From Christopher “CJ” Halverson comes a bibliophilic game of Secret Santa:
Every year, CJ’s family of five (including CJ) does a book exchange for Christmas. First, each person puts their name in a hat. The hat is shaken, and then each person draws a random name from the hat and gifts that person a book. However, if anyone draws their own name, they all put their names back into the hat and start over.
What is the probability that no one will draw their own name?
Assume that the five people in the family are AJ, BJ, CJ, DJ, and EJ.
WLOG, assume that BJ picks AJ's name.
AJ, CJ, DJ, and EJ have to now pick BJ, CJ, DJ, and EJ's names.
| BJ's name | CJ's name | DJ's name | EJ's name |
|---|---|---|---|
| AJ | DJ | EJ | CJ |
| AJ | EJ | CJ | DJ |
| CJ | AJ | EJ | DJ |
| CJ | DJ | EJ | AJ |
| CJ | EJ | AJ | DJ |
| DJ | AJ | EJ | CJ |
| DJ | EJ | AJ | CJ |
| DJ | EJ | CJ | AJ |
| EJ | AJ | CJ | DJ |
| EJ | DJ | AJ | CJ |
| EJ | DJ | CJ | AJ |
There are 11 ways this can be done without anyone picking their own name. There are a total of four family members who could have picked AJ's name in the beginning, so the answer is:
$$\dfrac{4 \cdot 11}{5!} = \dfrac{44}{120} \approx 36.7\%$$.
Note:
For large families, this probability approaches $\dfrac{1}{e} \approx 36.8\%$. Number of derangements.