Suppose that $X_1,\dots,X_n$ are fair $k$-sided dices and $Z = \sum_{i=1}^nX_i$. How could you determine the probability mass function of $Z$? Since each roll is independent this all comes down to calculating the number of combinations for any possible sum value $a \in \{n,\dots,nk\}$. If $n = 2$, then I am guessing that the probability of any one value $a$ is $P(Z = a) = \sum_{m = \max\{1,a - k\}}^{a - 1}P(X_1 = m)*P(X_2 = a - m) = \frac{1}{k^2}\sum_{m = \max\{1,a - k\}}^{a - 1}$, but it I can't see any immediate way of simplifying this expression. For small values of $n$ and $k$ one method is just to brute force the different combinations, but that doesn't really feel satifying.
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$P(Z=m)$ can be described as the coefficient of $x^m$ in polynomial $(x+x^2+\cdots+x^k)^n$ but AFAIK there is no nice expression for it. – drhab Feb 15 '22 at 15:28
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@drhab Do you mean in the case that $k = 6$? Or why do have $x^1,\dots,x^6$ in the parentheses? Also, where does this observation come from? – Epsilon Away Feb 15 '22 at 15:30
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I overlooked that you were talking about $k$-sided dice and repaired my former comment. It comes from generating functions. – drhab Feb 15 '22 at 15:30