In (one version of) the hat-check problem, https://proofwiki.org/wiki/Hat-Check_Problem the question is to find the probability that for $n$ hat-checkers, nobody gets their own hat. If this is called $p_n,$ the solution is to obtain the $n$th partial sum of the series for $1/e.$
So we have, for $n=1,2,3,\cdots,$ that $$p_n=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^n}{n!}.$$
For odd $n,$ the likelihood increases as $n$ does. This seems peculiar (to me) in that for more people checking their hats, it would become more likely that at least one person got their own hat, and so less likely that nobody did.
For even $n$ as $n$ goes up the likelihood $p_n$ decreases as expected by the vague (incorrect) intuition proposed above.
I'm looking for some reasonable intuitive reason why one would expect this even/odd $n$ behavior for this problem. Also from the answer, for any $m,n$ where $m$ even and $n$ odd, $p_m>p_n,$ which puzzles my intuition more.
Any intuitions (not proofs, which can be done and are known anyway) appreciated.
