Talk:Variance of Hat-Check Distribution

This theorem holds for $n \ge 2$. When $n = 1$, trivially $\var X = 0$. What's wrong? In fact, $\ds \sum_{k \mathop = 0}^{n - 1} \frac k {\paren {n - 1 - k}!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} = n - 2$. --Fake Proof (talk contribs) 09:53, 6 April 2022 (UTC)