I'm trying to work on a project in code that asks this question. I'm currently able to get the right answer by breaking it up, but I'm wondering if there's a mathematical formula I can address this problem to as I'm actually facing computational resource limitations.

For example, the amount of combinations for 50 dice is 50^6, which is 15,625,000,000 which is something my computer can't handle.

I'm trying to find the probability distribution of getting a certain value of n dice where no single value of any dice is 1.

For example, with a single dice, the probability of getting a 1,2,3,4,5, or 6 is 1/6.

For two dice, the probability of getting a total value of 4 or 12 is 1/36 (I ignore the case of 2 and 3 since one of the dice has to have a value of 1). The probability of a 5 or 11 WITHOUT a dice having a value of one is 1/18, 6 or 10 WITHOUT a dice having a value of 1 is 1/12, etc. etc.

Is anyone familiar with generating the probability distribution that all the possible values that can be rolled with n dice such that no dice contains a 1?

  • $\begingroup$ You have a poor computer indeed if it cannot handle an 11 digit number! Since most computers are binary, you might find it convenient to represent $50=10^2/2$, for then $$50^6=10^{12}/2^6=0.015625\times 10^{12}$$ shows that you only need to deal with a six-digit number separately from the exponent. Regardless, there are many solutions, including those given at stats.stackexchange.com/questions/3614 and stats.stackexchange.com/questions/24385. $\endgroup$
    – whuber
    Mar 11, 2018 at 14:38

1 Answer 1


Solve the problem for fifty 5-sided dices (and add 50 to the dice rolls).

You could do the first part (the standard dice roll problem) by

  • approximation with a normal distribution ( https://math.stackexchange.com/questions/406192/probability-distribution-of-rolling-multiple-dice )
  • by the formula for the exact solution from Whuber mentioned in the comments.

  • or by computation doing a convolution 49 times

    rolls <- rep(0,5*50)
    rolls[1:5] <- 1
    for (i in 1:49) {
        rolls <- c(0,rolls[1:245],0,0,0,0) +
                 c(0,0,rolls[1:245],0,0,0) +
                 c(0,0,0,rolls[1:245],0,0) +
                 c(0,0,0,0,rolls[1:245],0) +  

    The probability for 5-sided dice roll being x is rolls[x]/5^50. The probability for a 6-sided dice roll being x (and none of the dices is a one) is rolls[x-50]/5^60. Or rolls[x-50]/5^50 if you wish to condition on none of the dices being 1.

The above code can be improved by using 128 bit integers (if you find that necessary)

  • 1
    $\begingroup$ To what, exactly will you add 50? Certainly not to the probability. Given that the chance of the event in question is $(5/6)^{50}\approx 0.00011,$ there doesn't seem to be anything in the output of your code that directly corresponds to it. A little more explanation will help readers understand this solution. $\endgroup$
    – whuber
    Mar 11, 2018 at 14:33
  • $\begingroup$ I intended to add the 50 to the variable. The probability for X=x is rolls[x-50]/6^50. But I actually didn't wanted to give away too much, create a completely finished and polished solution, before it was clear that this is not a self-study question. $\endgroup$ Mar 11, 2018 at 15:00

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.