Three coats returned at random: nobody gets their own

Question

Three guests hand in their coats, and a cloakroom mix-up returns the three coats in a completely random order, so all $3! = 6$ arrangements are equally likely.

What is the probability that no guest gets their own coat back?

Accepted Answer

We want a derangement of 3 items. It is small enough to list. Label guests and coats 1, 2, 3; the six arrangements (as the coat given to guest 1, 2, 3) are: 123,\ 132,\ 213,\ 231,\ 312,\ 321. A derangement has no position matching its own label. Checking each, only 231 and 312 qualify, so D3 = 2. Inclusion-exclusion confirms this: D3 = 3!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) = 6\left(1 - 1 + 0.5 - 0.1667\right) = 2. The probability is \frac{D3}{3!} = \frac{2}{6} = \frac{1}{3} \approx 0.3333. Takeaway. With only three items you can list all arrangements, and the two deran…

Three coats returned at random: nobody gets their own

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