Secret Santa is a traditional event in December with family, friends, or office colleagues, in which each guest gives a gift to another guest without knowing who will give him/her a gift. Could you imagine that you want to organize it? Please let your guests know to choose a gift by keeping it secret. For maximum surprise, you would like to give each gift to a guest with a draw at the last minute. So your guests don’t know for whom the gift that he/she chose will be. However, several problems are raised that can be answered by applied mathematics.

🎅 Secret Santa: Hat-check problem

Let’s begin with the first problem: to whom each guest gives his/her gift. Let’s suppose each guest has well his/her gift for a unique other guest. Of course, the draw could have been done before and kept secret. But you want to do the draw at the last moment and by revealing simultaneously who gives his/her gift to whom. That is to say, you write the name of each guest on a different piece of paper for each and mix it in a Christmas hat. Then, each one takes a piece of paper randomly. Finally, all of them, at the same time, unfold the piece of paper and discover the name of the guest to whom he/she gives his/her gift.

Will a guest draw his/her name at random? If this is the case, the surprise effect will be broken for him/her, and what would imply redoing the draw or exchanging gifts? What we want to avoid.

This concrete problem can be reformulated: Is there at least a fixed point in the permutation of a set with $n$ elements? Where $n$ is the number of guests. It is the famous hat-check problem. You can use the probabilities to solve it.

The probability that at least one guest draws his/her name is the inverse of the probability that no one drew his/her name, that is to say:

\[p_n = 1 - \dfrac{\mathcal{D}_{n}}{n!} = 1 - u_n\]

$\mathcal{D}_{n}$ is the number of derangements, that is to say, the number of permutations without fixed points, knowing that $n!$ is the number of permutation with $n$ elements. You can prove with the inclusion-exclusion generalized principle that:

\[\mathcal{D}_{n} = n! \sum_{k=0}^{n} \dfrac{(-1)^k}{k!}\]

According to the Taylor-Maclaurin development, we have:

\[\displaystyle\lim_{n\mapsto +\infty}u_n = \displaystyle\lim_{n\mapsto +\infty}\sum_{k=0}^{n}\dfrac{(-1)^k}{k!} = \dfrac{1}{e}\]

with $e \approx 2.71828…$. In addition, this suite $(u_n)_{n\in \mathbb{N}}$ has a rate of convergence quick:

Inverse of the probability that a guest draws randomly his/her name

So if your number of guests is greater or equal to 4, you have around $0.63\%$ of chance, to have at least one guest who drew its name, that is a bit less than 2 out of 3 chances:

\[p_{n\geq 4} \approx \dfrac{2}{3}\]

It is risky to choose this simultaneous strategy, and it breaks the surprise phenomenon when everyone unfolds the paper if a guest randomly draws his/her name. An exchange of pieces of paper or a new drawing of lots must be made. Another sequential strategy could be considered: A guest randomly draws a piece of paper, keeps it, and gives the Christmas hat to the person whose name is on it. Only the first guest could draw her/his name, but all she/he had to do was redraw a piece of paper and then put the paper back with his/her name.

🎁 Gift types: Birthday problem

A second problem will be to know if 2 guests could bring the same type of gift. This is equivalent to another famous problem, the birthday problem. The probability that two people were born on the same day increases quickly depending on the number of people. It becomes almost certain, that is to say, more than $99\%$ of a chance to be valid for a group of 60 people.

More generally, the probability that a subset of size $n$ of a set $E$ of size $N$ contains at least 2 identical elements:

\[g_n = 1 - \frac{N}{N}\,\frac{N-1}{N}\,\cdots\,\frac{N-n+1}{N} = 1 - \dfrac{N!}{(N-n)!}\,\dfrac{1}{N^n}\]

It is the inverse of the probability that no $n$ element is the same.

The element $E$ is the set of possible types of Christmas gifts, and $n$ is the number of guests. Except for some guests with a very original gift, suppose that your guest chose a pre-defined list influenced by the constraints and the ads, with a few hundred types of items, let’s say $N=300$:

Probability to have 2 identical types of gifts among 300 possible types of Christmas gifts

With an approximation $e^{x} \approx 1 +x$ when $x$ is close to $0$, and the sum of young Gauss $\sum_{k=0}^{n}\,k = \frac{n(n+1)}{2}$, you can get:

\[g_n \approx 1 - e^{-\frac{n(n-1)}{2N}}\]

But in any case, you see:

\[\displaystyle\lim_{n\mapsto +\infty}g_n = 1\]

And:

\[g_{n\geq 20} \geq \dfrac{1}{2} \qquad g_{n\geq 60} \approx 1\]

If you have more than 20 guests, we have more than 1 out of 2 chances to have at least 2 identical types of gifts.

So after explaining these 2 problems, I wish you a good Secret Santa.