Question: Simply put, I am curious to find the likelyhood of each possibility (4-100) if I were to roll 4 25-sided die. It seems to start off simply, as 1, 4, 10, 20...following the inside diagonal of Pascal's triangle. But as soon as i reach the point where the sum is greater than the number of sides, Pascal's no longer seems to work. I would be greatly appreciated if I could be informed of the proper formula to find this answer, or perhaps if my theory of doing it is wrong.

Hi Dan,

On a roll each die will show a number between 1 and 25. Let k be the sum of the four numbers that appear on the dice, then 4 ≤ k ≤ 100.

We solved this using generating functions, so for each k we need the we need the coefficient of x^k in the polynomial

f(x) = (x¹ + x² + ··· + x²⁵)⁴

There is a common factor of x⁴ in this expression so

f(x) = (x¹ + x² + ··· + x²⁵)⁴
= x⁴ (1 + x¹ + ··· + x²⁴)⁴
= x⁴ [ ^{(1 - x25)} /_{(1 - x)} ]⁴
= x⁴ ( 1 - x²⁵ )⁴ ( ¹/_{1 - x} )⁴
= x⁴ ( 1 - x²⁵ )⁴ (1 - x)^-4

Next we expanded using the binomial theorem but first we should say something about notation. The symbol is called a binomial coefficient and it is the number of ways of selecting r things from n things. Some textbooks use the notation  _nC_r. The number of ways of selecting r things from n things is

where , for any positive integer k, k! = k (k-1) (k-2) ··· 2 1

So, for example

Now, back to expanding using the binomial theorem.

This is an infinite sum but, inside the square brackets we want the coefficient of x^k-4 where 0 ≤ k-4 ≤ 96.

We get the coefficient to be

Here we have used the convention that

if r < 0 then = 0 and if r = 0 then = 1.

Thus, for example, if k = 50 then the coefficient of x⁵⁰ is

Andrei and Shawn