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### Probability of a Sum on Multiple Dice

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Date: 03/26/2001 at 07:11:15
From: Regan
Subject: Probability of getting a sum s on n dice with x sides

Hi,

I want to be able to write a program for calculating the probability
of getting a sum s on n dice with x sides.

I have seen the answer before and it uses binomial coefficients, but
I can't find it again and I'm stuck.

Thanks.
```

```
Date: 03/26/2001 at 13:31:33
From: Doctor Rob
Subject: Re: Probability of getting a sum s on n dice with x sides

Thanks for writing to Ask Dr. Math, Regan.

You can figure this out by using generating functions.

The generating function for one die with x sides is

f(z) = z + z^2 + z^3 + ... + z^x.

The coefficient of a term tells you how many ways you can roll the
exponent of z in the term. In this case, for numbers from 1 to x, you
can roll each in one way. For other numbers, like 0 and numbers
greater than x, you can roll them in zero ways. Thus the above
generating function is right for one die. For two dice, square the
above function. For n dice, raise it to the nth power. Then you want
the coefficient of z^s in that expression. When calculating this
coefficient for some specific value of s, you can do the arithmetic
and ignore all powers of z with exponents bigger than s. If you want
all the probabilities, you can ignore all exponents bigger than
(n*x+1)/2, because the number of ways of rolling s is the same as the
number of ways of rolling n*x+1-s, and if one is larger than
(n*x+1)/2, then the other is smaller.

To express this in binomial coefficients, you can write:

f(z) = (z - z^[x+1]) / (1 - z)

f(z)^n = [(z -z^[x+1]) / (1 - z)]^n

= z^n * (1-z^x)^n * (1-z)^(-n)

n                               n
= z^n * SUM (-1)^k * C(n,k) * z^(x*k) * SUM C(n+i-1,i) * z^i
k=0                             i=0

n   n
= SUM SUM (-1)^k * C(n,k) * C(n+i-1,n-1) * z^(n+x*k+i)
k=0 i=0

Now the coefficient of z^s in this will be given by:

(s-n)/x
SUM  (-1)^k * C(n,k) * C(s-x*k-1,n-1)
k=0

This is the formula you seek.

For example, if x = 6, n = 4, and s = 13, then the coefficient of z^13
is:

(13-4)/6
SUM   (-1)^k * C(4,k) * C(12-4*k,3)
k=0

= (-1)^0 * C(4,0) * C(12,3)  +  (-1)^1 * C(4,1) * C(6,3)

= 220 - 4*20

= 140

This is the coefficient of z^13 in:

z^4
* (1 - 4*z^6 + ...)
* (1 + 4*z   + 10*z^2 +  20*z^3 +  35*z^4 +  56*z^5 + 84*z^6 +
+ 120*z^7 + 165*z^8 + 220*z^9 + ...)

Thus there are 140 ways to roll 13 with four 6-sided dice.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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