Expected number of right hats

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Six guests check identical-looking hats and, on the way out, each grabs a hat uniformly at random. On average, how many of the six guests end up with their own hat?

Accepted Answer

This looks like it needs the full matching machinery, but the average number of matches is easy thanks to linearity of expectation: the expected size of a sum is the sum of the individual expected pieces, whether or not the pieces are independent. For each guest i, define an indicator that is 1 if guest i gets their own hat and 0 otherwise. By symmetry, guest i is equally likely to end up with any of the six hats, so P(	ext{guest } i 	ext{ gets own hat}) = \frac{1}{6}. The total number of correct matches is the sum of these six indicators, so its expected value is E[	ext{matches}] = 6 	im…

Expected number of right hats

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