Formula for calculating probability on customized dice.












0












$begingroup$


Given the following:
Player may choose from among the following:



1-10 White dice [0,0,0,1,1,2]

0-40 Blue dice [0,0,1,1,1,2]

0-10 Red dice [0,1,1,1,1,2]


and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.



I know this can be brute forced for each combination, but is there a simple formula to do this?






probability dice







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  • 1




    $begingroup$
    What does 1-10 mean?
    $endgroup$
    – Henry
    Dec 4 '18 at 20:51










  • $begingroup$
    @Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
    $endgroup$
    – Ross Millikan
    Dec 4 '18 at 21:26












  • $begingroup$
    @ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
    $endgroup$
    – Jared Tritsch
    Dec 4 '18 at 23:07
















0












$begingroup$


Given the following:
Player may choose from among the following:



1-10 White dice [0,0,0,1,1,2]

0-40 Blue dice [0,0,1,1,1,2]

0-10 Red dice [0,1,1,1,1,2]


and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.



I know this can be brute forced for each combination, but is there a simple formula to do this?






probability dice







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does 1-10 mean?
    $endgroup$
    – Henry
    Dec 4 '18 at 20:51










  • $begingroup$
    @Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
    $endgroup$
    – Ross Millikan
    Dec 4 '18 at 21:26












  • $begingroup$
    @ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
    $endgroup$
    – Jared Tritsch
    Dec 4 '18 at 23:07














0












0








0





$begingroup$


Given the following:
Player may choose from among the following:



1-10 White dice [0,0,0,1,1,2]

0-40 Blue dice [0,0,1,1,1,2]

0-10 Red dice [0,1,1,1,1,2]


and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.



I know this can be brute forced for each combination, but is there a simple formula to do this?






probability dice







share|cite|improve this question











$endgroup$




Given the following:
Player may choose from among the following:



1-10 White dice [0,0,0,1,1,2]

0-40 Blue dice [0,0,1,1,1,2]

0-10 Red dice [0,1,1,1,1,2]


and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.



I know this can be brute forced for each combination, but is there a simple formula to do this?






probability dice




probability dice






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 23:15









Ross Millikan

293k23197371




293k23197371










asked Dec 4 '18 at 19:42









Jared TritschJared Tritsch

1085




1085








  • 1




    $begingroup$
    What does 1-10 mean?
    $endgroup$
    – Henry
    Dec 4 '18 at 20:51










  • $begingroup$
    @Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
    $endgroup$
    – Ross Millikan
    Dec 4 '18 at 21:26












  • $begingroup$
    @ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
    $endgroup$
    – Jared Tritsch
    Dec 4 '18 at 23:07














  • 1




    $begingroup$
    What does 1-10 mean?
    $endgroup$
    – Henry
    Dec 4 '18 at 20:51










  • $begingroup$
    @Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
    $endgroup$
    – Ross Millikan
    Dec 4 '18 at 21:26












  • $begingroup$
    @ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
    $endgroup$
    – Jared Tritsch
    Dec 4 '18 at 23:07








1




1




$begingroup$
What does 1-10 mean?
$endgroup$
– Henry
Dec 4 '18 at 20:51




$begingroup$
What does 1-10 mean?
$endgroup$
– Henry
Dec 4 '18 at 20:51












$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26






$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26














$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07




$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07










2 Answers
2






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$begingroup$

For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Worked it out using Generating functions.



    These dice can be expressed as:




    (1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R




    where W, B, and R are the quantities of each dice.



    Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      0












      $begingroup$

      For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.






          share|cite|improve this answer











          $endgroup$



          For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 23:12

























          answered Dec 4 '18 at 19:51









          Ross MillikanRoss Millikan

          293k23197371




          293k23197371























              0












              $begingroup$

              Worked it out using Generating functions.



              These dice can be expressed as:




              (1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R




              where W, B, and R are the quantities of each dice.



              Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Worked it out using Generating functions.



                These dice can be expressed as:




                (1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R




                where W, B, and R are the quantities of each dice.



                Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Worked it out using Generating functions.



                  These dice can be expressed as:




                  (1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R




                  where W, B, and R are the quantities of each dice.



                  Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.






                  share|cite|improve this answer









                  $endgroup$



                  Worked it out using Generating functions.



                  These dice can be expressed as:




                  (1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R




                  where W, B, and R are the quantities of each dice.



                  Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 0:39









                  Jared TritschJared Tritsch

                  1085




                  1085






























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