I don't know some special name for exactly that sum, but it is closely related to what are known as polynomial coefficients or extended binomial coefficients. These are often written as ${(\genfrac{}{}{0ex}{}{n}{j})}_{k+1}$ and are defined indirectly via: $${\left(\sum _{i=0}^k{x}^i\right)}^n=\sum _{j=0}^{nk}{(\genfrac{}{}{0ex}{}{n}{j})}_{k+1}{x}^j$$ with ${(\genfrac{}{}{0ex}{}{n}{0})}_0=1$ and ${(\genfrac{}{}{0ex}{}{n}{j})}_{k+1}=0$ for $j\notin \{0,\dots ,nk\}$.

Polynomial coefficients and distribution of the sum of discrete uniform variables by Caiado, C.C.S. and Rathie, P.N. provides this definition and a recurrence relation to compute this number. As the title suggests, it also provides a direct solution to your probability problem. If $Y=\sum _{i=0}^n{X}_i$ where each $X_i$ is a discrete random variable uniformly sampled from $\{0,\dots ,k\}$, then $$P(Y=y)=\frac{1}{(k+1{)}^n}{(\genfrac{}{}{0ex}{}{n}{y})}_{k+1}$$ This is the probability that rolling $n$ dice with values from $0$ to $k$ produces the sum $y$. The more usual case of dice with values $1$ to $k$ can be reproduced by noting this is equivalent to rolling $n$ dice with values from $0$ to $k-1$ to produce the sum $y-n$.

The aforementioned paper gives a summation expression in terms of gamma functions which probably does reduce to your sum, but a clearer expression is in Restricted Weighted Integer Compositions and Extended Binomial Coefficients by Steffen Eger. Table 1 in the conclusion (page 22) list various binomial identities and the corresponding extended binomial identities derived in that paper. (See also equation 15 in Example 33.) One of which is: $${(\genfrac{}{}{0ex}{}{k}{n})}_{l+1}=\sum _{j\ge 0}(-1{)}^j(\genfrac{}{}{0ex}{}{k}{j})(\genfrac{}{}{0ex}{}{n+k-(l+1)j-1}{k-1})$$

Matching your notation and performing the adjustment as above gives: $${(\genfrac{}{}{0ex}{}{z}{n-z})}_a=\sum _{k\ge 0}(-1{)}^k(\genfrac{}{}{0ex}{}{z}{k})(\genfrac{}{}{0ex}{}{n-ak-1}{z-1})$$ This is almost exactly the sum you wrote except you have $(n-a)(k-1)$ where I have $n-ak-1$. Checking the $n=3$ and $a=z=2$ case, the above formula gives the correct result of $2$ whereas your variant is $0$, so I'm going to assume that was just a typo.

Just for completeness, for the second binomial coefficient factor to be non-zero we need $n-ak-1\ge z-1$ or $k\le (n-z)/a$ giving $\lfloor (n-z)/a\rfloor$ as an upperbound to the summation assuming the other numbers make sense, i.e. that $z\le n\le az$.

posted over 2 years ago

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Derek Elkins‭

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