Exactly one of six people gets their own gift

Question

Six people place gifts in a pile and each takes one back at random, giving a uniformly random assignment of the six gifts, all $6! = 720$ equally likely.

What is the probability that exactly one person gets their own gift?

Accepted Answer

Exactly one match means one person is matched and the other five are all mismatched. Count in two steps: 1. Choose the single matched person: \binom{6}{1} = 6 ways. 2. Derange the remaining five gifts so nobody else matches: D5 = 44 ways. So there are 6 	imes 44 = 264 favorable arrangements, and P(	ext{exactly one match}) = \frac{6 \cdot D5}{6!} = \frac{264}{720} = \frac{11}{30} \approx 0.3667. Takeaway. As n grows, "exactly one match" tends to 1/e \approx 0.368, the same limit as "no match," a signature of the Poisson(1) behaviour of fixed points.

Exactly one of six people gets their own gift

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