4
$\begingroup$

If I roll a 6-sided die x times (or roll x 6-sided dice at once), what is the probability that the sum of the result is greater than another y rolls (or one roll of y dice)?

Improve this question
$\endgroup$
5
  • $\begingroup$ Parts of the solution are given at stats.stackexchange.com/questions/291549, stats.stackexchange.com/questions/392943, and stats.stackexchange.com/questions/3614. stats.stackexchange.com/a/116913/919 includes software to (easily and naturally) compute the answer. Once x and y grow large (around 10 or so), Normal approximations work well (provided some care is taken also to estimate the chance of a tie). $\endgroup$
    – whuber ♦
    Commented Oct 9, 2021 at 15:08
  • 2
    $\begingroup$ Note that for fair six-sided dice, if V is the outcome on one die, 7-V has the same probability distribution as V. So you can use this to get all the dice terms on one side of the inequality. Let $S_x$ and $T_y$ be the totals for the x and y dice respectively. Then $P(S_x>T_y)$ $=$ $P(S_x>7y-T_y)$ $=$ $P(S_x+T_y>7y)$. That is, it's the same as the probability that the total on $x+y$ dice exceeds $7y$. So for example, the chance that a sum on 3 dice exceeds a sum on 2 dice is the same as the sum of 5 dice exceeding 14. ... ctd $\endgroup$
    – Glen_b
    Commented Oct 9, 2021 at 23:35
  • $\begingroup$ ... ctd ... Now the probabilities of sums on small numbers of dice are easy to generate (e.g. in a spreadsheet, such as described here: stats.stackexchange.com/questions/3614/…). This is fairly quick and easy up to about 10 total dice or so. Beyond that a normal approximation will usually be adequate, but some of the methods whuber points to are better if you need good accuracy further into the tails. $\endgroup$
    – Glen_b
    Commented Oct 9, 2021 at 23:37
  • $\begingroup$ Thank you that helped me very much. I guess using Troll software to determine the probability distribution of sum of dice is the best way to go. Very interesting that probability of x dice exceeds y dice is the same as x+y dice exceeds 7y, I would never have guessed that $\endgroup$
    – Fabian Allendorf
    Commented Oct 10, 2021 at 11:27
  • $\begingroup$ You misread Glen_b's comment: please distinguish "$T_y$" from "$y.$" $\endgroup$
    – whuber ♦
    Commented Mar 27 at 16:38

1 Answer 1

Reset to default
1
$\begingroup$

To give an answer, here is some R code, both doing the calculation exactly and with a normal approximation with continuity correction.

probdice <- function(numberdice, sides=6){
  probmatrix <- matrix(0, ncol=numberdice, nrow=numberdice*sides)
  probmatrix[1:sides, 1] <- 1 / sides
  for (d in 2:numberdice){
    for (s in 1:sides){
      probmatrix[(d-1+s):((d-1)*sides+s), d] <- 
          probmatrix[(d-1+s):((d-1)*sides+s), d] + 
          probmatrix[(d-1):((d-1)*sides), d-1] / sides
      }
    }
  return(probmatrix)
  }
 
probXgtY <- function(X, Y, sides=6){
  distXplusY <- probdice(X+Y, sides)[, X+Y]
  XgtY <- c(sum(distXplusY[(1:((X+Y)*sides)) >  7*Y]),  
            sum(distXplusY[(1:((X+Y)*sides)) == 7*Y]),
            sum(distXplusY[(1:((X+Y)*sides)) <  7*Y]))
  names(XgtY) <- c("P[sum(X)>sum(Y)]", "P[sum(X)=sum(Y)]", "P[sum(X)<sum(Y)]")

  return(XgtY) 
  }

pnormalapprox <- function(X, Y, sides=6){
  meandiff <- (X-Y) * (sides+1) / 2
  sddiff <- sqrt((X+Y) * (sides^2-1) / 12)
  PXgtY <- c(1 - pnorm(1/2, meandiff, sddiff), 
      pnorm(1/2, meandiff, sddiff) - pnorm(-1/2, meandiff, sddiff),
      pnorm(-1/2, meandiff, sddiff))
  names(PXgtY) <- c("P[sum(X)>sum(Y)]","P[sum(X)=sum(Y)]","P[sum(X)<sum(Y)]")
  return(PXgtY)
  }

Trying with $X=51$ dice and $Y=49$ dice gives these probabilities. The normal approximation with continuity correction is close.

probXgtY(51,49)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#       0.64805704       0.02145462       0.33048834 
 
pnormalapprox(51,49)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#       0.64825034       0.02147506       0.33027460

With small numbers of dice, the normal approximation is not bad, for example with $X=3,Y=2$:

probXgtY(3,2)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#       0.77854938       0.06944444       0.15200617 

pnormalapprox(3,2)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#       0.78394450       0.06860851       0.14744699

In a more extreme cases in the tail of the distribution, the normal approximation is good in absolute terms but can be poor in relative terms, for example with $X=40,Y=80$:

probXgtY(40,80)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#     6.444190e-15     3.694881e-15     1.000000e+00 

pnormalapprox(40,80)
# P[sum(X)>sum(Y)] P[sum(X)=sum(Y)] P[sum(X)<sum(Y)] 
#     2.953193e-14     1.487699e-14     1.000000e+00
Improve this answer
$\endgroup$
1
  • $\begingroup$ Alternatively, you could use the code I posted at stats.stackexchange.com/a/116913/919 to compute (say) mean(d(40,6) > d(80, 6)) or mean(d(40,6) == d(80, 6)) or whatever: that's fairly convenient. $\endgroup$
    – whuber ♦
    Commented Mar 27 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.