At least one of six workers gets their own key

Question

Six workers receive their six locker keys back in a completely random order, so all 6! = 720 assignments are equally likely. What is the probability that at least one worker gets their own key?

Accepted Answer

Use the complement of a full mismatch. A no-match arrangement is a derangement of 6 items. Using Dn = (n-1)(D{n-1}+D{n-2}) with D4=9 and D5=44, D6 = 5\,(44 + 9) = 265. So the chance nobody matches is 265/720, and P(	ext{at least one match}) = 1 - \frac{265}{720} = \frac{455}{720} = \frac{91}{144} \approx 0.6319. Takeaway. As the group grows, the at-least-one-match probability settles near 1 - 1/e \approx 0.632, which six workers already show.

At least one of six workers gets their own key

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