Hat-Check Problem/Examples/3

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Example of Hat-Check Problem

$p_3 = \dfrac 1 3$


Proof

When $n = 3$, there are only three hats to hand back.

Hence:

\(\ds p_3\) \(=\) \(\ds \dfrac {!3} {3!}\)
\(\ds \) \(=\) \(\ds \dfrac {\ds 3! \sum_{k \mathop = 0}^3 \dfrac {\paren {-1}^k } {k!} } {3!}\) Definition of Subfactorial
\(\ds \) \(=\) \(\ds \dfrac {3! \paren {1 - 1 + \dfrac 1 {2!} - \dfrac 1 {3!} } } {3! }\)
\(\ds \) \(=\) \(\ds \dfrac 2 6\)
\(\ds \) \(=\) \(\ds \dfrac 1 3\)
$p_3$ is roughly $0.0345$ (or less than $\dfrac 1 {4!}$) away from the estimate of $\dfrac 1 e$.

$\blacksquare$