Seven papers handed back at random: nobody gets their own

Question

A teacher hands back seven graded papers to seven students in a completely random order, so all $7! = 5040$ arrangements are equally likely.

What is the probability that no student receives their own paper?

Accepted Answer

We need a derangement of 7 items. Build up the count with the recurrence Dn = (n-1)(D{n-1} + D{n-2}), using D5 = 44 and D6 = 265: D7 = 6\,(265 + 44) = 6 	imes 309 = 1854. So the probability is \frac{D7}{7!} = \frac{1854}{5040} = \frac{103}{280} \approx 0.3679. This is the same setup inclusion-exclusion would give through D7 = 7!\sum{k=0}^{7}(-1)^k/k!; the recurrence is just faster. Takeaway. By seven items the full-mismatch probability 103/280 \approx 0.3679 is nearly indistinguishable from 1/e \approx 0.36788.

Seven papers handed back at random: nobody gets their own

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