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)

I did not notice this - great observation.

The Hat-Check distribution with parameter $n$ is currently defined (where $n > 0$) I think it should be changed for where $n > 1$

The reason - shouldn't a totally ordered set require a set with at least 2 elements?

The definition - Let $X$ represent the number of elements in a a totally ordered set with $n$ elements that are not in the correct order.

I need help on this - I'm certain that I don't know what I'm talking about - (intuitively) I think any work done on totally ordered sets should mean that the aforementioned set MUST have 2 or more elements. Am I wrong? (What does order even mean on a set with one or fewer objects?). --Robkahn131 (talk) 12:04, 7 April 2022 (UTC)