In a secret Santa game, each of 6 participants randomly draws a name from a hat containing all 6 participants' names, including their own. What is the probability that at least one person draws their own name?

Instruction: Assume six people receive a uniformly random permutation of all six names, without replacement or redraws. Use inclusion–exclusion to calculate the probability that at least one receives their own name.

Context: This question challenges the candidate to apply a complex combinatorial principle to calculate the probability of a common holiday game scenario.

Updated

Official answer

Assume names are drawn uniformly without replacement, each name goes to exactly one person, and nobody redraws after receiving their own name. All 6!=720 assignments are equally likely.

For a particular set of k people to receive their own names, the remaining names have (6−k)! assignments. Inclusion–exclusion gives the count with at least one self-draw:

C(6,1)5! − C(6,2)4! + C(6,3)3! − C(6,4)2! + C(6,5)1! − C(6,6)0!

=720−360+120−30+6−1=455.

Thus P(at least one)=455/720=91/144≈63.19%. The remaining 265 assignments are derangements. Rejecting self-draws or restarting the game changes the sampling process and the question.

Related Questions