At least one of four people grabs their own umbrella

Question

Four people leave a cafe and each grabs one of the four identical-looking umbrellas at random, so all $4! = 24$ assignments are equally likely.

What is the probability that at least one person grabs their own umbrella?

Accepted Answer

"At least one match" is the opposite of "no match at all," so use the complement. A no-match assignment is a derangement of 4 items, counted by inclusion-exclusion: D4 = 4!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}ight) = 9. So the chance of no match is D4/4! = 9/24, and P(	ext{at least one match}) = 1 - \frac{9}{24} = \frac{15}{24} = \frac{5}{8} = 0.625. Takeaway. For "at least one gets their own," compute the derangement (no-match) probability and subtract from 1.

At least one of four people grabs their own umbrella

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