Four cars returned at random: no owner gets their own

Question

A careless valet returns four cars to their four owners in a completely random order, so every one of the $4! = 24$ assignments is equally likely.

What is the probability that no owner receives their own car?

Accepted Answer

We want a derangement of 4 items: an assignment with no owner matched to their own car. Count derangements with inclusion-exclusion, starting from all 24 arrangements and removing those where at least one owner is matched. Fixing j specific owners to get their own car leaves (4-j)! ways for the rest, with \binom{4}{j} choices of which are fixed. So D4 = 4!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24\left(1 - 1 + 0.5 - 0.1667 + 0.0417\right) = 9. The probability is \frac{D4}{4!} = \frac{9}{24} = \frac{3}{8} = 0.375. Takeaway. For four items the full-mismatch …

Four cars returned at random: no owner gets their own

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