In Folha, Marcelo Viana talks about the Secret Santa exchange.
Reproduction of Marcelo Viana's column in Folha de S.Paulo.
The strange year of 2020 is drawing to a close, and soon it will be time for the Christmas festivities. One of the most charming aspects of our tradition is the Secret Santa game: for each person in the group – family or work colleagues – someone in the group is secretly drawn to whom that person will give a gift.
Here at home, I was the one who handled the drawing. The simplest way is to write the names on slips of paper, which are then folded. Each person chooses a slip and secretly checks their friend's name. Each possible result is called a permutation. The total number of permutations of N people is equal to N! (read N factorial), which is N times (N-1) times … times 3 times 2 times 1.
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But there's a problem with this procedure: however pleasurable it may be to give gifts to oneself, the secret friend cannot be oneself! Therefore, only those permutations in which no one chooses the paper with their own name are valid; these are called group misalignments.
When I was organizing the family's Secret Santa, I was aware of this problem, but I hoped something would go wrong on the first try, and if not, we'd try again. But after four failed attempts, the family got tired, and I think their confidence in my math skills plummeted… I was replaced by an app! That's when I decided to study the subject properly.
The problem of determining how many derangements exist was formulated in 1708 by the Frenchman Pierre Rémond de Montmort (1678–1719) and was solved by himself, and also by his friend Nicolas Bernoulli (1687–1759), around 1713. What they found was that the total number of derangements in a group with N members is the integer closest to N factorial divided by Euler's constant e = 2.718281828459045… (Readers would be interested in verifying this fact. Responses are welcome via email at viana.folhasp@gmail.com)
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