Exactly one friend gets their own gift back

Question

Four friends each bring one wrapped gift to a party. The gifts are collected, shuffled, and handed back so that each friend receives exactly one gift, uniformly at random among all possible assignments.

What is the probability that exactly one friend ends up with the gift they brought?

Accepted Answer

There are 4! = 24 equally likely ways to hand the gifts back. For exactly one correct match, first choose which friend gets their own gift: \binom{4}{1} = 4 ways. The remaining 3 friends must all get a wrong gift, an arrangement with no matches, called a derangement. For 3 items the number of derangements is D3 = 2 (the two valid rearrangements of three items with nobody fixed). So the number of favorable assignments is \binom{4}{1}	imes D3 = 4 	imes 2 = 8, giving P(	ext{exactly one match}) = \frac{8}{24} = \frac{1}{3}. Takeaway. "Exactly k matches" splits into two independent choices: whi…

Exactly one friend gets their own gift back

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