Name: dan Who is asking: Student Level of the question: Secondary 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 xk 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  nCr. 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 xk-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