TITLE: Secret-Santa: Probability of two people drawing each other.
QUESTION [8 upvotes]: When playing Secret Santa this year, where a group of $n$ people buy presents and these $n$ presents get randomly distributed to the other people, excluding the possibility of someone getting his or her own present (we did it with a online-distributer, so this possibility really is excluded), someone in this round asked the question about the probabilty of two people giving presents to each other. Since then I've been thinking about this problem, but I didnt get very far. The only thing that comes to my mind is working with the symmetric group $S_{n}$, and considering the possibilities $S_{n}\setminus F$, with $F$ being the set of all permuations with at least one fixpoint, as this is not possible in the game of Secret Santa. The probabilty I'm interested in now can be calculated by coming up with the cardinality of the set $\{\pi \in S_{n}\setminus F\ \mid \pi $ contains at least a two-cycle$\}$. Any futher ideas?

REPLY [4 votes]: The numbers of ways to distribute $n$ gifts according to the rules are OEIS A000166. The numbers of those ways where two people give gifts to each other (i.e. there is a $2$-cycle) is OEIS A158243. The probability you seek is then an entry of the first sequence divided by that of the second.
Asymptotically, as given in the formulas section of both sequences, the numerator tends to $n!(e^{-1}-e^{-3/2})$ and the denominator to $n!(e^{-1})$, leading to a limiting probability of $1-\frac1{\sqrt e}=0.393469\dots$