Four gifts handed back at random

Question

Four people each bring one wrapped gift to a party. The gifts are pooled and handed back so that each person receives a uniformly random gift.

What is the probability that no one receives the gift they brought?

Accepted Answer

We want the fraction of the 4! = 24 orderings in which nobody is matched to their own gift, that is, the derangements of four items. Using the derangement formula, D4 = 4!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}ight) = 24\left(1 - 1 + 	frac12 - 	frac16 + 	frac{1}{24}ight) = 24 \cdot 	frac{9}{24} = 9. You can also list them by hand to confirm there are exactly 9 ways for four labeled items to avoid every fixed position. So \frac{D4}{4!} = \frac{9}{24} = \frac{3}{8} = 0.375. This is already close to the limiting value 	frac1e \approx 0.368, showing how quickl…

Four gifts handed back at random

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