At least one of seven people grabs their own phone

Question

Seven people at a meeting pick up their seven identical-looking phones in a completely random order, so all $7! = 5040$ assignments are equally likely.

What is the probability that at least one person picks up their own phone?

Accepted Answer

Use the complement of a full mismatch. A no-match arrangement is a derangement of 7 items. Using Dn = (n-1)(D{n-1}+D{n-2}) with D5 = 44 and D6 = 265, D7 = 6\,(265 + 44) = 1854. So the chance nobody matches is 1854/5040, and P(	ext{at least one match}) = 1 - \frac{1854}{5040} = \frac{3186}{5040} = \frac{177}{280} \approx 0.6321. Takeaway. The at-least-one-match probability converges to 1 - 1/e \approx 0.63212, and seven items give 177/280 \approx 0.63214, essentially the limit.

At least one of seven people grabs their own phone

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