Hats picked at random

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At a dinner $4$ people leave their hat at the entrance and on leaving pick up one at random. What is the probability that no one picks up their own hat?

Solution

The random returns correspond to a permutation of the $4$ hats.

• Possible cases: $4! = 24$.
• Favorable cases: the permutations in which no hat returns to its owner, i.e. the derangements $D_{4}$. By the formula $D_{4} = 4!\left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 9.$

Hence $P = \frac{D_{4}}{4!} = \frac{9}{24} = \frac{3}{8} = \boxed{0.375}.$