answer












                                                                $endgroup$




                                                                Jelly, 9 bytes



                                                                6×X$X’$¡X


                                                                Try it online!



                                                                Jonathan Allan's answer claims that it's




                                                                Saving a byte over the more obvious 6X×$6X’¤¡X




                                                                . In fact, we don't need to make such a big modification. Therefore, this is an alternative approach to Jonathan Allan's answer, and, also, a resting place for my initial invalid 6-byter. :(







                                                                share|improve this answer















                                                                share|improve this answer




                                                                share|improve this answer








                                                                edited Jun 2 at 0:55

























                                                                answered Jun 2 at 0:06









                                                                Erik the OutgolferErik the Outgolfer

                                                                36.6k4 gold badges31 silver badges113 bronze badges




                                                                36.6k4 gold badges31 silver badges113 bronze badges
























                                                                    2
















                                                                    $begingroup$


                                                                    Perl 5, 54 bytes





                                                                    $a=6;map$a*=$r=1+int rand$a,0..rand 6;say 1+int rand$r


                                                                    Try it online!






                                                                    share|improve this answer










                                                                    $endgroup$














                                                                    • $begingroup$
                                                                      That's one rand too many, you should just say$r
                                                                      $endgroup$
                                                                      – Grimy
                                                                      Jun 1 at 19:25















                                                                    2
















                                                                    $begingroup$


                                                                    Perl 5, 54 bytes





                                                                    $a=6;map$a*=$r=1+int rand$a,0..rand 6;say 1+int rand$r


                                                                    Try it online!






                                                                    share|improve this answer










                                                                    $endgroup$














                                                                    • $begingroup$
                                                                      That's one rand too many, you should just say$r
                                                                      $endgroup$
                                                                      – Grimy
                                                                      Jun 1 at 19:25













                                                                    2














                                                                    2










                                                                    2







                                                                    $begingroup$


                                                                    Perl 5, 54 bytes





                                                                    $a=6;map$a*=$r=1+int rand$a,0..rand 6;say 1+int rand$r


                                                                    Try it online!






                                                                    share|improve this answer










                                                                    $endgroup$




                                                                    Perl 5, 54 bytes





                                                                    $a=6;map$a*=$r=1+int rand$a,0..rand 6;say 1+int rand$r


                                                                    Try it online!







                                                                    share|improve this answer













                                                                    share|improve this answer




                                                                    share|improve this answer










                                                                    answered Jun 1 at 4:01









                                                                    XcaliXcali

                                                                    6,9951 gold badge7 silver badges24 bronze badges




                                                                    6,9951 gold badge7 silver badges24 bronze badges














                                                                    • $begingroup$
                                                                      That's one rand too many, you should just say$r
                                                                      $endgroup$
                                                                      – Grimy
                                                                      Jun 1 at 19:25
















                                                                    • $begingroup$
                                                                      That's one rand too many, you should just say$r
                                                                      $endgroup$
                                                                      – Grimy
                                                                      Jun 1 at 19:25















                                                                    $begingroup$
                                                                    That's one rand too many, you should just say$r
                                                                    $endgroup$
                                                                    – Grimy
                                                                    Jun 1 at 19:25




                                                                    $begingroup$
                                                                    That's one rand too many, you should just say$r
                                                                    $endgroup$
                                                                    – Grimy
                                                                    Jun 1 at 19:25











                                                                    2
















                                                                    $begingroup$


                                                                    Charcoal, 16 bytes



                                                                    ⊞υ⁶F⊕‽⁶⊞υ⊕‽ΠυI⊟υ


                                                                    Try it online! Link is to verbose version of code. Explanation:



                                                                    ⊞υ⁶


                                                                    Push 6 to the predefined list.



                                                                    F⊕‽⁶


                                                                    Repeat a random number of times from 1 to 6...



                                                                    ⊞υ⊕‽Πυ


                                                                    ... push a random number between 1 and the product of the list to the list.



                                                                    I⊟υ


                                                                    Output the last number pushed to the list.



                                                                    Alternative approach, also 16 bytes



                                                                    ≔⁶θF‽⁶≧×⊕‽θθI⊕‽θ


                                                                    Try it online! Link is to verbose version of code. Explanation:



                                                                    ≔⁶θ


                                                                    Set the number of sides to 6.



                                                                    F‽⁶


                                                                    Repeat a random number between 0 and 5 times...



                                                                    ≧×⊕‽θθ


                                                                    ... multiply the number of sides by a random number from 1 to the number of sides.



                                                                    I⊕‽θ


                                                                    Print a random number from 1 to the number of sides.






                                                                    share|improve this answer










                                                                    $endgroup$



















                                                                      2
















                                                                      $begingroup$


                                                                      Charcoal, 16 bytes



                                                                      ⊞υ⁶F⊕‽⁶⊞υ⊕‽ΠυI⊟υ


                                                                      Try it online! Link is to verbose version of code. Explanation:



                                                                      ⊞υ⁶


                                                                      Push 6 to the predefined list.



                                                                      F⊕‽⁶


                                                                      Repeat a random number of times from 1 to 6...



                                                                      ⊞υ⊕‽Πυ


                                                                      ... push a random number between 1 and the product of the list to the list.



                                                                      I⊟υ


                                                                      Output the last number pushed to the list.



                                                                      Alternative approach, also 16 bytes



                                                                      ≔⁶θF‽⁶≧×⊕‽θθI⊕‽θ


                                                                      Try it online! Link is to verbose version of code. Explanation:



                                                                      ≔⁶θ


                                                                      Set the number of sides to 6.



                                                                      F‽⁶


                                                                      Repeat a random number between 0 and 5 times...



                                                                      ≧×⊕‽θθ


                                                                      ... multiply the number of sides by a random number from 1 to the number of sides.



                                                                      I⊕‽θ


                                                                      Print a random number from 1 to the number of sides.






                                                                      share|improve this answer










                                                                      $endgroup$

















                                                                        2














                                                                        2










                                                                        2







                                                                        $begingroup$


                                                                        Charcoal, 16 bytes



                                                                        ⊞υ⁶F⊕‽⁶⊞υ⊕‽ΠυI⊟υ


                                                                        Try it online! Link is to verbose version of code. Explanation:



                                                                        ⊞υ⁶


                                                                        Push 6 to the predefined list.



                                                                        F⊕‽⁶


                                                                        Repeat a random number of times from 1 to 6...



                                                                        ⊞υ⊕‽Πυ


                                                                        ... push a random number between 1 and the product of the list to the list.



                                                                        I⊟υ


                                                                        Output the last number pushed to the list.



                                                                        Alternative approach, also 16 bytes



                                                                        ≔⁶θF‽⁶≧×⊕‽θθI⊕‽θ


                                                                        Try it online! Link is to verbose version of code. Explanation:



                                                                        ≔⁶θ


                                                                        Set the number of sides to 6.



                                                                        F‽⁶


                                                                        Repeat a random number between 0 and 5 times...



                                                                        ≧×⊕‽θθ


                                                                        ... multiply the number of sides by a random number from 1 to the number of sides.



                                                                        I⊕‽θ


                                                                        Print a random number from 1 to the number of sides.






                                                                        share|improve this answer










                                                                        $endgroup$




                                                                        Charcoal, 16 bytes



                                                                        ⊞υ⁶F⊕‽⁶⊞υ⊕‽ΠυI⊟υ


                                                                        Try it online! Link is to verbose version of code. Explanation:



                                                                        ⊞υ⁶


                                                                        Push 6 to the predefined list.



                                                                        F⊕‽⁶


                                                                        Repeat a random number of times from 1 to 6...



                                                                        ⊞υ⊕‽Πυ


                                                                        ... push a random number between 1 and the product of the list to the list.



                                                                        I⊟υ


                                                                        Output the last number pushed to the list.



                                                                        Alternative approach, also 16 bytes



                                                                        ≔⁶θF‽⁶≧×⊕‽θθI⊕‽θ


                                                                        Try it online! Link is to verbose version of code. Explanation:



                                                                        ≔⁶θ


                                                                        Set the number of sides to 6.



                                                                        F‽⁶


                                                                        Repeat a random number between 0 and 5 times...



                                                                        ≧×⊕‽θθ


                                                                        ... multiply the number of sides by a random number from 1 to the number of sides.



                                                                        I⊕‽θ


                                                                        Print a random number from 1 to the number of sides.







                                                                        share|improve this answer













                                                                        share|improve this answer




                                                                        share|improve this answer










                                                                        answered Jun 1 at 9:29









                                                                        NeilNeil

                                                                        90.2k8 gold badges46 silver badges190 bronze badges




                                                                        90.2k8 gold badges46 silver badges190 bronze badges
























                                                                            2
















                                                                            $begingroup$


                                                                            Python 3, 76 bytes





                                                                            from random import*
                                                                            r=randint
                                                                            a=6
                                                                            for i in" "*r(0,5):a*=r(1,a)
                                                                            print(r(1,a))


                                                                            Try it online!



                                                                            -2 bytes thanks to TFeld






                                                                            share|improve this answer












                                                                            $endgroup$










                                                                            • 1




                                                                              $begingroup$
                                                                              -2 bytes by changing the import
                                                                              $endgroup$
                                                                              – TFeld
                                                                              Jun 1 at 7:34










                                                                            • $begingroup$
                                                                              @TFeld thanks [filler]
                                                                              $endgroup$
                                                                              – HyperNeutrino
                                                                              Jun 1 at 19:23















                                                                            2
















                                                                            $begingroup$


                                                                            Python 3, 76 bytes





                                                                            from random import*
                                                                            r=randint
                                                                            a=6
                                                                            for i in" "*r(0,5):a*=r(1,a)
                                                                            print(r(1,a))


                                                                            Try it online!



                                                                            -2 bytes thanks to TFeld






                                                                            share|improve this answer












                                                                            $endgroup$










                                                                            • 1




                                                                              $begingroup$
                                                                              -2 bytes by changing the import
                                                                              $endgroup$
                                                                              – TFeld
                                                                              Jun 1 at 7:34










                                                                            • $begingroup$
                                                                              @TFeld thanks [filler]
                                                                              $endgroup$
                                                                              – HyperNeutrino
                                                                              Jun 1 at 19:23













                                                                            2














                                                                            2










                                                                            2







                                                                            $begingroup$


                                                                            Python 3, 76 bytes





                                                                            from random import*
                                                                            r=randint
                                                                            a=6
                                                                            for i in" "*r(0,5):a*=r(1,a)
                                                                            print(r(1,a))


                                                                            Try it online!



                                                                            -2 bytes thanks to TFeld






                                                                            share|improve this answer












                                                                            $endgroup$




                                                                            Python 3, 76 bytes





                                                                            from random import*
                                                                            r=randint
                                                                            a=6
                                                                            for i in" "*r(0,5):a*=r(1,a)
                                                                            print(r(1,a))


                                                                            Try it online!



                                                                            -2 bytes thanks to TFeld







                                                                            share|improve this answer















                                                                            share|improve this answer




                                                                            share|improve this answer








                                                                            edited Jun 1 at 19:23

























                                                                            answered Jun 1 at 2:54









                                                                            HyperNeutrinoHyperNeutrino

                                                                            20.7k4 gold badges41 silver badges155 bronze badges




                                                                            20.7k4 gold badges41 silver badges155 bronze badges










                                                                            • 1




                                                                              $begingroup$
                                                                              -2 bytes by changing the import
                                                                              $endgroup$
                                                                              – TFeld
                                                                              Jun 1 at 7:34










                                                                            • $begingroup$
                                                                              @TFeld thanks [filler]
                                                                              $endgroup$
                                                                              – HyperNeutrino
                                                                              Jun 1 at 19:23












                                                                            • 1




                                                                              $begingroup$
                                                                              -2 bytes by changing the import
                                                                              $endgroup$
                                                                              – TFeld
                                                                              Jun 1 at 7:34










                                                                            • $begingroup$
                                                                              @TFeld thanks [filler]
                                                                              $endgroup$
                                                                              – HyperNeutrino
                                                                              Jun 1 at 19:23







                                                                            1




                                                                            1




                                                                            $begingroup$
                                                                            -2 bytes by changing the import
                                                                            $endgroup$
                                                                            – TFeld
                                                                            Jun 1 at 7:34




                                                                            $begingroup$
                                                                            -2 bytes by changing the import
                                                                            $endgroup$
                                                                            – TFeld
                                                                            Jun 1 at 7:34












                                                                            $begingroup$
                                                                            @TFeld thanks [filler]
                                                                            $endgroup$
                                                                            – HyperNeutrino
                                                                            Jun 1 at 19:23




                                                                            $begingroup$
                                                                            @TFeld thanks [filler]
                                                                            $endgroup$
                                                                            – HyperNeutrino
                                                                            Jun 1 at 19:23











                                                                            2
















                                                                            $begingroup$


                                                                            R, 43 bytes





                                                                            s=sample
                                                                            for(i in 1:s(k<-6))T=s(k<-k*T,1)
                                                                            T


                                                                            Try it online!



                                                                            k keeps track of the current number of faces on the die. Uses the fact that T is initialized as 1.



                                                                            I tried a few other things, but couldn't beat this simple, straightforward approach.






                                                                            share|improve this answer










                                                                            $endgroup$



















                                                                              2
















                                                                              $begingroup$


                                                                              R, 43 bytes





                                                                              s=sample
                                                                              for(i in 1:s(k<-6))T=s(k<-k*T,1)
                                                                              T


                                                                              Try it online!



                                                                              k keeps track of the current number of faces on the die. Uses the fact that T is initialized as 1.



                                                                              I tried a few other things, but couldn't beat this simple, straightforward approach.






                                                                              share|improve this answer










                                                                              $endgroup$

















                                                                                2














                                                                                2










                                                                                2







                                                                                $begingroup$


                                                                                R, 43 bytes





                                                                                s=sample
                                                                                for(i in 1:s(k<-6))T=s(k<-k*T,1)
                                                                                T


                                                                                Try it online!



                                                                                k keeps track of the current number of faces on the die. Uses the fact that T is initialized as 1.



                                                                                I tried a few other things, but couldn't beat this simple, straightforward approach.






                                                                                share|improve this answer










                                                                                $endgroup$




                                                                                R, 43 bytes





                                                                                s=sample
                                                                                for(i in 1:s(k<-6))T=s(k<-k*T,1)
                                                                                T


                                                                                Try it online!



                                                                                k keeps track of the current number of faces on the die. Uses the fact that T is initialized as 1.



                                                                                I tried a few other things, but couldn't beat this simple, straightforward approach.







                                                                                share|improve this answer













                                                                                share|improve this answer




                                                                                share|improve this answer










                                                                                answered Jun 3 at 21:29









                                                                                Robin RyderRobin Ryder

                                                                                4,3625 silver badges29 bronze badges




                                                                                4,3625 silver badges29 bronze badges
























                                                                                    1
















                                                                                    $begingroup$


                                                                                    Jelly, 10 bytes



                                                                                    6×X$6X’¤¡X


                                                                                    Try it online!



                                                                                    Explanation



                                                                                     ¤ | Following as a nilad:
                                                                                    6X | - A random number between 1 and 6
                                                                                    ’ | - Decrease by 1 (call this N)
                                                                                    6 | Now, start with 6
                                                                                    $ ¡ | Repeat the following N times, as a monad
                                                                                    × | - Multiply by:
                                                                                    X | - A random number between 1 and the current total
                                                                                    X | Finally, generate a random number between 1 and the output of the above loop





                                                                                    share|improve this answer










                                                                                    $endgroup$



















                                                                                      1
















                                                                                      $begingroup$


                                                                                      Jelly, 10 bytes



                                                                                      6×X$6X’¤¡X


                                                                                      Try it online!



                                                                                      Explanation



                                                                                       ¤ | Following as a nilad:
                                                                                      6X | - A random number between 1 and 6
                                                                                      ’ | - Decrease by 1 (call this N)
                                                                                      6 | Now, start with 6
                                                                                      $ ¡ | Repeat the following N times, as a monad
                                                                                      × | - Multiply by:
                                                                                      X | - A random number between 1 and the current total
                                                                                      X | Finally, generate a random number between 1 and the output of the above loop





                                                                                      share|improve this answer










                                                                                      $endgroup$

















                                                                                        1














                                                                                        1










                                                                                        1







                                                                                        $begingroup$


                                                                                        Jelly, 10 bytes



                                                                                        6×X$6X’¤¡X


                                                                                        Try it online!



                                                                                        Explanation



                                                                                         ¤ | Following as a nilad:
                                                                                        6X | - A random number between 1 and 6
                                                                                        ’ | - Decrease by 1 (call this N)
                                                                                        6 | Now, start with 6
                                                                                        $ ¡ | Repeat the following N times, as a monad
                                                                                        × | - Multiply by:
                                                                                        X | - A random number between 1 and the current total
                                                                                        X | Finally, generate a random number between 1 and the output of the above loop





                                                                                        share|improve this answer










                                                                                        $endgroup$




                                                                                        Jelly, 10 bytes



                                                                                        6×X$6X’¤¡X


                                                                                        Try it online!



                                                                                        Explanation



                                                                                         ¤ | Following as a nilad:
                                                                                        6X | - A random number between 1 and 6
                                                                                        ’ | - Decrease by 1 (call this N)
                                                                                        6 | Now, start with 6
                                                                                        $ ¡ | Repeat the following N times, as a monad
                                                                                        × | - Multiply by:
                                                                                        X | - A random number between 1 and the current total
                                                                                        X | Finally, generate a random number between 1 and the output of the above loop






                                                                                        share|improve this answer













                                                                                        share|improve this answer




                                                                                        share|improve this answer










                                                                                        answered Jun 1 at 6:35









                                                                                        Nick KennedyNick Kennedy

                                                                                        7,3091 gold badge9 silver badges19 bronze badges




                                                                                        7,3091 gold badge9 silver badges19 bronze badges
























                                                                                            1
















                                                                                            $begingroup$


                                                                                            Ruby, 41 bytes





                                                                                            r=6;rand(2..7).timesr*=$s=rand 1..r;p$s


                                                                                            Try it online!



                                                                                            Explanation



                                                                                            r=6 # Set dice number to 6

                                                                                            rand(2..7).times # Repeat X times, where X=dice roll+1
                                                                                            r*=$s=rand 1..r # Multiply dice number by a dice roll
                                                                                            # Save the most recent dice roll

                                                                                            p$s # Print last dice roll (this is why
                                                                                            # we did the last step one extra time)





                                                                                            share|improve this answer










                                                                                            $endgroup$



















                                                                                              1
















                                                                                              $begingroup$


                                                                                              Ruby, 41 bytes





                                                                                              r=6;rand(2..7).timesr*=$s=rand 1..r;p$s


                                                                                              Try it online!



                                                                                              Explanation



                                                                                              r=6 # Set dice number to 6

                                                                                              rand(2..7).times # Repeat X times, where X=dice roll+1
                                                                                              r*=$s=rand 1..r # Multiply dice number by a dice roll
                                                                                              # Save the most recent dice roll

                                                                                              p$s # Print last dice roll (this is why
                                                                                              # we did the last step one extra time)





                                                                                              share|improve this answer










                                                                                              $endgroup$

















                                                                                                1














                                                                                                1










                                                                                                1







                                                                                                $begingroup$


                                                                                                Ruby, 41 bytes





                                                                                                r=6;rand(2..7).timesr*=$s=rand 1..r;p$s


                                                                                                Try it online!



                                                                                                Explanation



                                                                                                r=6 # Set dice number to 6

                                                                                                rand(2..7).times # Repeat X times, where X=dice roll+1
                                                                                                r*=$s=rand 1..r # Multiply dice number by a dice roll
                                                                                                # Save the most recent dice roll

                                                                                                p$s # Print last dice roll (this is why
                                                                                                # we did the last step one extra time)





                                                                                                share|improve this answer










                                                                                                $endgroup$




                                                                                                Ruby, 41 bytes





                                                                                                r=6;rand(2..7).timesr*=$s=rand 1..r;p$s


                                                                                                Try it online!



                                                                                                Explanation



                                                                                                r=6 # Set dice number to 6

                                                                                                rand(2..7).times # Repeat X times, where X=dice roll+1
                                                                                                r*=$s=rand 1..r # Multiply dice number by a dice roll
                                                                                                # Save the most recent dice roll

                                                                                                p$s # Print last dice roll (this is why
                                                                                                # we did the last step one extra time)






                                                                                                share|improve this answer













                                                                                                share|improve this answer




                                                                                                share|improve this answer










                                                                                                answered Jun 4 at 0:45









                                                                                                Value InkValue Ink

                                                                                                9,5288 silver badges37 bronze badges




                                                                                                9,5288 silver badges37 bronze badges
























                                                                                                    1
















                                                                                                    $begingroup$

                                                                                                    Java 10, 214 93 86 bytes





                                                                                                    v->int r=6,n=0;for(var d=Math.random()*6;d-->0;n*=Math.random(),r*=++n)n=r;return n;


                                                                                                    Try it online or try it online with additional print-lines to see the steps.



                                                                                                    -128 bytes by using int instead of java.math.BigInteger. $6^32$, the largest possible result, doesn't fit inside an int nor long, which is why BigInteger was used initially. OP allowed to use int and assuming it's infinitely large, so that saved more than 125 bytes here. :) (Here the previous 214 bytes version using BigIntegers.)



                                                                                                    Explanation:



                                                                                                    v-> // Method with empty unused parameter & integer return-type
                                                                                                    int r=6, // The range in which to roll, starting at 6
                                                                                                    n=0; // The roll itself (which must be initialized, therefor is 0)
                                                                                                    for(var d=Math.random()*6;// Roll the Prime Bubble Dice
                                                                                                    d-->0 // Loop that many times:
                                                                                                    ; // After every iteration:
                                                                                                    n*=Math.random(), // Roll a random dice in the range [0, n)
                                                                                                    r*=++n) // Increase `n` by 1 first with `++n`, so the range is [1,n]
                                                                                                    // And then multiply `r` by `n` for the new range
                                                                                                    n=r; // Set `n` to `r`
                                                                                                    return n; // After the loop, return `n` as result





                                                                                                    share|improve this answer












                                                                                                    $endgroup$














                                                                                                    • $begingroup$
                                                                                                      Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:16










                                                                                                    • $begingroup$
                                                                                                      @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 3 at 16:26










                                                                                                    • $begingroup$
                                                                                                      I'll see if I can find anything on meta about this.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:29






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      OP says don't worry about integer size limits. Use the int type.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 8 at 1:26






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 8 at 12:00















                                                                                                    1
















                                                                                                    $begingroup$

                                                                                                    Java 10, 214 93 86 bytes





                                                                                                    v->int r=6,n=0;for(var d=Math.random()*6;d-->0;n*=Math.random(),r*=++n)n=r;return n;


                                                                                                    Try it online or try it online with additional print-lines to see the steps.



                                                                                                    -128 bytes by using int instead of java.math.BigInteger. $6^32$, the largest possible result, doesn't fit inside an int nor long, which is why BigInteger was used initially. OP allowed to use int and assuming it's infinitely large, so that saved more than 125 bytes here. :) (Here the previous 214 bytes version using BigIntegers.)



                                                                                                    Explanation:



                                                                                                    v-> // Method with empty unused parameter & integer return-type
                                                                                                    int r=6, // The range in which to roll, starting at 6
                                                                                                    n=0; // The roll itself (which must be initialized, therefor is 0)
                                                                                                    for(var d=Math.random()*6;// Roll the Prime Bubble Dice
                                                                                                    d-->0 // Loop that many times:
                                                                                                    ; // After every iteration:
                                                                                                    n*=Math.random(), // Roll a random dice in the range [0, n)
                                                                                                    r*=++n) // Increase `n` by 1 first with `++n`, so the range is [1,n]
                                                                                                    // And then multiply `r` by `n` for the new range
                                                                                                    n=r; // Set `n` to `r`
                                                                                                    return n; // After the loop, return `n` as result





                                                                                                    share|improve this answer












                                                                                                    $endgroup$














                                                                                                    • $begingroup$
                                                                                                      Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:16










                                                                                                    • $begingroup$
                                                                                                      @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 3 at 16:26










                                                                                                    • $begingroup$
                                                                                                      I'll see if I can find anything on meta about this.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:29






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      OP says don't worry about integer size limits. Use the int type.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 8 at 1:26






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 8 at 12:00













                                                                                                    1














                                                                                                    1










                                                                                                    1







                                                                                                    $begingroup$

                                                                                                    Java 10, 214 93 86 bytes





                                                                                                    v->int r=6,n=0;for(var d=Math.random()*6;d-->0;n*=Math.random(),r*=++n)n=r;return n;


                                                                                                    Try it online or try it online with additional print-lines to see the steps.



                                                                                                    -128 bytes by using int instead of java.math.BigInteger. $6^32$, the largest possible result, doesn't fit inside an int nor long, which is why BigInteger was used initially. OP allowed to use int and assuming it's infinitely large, so that saved more than 125 bytes here. :) (Here the previous 214 bytes version using BigIntegers.)



                                                                                                    Explanation:



                                                                                                    v-> // Method with empty unused parameter & integer return-type
                                                                                                    int r=6, // The range in which to roll, starting at 6
                                                                                                    n=0; // The roll itself (which must be initialized, therefor is 0)
                                                                                                    for(var d=Math.random()*6;// Roll the Prime Bubble Dice
                                                                                                    d-->0 // Loop that many times:
                                                                                                    ; // After every iteration:
                                                                                                    n*=Math.random(), // Roll a random dice in the range [0, n)
                                                                                                    r*=++n) // Increase `n` by 1 first with `++n`, so the range is [1,n]
                                                                                                    // And then multiply `r` by `n` for the new range
                                                                                                    n=r; // Set `n` to `r`
                                                                                                    return n; // After the loop, return `n` as result





                                                                                                    share|improve this answer












                                                                                                    $endgroup$



                                                                                                    Java 10, 214 93 86 bytes





                                                                                                    v->int r=6,n=0;for(var d=Math.random()*6;d-->0;n*=Math.random(),r*=++n)n=r;return n;


                                                                                                    Try it online or try it online with additional print-lines to see the steps.



                                                                                                    -128 bytes by using int instead of java.math.BigInteger. $6^32$, the largest possible result, doesn't fit inside an int nor long, which is why BigInteger was used initially. OP allowed to use int and assuming it's infinitely large, so that saved more than 125 bytes here. :) (Here the previous 214 bytes version using BigIntegers.)



                                                                                                    Explanation:



                                                                                                    v-> // Method with empty unused parameter & integer return-type
                                                                                                    int r=6, // The range in which to roll, starting at 6
                                                                                                    n=0; // The roll itself (which must be initialized, therefor is 0)
                                                                                                    for(var d=Math.random()*6;// Roll the Prime Bubble Dice
                                                                                                    d-->0 // Loop that many times:
                                                                                                    ; // After every iteration:
                                                                                                    n*=Math.random(), // Roll a random dice in the range [0, n)
                                                                                                    r*=++n) // Increase `n` by 1 first with `++n`, so the range is [1,n]
                                                                                                    // And then multiply `r` by `n` for the new range
                                                                                                    n=r; // Set `n` to `r`
                                                                                                    return n; // After the loop, return `n` as result






                                                                                                    share|improve this answer















                                                                                                    share|improve this answer




                                                                                                    share|improve this answer








                                                                                                    edited Jun 8 at 12:00

























                                                                                                    answered Jun 3 at 9:32









                                                                                                    Kevin CruijssenKevin Cruijssen

                                                                                                    52.4k7 gold badges85 silver badges252 bronze badges




                                                                                                    52.4k7 gold badges85 silver badges252 bronze badges














                                                                                                    • $begingroup$
                                                                                                      Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:16










                                                                                                    • $begingroup$
                                                                                                      @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 3 at 16:26










                                                                                                    • $begingroup$
                                                                                                      I'll see if I can find anything on meta about this.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:29






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      OP says don't worry about integer size limits. Use the int type.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 8 at 1:26






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 8 at 12:00
















                                                                                                    • $begingroup$
                                                                                                      Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:16










                                                                                                    • $begingroup$
                                                                                                      @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 3 at 16:26










                                                                                                    • $begingroup$
                                                                                                      I'll see if I can find anything on meta about this.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 3 at 16:29






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      OP says don't worry about integer size limits. Use the int type.
                                                                                                      $endgroup$
                                                                                                      – Stackstuck
                                                                                                      Jun 8 at 1:26






                                                                                                    • 1




                                                                                                      $begingroup$
                                                                                                      @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                      $endgroup$
                                                                                                      – Kevin Cruijssen
                                                                                                      Jun 8 at 12:00















                                                                                                    $begingroup$
                                                                                                    Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 3 at 16:16




                                                                                                    $begingroup$
                                                                                                    Post the solution that doesn't do BigInteger as your competing solution.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 3 at 16:16












                                                                                                    $begingroup$
                                                                                                    @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                    $endgroup$
                                                                                                    – Kevin Cruijssen
                                                                                                    Jun 3 at 16:26




                                                                                                    $begingroup$
                                                                                                    @Stackstuck Will await what OP has to say to your comment. Was about to make one myself to confirm. I personally use BigIntegers/BigDecimals for challenges where it's necessary, unless OP explicitly states you can use your native max integer size (which is usually $2^32$, so int for both Java and .NET C#).
                                                                                                    $endgroup$
                                                                                                    – Kevin Cruijssen
                                                                                                    Jun 3 at 16:26












                                                                                                    $begingroup$
                                                                                                    I'll see if I can find anything on meta about this.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 3 at 16:29




                                                                                                    $begingroup$
                                                                                                    I'll see if I can find anything on meta about this.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 3 at 16:29




                                                                                                    1




                                                                                                    1




                                                                                                    $begingroup$
                                                                                                    OP says don't worry about integer size limits. Use the int type.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 8 at 1:26




                                                                                                    $begingroup$
                                                                                                    OP says don't worry about integer size limits. Use the int type.
                                                                                                    $endgroup$
                                                                                                    – Stackstuck
                                                                                                    Jun 8 at 1:26




                                                                                                    1




                                                                                                    1




                                                                                                    $begingroup$
                                                                                                    @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                    $endgroup$
                                                                                                    – Kevin Cruijssen
                                                                                                    Jun 8 at 12:00




                                                                                                    $begingroup$
                                                                                                    @Stackstuck Done, and golfed 7 bytes in the process. :)
                                                                                                    $endgroup$
                                                                                                    – Kevin Cruijssen
                                                                                                    Jun 8 at 12:00











                                                                                                    0
















                                                                                                    $begingroup$


                                                                                                    PHP, 59 bytes



                                                                                                    $r=$q=rand(1,$s=6);while($l++<$q)$r=rand(1,$s*=$r);print$r;


                                                                                                    expanded:



                                                                                                    $r=$q=rand(1,$s=6);
                                                                                                    while($l++<$q)$
                                                                                                    r=rand(1,$s*=$r);
                                                                                                    print$r;


                                                                                                    Not sure if I'm supposed to include the open tag.



                                                                                                    On my machine, it crashes if $s*$r is too large, so it doesn't print on $q>=5 sometimes... because the numbers get so big. Not sure of a fix.






                                                                                                    share|improve this answer










                                                                                                    $endgroup$



















                                                                                                      0
















                                                                                                      $begingroup$


                                                                                                      PHP, 59 bytes



                                                                                                      $r=$q=rand(1,$s=6);while($l++<$q)$r=rand(1,$s*=$r);print$r;


                                                                                                      expanded:



                                                                                                      $r=$q=rand(1,$s=6);
                                                                                                      while($l++<$q)$
                                                                                                      r=rand(1,$s*=$r);
                                                                                                      print$r;


                                                                                                      Not sure if I'm supposed to include the open tag.



                                                                                                      On my machine, it crashes if $s*$r is too large, so it doesn't print on $q>=5 sometimes... because the numbers get so big. Not sure of a fix.






                                                                                                      share|improve this answer










                                                                                                      $endgroup$

















                                                                                                        0














                                                                                                        0










                                                                                                        0







                                                                                                        $begingroup$


                                                                                                        PHP, 59 bytes



                                                                                                        $r=$q=rand(1,$s=6);while($l++<$q)$r=rand(1,$s*=$r);print$r;


                                                                                                        expanded:



                                                                                                        $r=$q=rand(1,$s=6);
                                                                                                        while($l++<$q)$
                                                                                                        r=rand(1,$s*=$r);
                                                                                                        print$r;


                                                                                                        Not sure if I'm supposed to include the open tag.



                                                                                                        On my machine, it crashes if $s*$r is too large, so it doesn't print on $q>=5 sometimes... because the numbers get so big. Not sure of a fix.






                                                                                                        share|improve this answer










                                                                                                        $endgroup$




                                                                                                        PHP, 59 bytes



                                                                                                        $r=$q=rand(1,$s=6);while($l++<$q)$r=rand(1,$s*=$r);print$r;


                                                                                                        expanded:



                                                                                                        $r=$q=rand(1,$s=6);
                                                                                                        while($l++<$q)$
                                                                                                        r=rand(1,$s*=$r);
                                                                                                        print$r;


                                                                                                        Not sure if I'm supposed to include the open tag.



                                                                                                        On my machine, it crashes if $s*$r is too large, so it doesn't print on $q>=5 sometimes... because the numbers get so big. Not sure of a fix.







                                                                                                        share|improve this answer













                                                                                                        share|improve this answer




                                                                                                        share|improve this answer










                                                                                                        answered Jun 2 at 5:00









                                                                                                        ReedReed

                                                                                                        1414 bronze badges




                                                                                                        1414 bronze badges
























                                                                                                            0
















                                                                                                            $begingroup$


                                                                                                            Pyth, 14 bytes



                                                                                                            uhO=*|Z6GO6hO6


                                                                                                            Try it online!



                                                                                                            uhO=*|Z6GO6hO6 
                                                                                                            O6 Random number in range [0-6)
                                                                                                            u Perform the following the above number of times...
                                                                                                            hO6 ... with starting value G a random number in range [1-6]:
                                                                                                            * G Multiply G with...
                                                                                                            |Z6 The value of Z, or 6 if it's the first time through (Z is 0 at program start)
                                                                                                            = Z Assign the above back into Z
                                                                                                            O Random number in range [0-Z)
                                                                                                            h Increment
                                                                                                            Implicit print result of final iteration





                                                                                                            share|improve this answer












                                                                                                            $endgroup$



















                                                                                                              0
















                                                                                                              $begingroup$


                                                                                                              Pyth, 14 bytes



                                                                                                              uhO=*|Z6GO6hO6


                                                                                                              Try it online!



                                                                                                              uhO=*|Z6GO6hO6 
                                                                                                              O6 Random number in range [0-6)
                                                                                                              u Perform the following the above number of times...
                                                                                                              hO6 ... with starting value G a random number in range [1-6]:
                                                                                                              * G Multiply G with...
                                                                                                              |Z6 The value of Z, or 6 if it's the first time through (Z is 0 at program start)
                                                                                                              = Z Assign the above back into Z
                                                                                                              O Random number in range [0-Z)
                                                                                                              h Increment
                                                                                                              Implicit print result of final iteration





                                                                                                              share|improve this answer












                                                                                                              $endgroup$

















                                                                                                                0














                                                                                                                0










                                                                                                                0







                                                                                                                $begingroup$


                                                                                                                Pyth, 14 bytes



                                                                                                                uhO=*|Z6GO6hO6


                                                                                                                Try it online!



                                                                                                                uhO=*|Z6GO6hO6 
                                                                                                                O6 Random number in range [0-6)
                                                                                                                u Perform the following the above number of times...
                                                                                                                hO6 ... with starting value G a random number in range [1-6]:
                                                                                                                * G Multiply G with...
                                                                                                                |Z6 The value of Z, or 6 if it's the first time through (Z is 0 at program start)
                                                                                                                = Z Assign the above back into Z
                                                                                                                O Random number in range [0-Z)
                                                                                                                h Increment
                                                                                                                Implicit print result of final iteration





                                                                                                                share|improve this answer












                                                                                                                $endgroup$




                                                                                                                Pyth, 14 bytes



                                                                                                                uhO=*|Z6GO6hO6


                                                                                                                Try it online!



                                                                                                                uhO=*|Z6GO6hO6 
                                                                                                                O6 Random number in range [0-6)
                                                                                                                u Perform the following the above number of times...
                                                                                                                hO6 ... with starting value G a random number in range [1-6]:
                                                                                                                * G Multiply G with...
                                                                                                                |Z6 The value of Z, or 6 if it's the first time through (Z is 0 at program start)
                                                                                                                = Z Assign the above back into Z
                                                                                                                O Random number in range [0-Z)
                                                                                                                h Increment
                                                                                                                Implicit print result of final iteration






                                                                                                                share|improve this answer















                                                                                                                share|improve this answer




                                                                                                                share|improve this answer








                                                                                                                edited Jun 3 at 8:53

























                                                                                                                answered Jun 3 at 8:12









                                                                                                                SokSok

                                                                                                                4,87711 silver badges28 bronze badges




                                                                                                                4,87711 silver badges28 bronze badges
























                                                                                                                    0
















                                                                                                                    $begingroup$


                                                                                                                    C# (.NET Core), 136 bytes





                                                                                                                    class Astatic void Main()var r=new System.Random();int i=r.Next(6),j=6;while(i-->0)j*=1+r.Next(j);System.Console.Write(r.Next(j)+1);


                                                                                                                    Try it online!



                                                                                                                    I'm pretty sure this works, given the assumption of infinite integer length that we're fond of here. If I have to actually handle the overflow, I'd need to bust out an entirely different class.






                                                                                                                    share|improve this answer












                                                                                                                    $endgroup$














                                                                                                                    • $begingroup$
                                                                                                                      It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                      $endgroup$
                                                                                                                      – Kevin Cruijssen
                                                                                                                      Jun 3 at 9:34











                                                                                                                    • $begingroup$
                                                                                                                      @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                      $endgroup$
                                                                                                                      – Stackstuck
                                                                                                                      Jun 3 at 16:12















                                                                                                                    0
















                                                                                                                    $begingroup$


                                                                                                                    C# (.NET Core), 136 bytes





                                                                                                                    class Astatic void Main()var r=new System.Random();int i=r.Next(6),j=6;while(i-->0)j*=1+r.Next(j);System.Console.Write(r.Next(j)+1);


                                                                                                                    Try it online!



                                                                                                                    I'm pretty sure this works, given the assumption of infinite integer length that we're fond of here. If I have to actually handle the overflow, I'd need to bust out an entirely different class.






                                                                                                                    share|improve this answer












                                                                                                                    $endgroup$














                                                                                                                    • $begingroup$
                                                                                                                      It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                      $endgroup$
                                                                                                                      – Kevin Cruijssen
                                                                                                                      Jun 3 at 9:34











                                                                                                                    • $begingroup$
                                                                                                                      @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                      $endgroup$
                                                                                                                      – Stackstuck
                                                                                                                      Jun 3 at 16:12













                                                                                                                    0














                                                                                                                    0










                                                                                                                    0







                                                                                                                    $begingroup$


                                                                                                                    C# (.NET Core), 136 bytes





                                                                                                                    class Astatic void Main()var r=new System.Random();int i=r.Next(6),j=6;while(i-->0)j*=1+r.Next(j);System.Console.Write(r.Next(j)+1);


                                                                                                                    Try it online!



                                                                                                                    I'm pretty sure this works, given the assumption of infinite integer length that we're fond of here. If I have to actually handle the overflow, I'd need to bust out an entirely different class.






                                                                                                                    share|improve this answer












                                                                                                                    $endgroup$




                                                                                                                    C# (.NET Core), 136 bytes





                                                                                                                    class Astatic void Main()var r=new System.Random();int i=r.Next(6),j=6;while(i-->0)j*=1+r.Next(j);System.Console.Write(r.Next(j)+1);


                                                                                                                    Try it online!



                                                                                                                    I'm pretty sure this works, given the assumption of infinite integer length that we're fond of here. If I have to actually handle the overflow, I'd need to bust out an entirely different class.







                                                                                                                    share|improve this answer















                                                                                                                    share|improve this answer




                                                                                                                    share|improve this answer








                                                                                                                    edited Jun 3 at 16:18

























                                                                                                                    answered Jun 2 at 19:54









                                                                                                                    StackstuckStackstuck

                                                                                                                    2091 silver badge7 bronze badges




                                                                                                                    2091 silver badge7 bronze badges














                                                                                                                    • $begingroup$
                                                                                                                      It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                      $endgroup$
                                                                                                                      – Kevin Cruijssen
                                                                                                                      Jun 3 at 9:34











                                                                                                                    • $begingroup$
                                                                                                                      @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                      $endgroup$
                                                                                                                      – Stackstuck
                                                                                                                      Jun 3 at 16:12
















                                                                                                                    • $begingroup$
                                                                                                                      It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                      $endgroup$
                                                                                                                      – Kevin Cruijssen
                                                                                                                      Jun 3 at 9:34











                                                                                                                    • $begingroup$
                                                                                                                      @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                      $endgroup$
                                                                                                                      – Stackstuck
                                                                                                                      Jun 3 at 16:12















                                                                                                                    $begingroup$
                                                                                                                    It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                    $endgroup$
                                                                                                                    – Kevin Cruijssen
                                                                                                                    Jun 3 at 9:34





                                                                                                                    $begingroup$
                                                                                                                    It can now give a System.ArgumentOutOfRangeException: 'maxValue' must be greater than zero error. The max result can be $6^32$, which is larger than the max size of both int and long, so you'll have to use BigIntegers instead.
                                                                                                                    $endgroup$
                                                                                                                    – Kevin Cruijssen
                                                                                                                    Jun 3 at 9:34













                                                                                                                    $begingroup$
                                                                                                                    @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                    $endgroup$
                                                                                                                    – Stackstuck
                                                                                                                    Jun 3 at 16:12




                                                                                                                    $begingroup$
                                                                                                                    @KevinCruijssen yes, that's the whole point of my comment.
                                                                                                                    $endgroup$
                                                                                                                    – Stackstuck
                                                                                                                    Jun 3 at 16:12











                                                                                                                    0
















                                                                                                                    $begingroup$


                                                                                                                    Julia 1.0, 60 bytes





                                                                                                                    g(b=big(6),r=rand)=(for i in 1:r(0:5) b=b*r(1:b) end;r(1:b))


                                                                                                                    b=big(6) makes it work with arbitrary sized integers
                                                                                                                    Try it online!






                                                                                                                    share|improve this answer










                                                                                                                    $endgroup$



















                                                                                                                      0
















                                                                                                                      $begingroup$


                                                                                                                      Julia 1.0, 60 bytes





                                                                                                                      g(b=big(6),r=rand)=(for i in 1:r(0:5) b=b*r(1:b) end;r(1:b))


                                                                                                                      b=big(6) makes it work with arbitrary sized integers
                                                                                                                      Try it online!






                                                                                                                      share|improve this answer










                                                                                                                      $endgroup$

















                                                                                                                        0














                                                                                                                        0










                                                                                                                        0







                                                                                                                        $begingroup$


                                                                                                                        Julia 1.0, 60 bytes





                                                                                                                        g(b=big(6),r=rand)=(for i in 1:r(0:5) b=b*r(1:b) end;r(1:b))


                                                                                                                        b=big(6) makes it work with arbitrary sized integers
                                                                                                                        Try it online!






                                                                                                                        share|improve this answer










                                                                                                                        $endgroup$




                                                                                                                        Julia 1.0, 60 bytes





                                                                                                                        g(b=big(6),r=rand)=(for i in 1:r(0:5) b=b*r(1:b) end;r(1:b))


                                                                                                                        b=big(6) makes it work with arbitrary sized integers
                                                                                                                        Try it online!







                                                                                                                        share|improve this answer













                                                                                                                        share|improve this answer




                                                                                                                        share|improve this answer










                                                                                                                        answered Jun 3 at 17:36









                                                                                                                        gggggggg

                                                                                                                        1,3946 silver badges8 bronze badges




                                                                                                                        1,3946 silver badges8 bronze badges
























                                                                                                                            0
















                                                                                                                            $begingroup$


                                                                                                                            Gaia, 13 bytes



                                                                                                                            :(ṛ×
                                                                                                                            6₅ṛ(↑ₓ(ṛ


                                                                                                                            Try it online!






                                                                                                                            share|improve this answer












                                                                                                                            $endgroup$



















                                                                                                                              0
















                                                                                                                              $begingroup$


                                                                                                                              Gaia, 13 bytes



                                                                                                                              :(ṛ×
                                                                                                                              6₅ṛ(↑ₓ(ṛ


                                                                                                                              Try it online!






                                                                                                                              share|improve this answer












                                                                                                                              $endgroup$

















                                                                                                                                0














                                                                                                                                0










                                                                                                                                0







                                                                                                                                $begingroup$


                                                                                                                                Gaia, 13 bytes



                                                                                                                                :(ṛ×
                                                                                                                                6₅ṛ(↑ₓ(ṛ


                                                                                                                                Try it online!






                                                                                                                                share|improve this answer












                                                                                                                                $endgroup$




                                                                                                                                Gaia, 13 bytes



                                                                                                                                :(ṛ×
                                                                                                                                6₅ṛ(↑ₓ(ṛ


                                                                                                                                Try it online!







                                                                                                                                share|improve this answer















                                                                                                                                share|improve this answer




                                                                                                                                share|improve this answer








                                                                                                                                edited Jun 3 at 19:12

























                                                                                                                                answered Jun 3 at 15:53









                                                                                                                                GiuseppeGiuseppe

                                                                                                                                19.4k3 gold badges16 silver badges71 bronze badges




                                                                                                                                19.4k3 gold badges16 silver badges71 bronze badges































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                                                                                                                                    • …Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
                                                                                                                                      Explanations of your answer make it more interesting to read and are very much encouraged.


                                                                                                                                    • …Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.


                                                                                                                                    More generally…



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