At least one of five guests gets their own coat

Question

Five guests receive their five coats back in a completely random order, so all 5! = 120 assignments are equally likely. What is the probability that at least one guest gets their own coat?

Accepted Answer

Use the complement of a full mismatch. A no-match arrangement is a derangement of 5 items: D5 = 5!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}ight) = 44. So the chance nobody matches is 44/120, and P(	ext{at least one match}) = 1 - \frac{44}{120} = \frac{76}{120} = \frac{19}{30} \approx 0.6333. Takeaway. At least one match equals 1 minus the derangement probability D5/5!.

At least one of five guests gets their own coat

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