Probability of successful Secret Santa selection

Question

In the following Secret Santa Scenario:

There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.

What is the probability that no one picks out their own name?

At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.

Answer 1

There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)

Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by

$$\text{# of derangements} = n! \cdot \sum_{k=1}^n \frac{(-1)^k}{k!}$$

among other potential relations. This is commonly denoted $!n$.

Answer 2

Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.

The answer is

$$\frac{!12}{12!}=\frac{\lfloor \frac{12!}e + \frac12 \rfloor}{12!}=\frac{176214841}{12!}\approx 0.3679$$