Derangements: Nobody Gets Their Own Item

A classic setup: n people write their names on gifts, the gifts get shuffled randomly, and everyone receives one gift. What's the probability that nobody receives their own gift? A permutation with this property, nothing ends up in its original spot, is called a derangement.

Counting derangements directly is fiddly, but inclusion-exclusion gives a clean formula: the number of derangements of n items is n! times the alternating sum $1 - 1/1! + 1/2! - 1/3! + \dots \pm 1/n!$. For n = 4, that's $24 \times (1 - 1 + 0.5 - 0.1667 + 0.0417) \approx 24 \times 0.375 = 9$ derangements out of 24 total permutations, so the probability nobody gets their own gift is 9/24 = 0.375.

What makes the puzzle memorable is that as n grows, that alternating sum converges extremely fast to $1/e \approx 0.3679$, so the probability of a total mismatch settles near 37% regardless of whether there are 10 people or 10,000, a rare case where a combinatorics answer stops depending on the size of the problem almost as soon as n passes single digits.