Sum of n elements of a finite set of integers, 1 through s

AI Thread Summary
The discussion focuses on calculating the probability of rolling a total t with n s-sided dice, specifically how to determine the number of combinations that yield this sum. The challenge arises in the range where n+s ≤ t ≤ (n-1)s, as traditional combinatorial methods become ineffective. Participants explore the convergence of the sum of rolls to a normal distribution, referencing the weak law of large numbers. They seek clarification on the parameters σ and μ in the context of the normal distribution related to n, s, and t. The conversation highlights the complexity of the problem and the need for an efficient solution.
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The general problem I'm trying to solve is the probability of rolling a total t on n s-sided dice. A good chunk of the problem is easy enough, but where I run into difficulty is this:

How many combinations of dice will yield a sum total of t? Because the number set is limited, {a \choose n-1} (where a={n(s+1) \over 2} - \left|{n(s+1) \over 2} - t\right|) no longer works when n+s \leq t \leq (n-1)s. It is this region in the middle that interests me. Enumerating all combinations could be time-consuming, and, I expect, is entirely unnecessary. Is there a known formula for computing these numbers?
 
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sum of the m rolls diverges as m gets large. but we can use standardize to make it converge to standard normal distribution. this the weak law of large number says.
 
I'm not sure I understand what you're saying. The total t must be in the range [n, ns]. I am aware of the idea that it would probably converge to a normal distribution, but for {1 \over {\sqrt{2\pi\sigma^2}}} e^{-{{\left((x-{n(s+1) \over 2})-\mu\right)^2} \over {2\sigma^2}}}, what would \sigma and \mu be in terms of n, s and t?
 
That's exactly what I was looking for, thanks!
 
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