Secret Santa self-draw

Question

Three friends run a Secret Santa by writing their names on slips, mixing them in a hat, and each drawing one slip at random. What is the probability that at least one person draws their own name?

Accepted Answer

"At least one match" is cleanest through its complement: 1 minus the probability that nobody draws their own name, which is a derangement of three items. The number of derangements of three items is D3 = 3!\left(1 - 1 + 	frac12 - 	frac16ight) = 6 \cdot 	frac{2}{6} = 2, so the probability that nobody draws their own name is 	frac{2}{6} = 	frac13. Therefore P(	ext{at least one self-draw}) = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667. You can double-check by listing all 6 orderings of three slips: exactly 2 are complete mismatches and the other 4 have someone drawing their own name, giv…

Secret Santa self-draw

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