# What is the hat number problem?

I believe I remember the answer is surprisingly $\displaystyle \frac{1}{e}$ when calculating some permutation when people are switching hats. Do you know what I'm talking about?

It's supposedly applied mathematics where a number of people are switching hats and surprisingly a probability turn out to be $\displaystyle \frac{1}{e}$ or likewise.

## Update

Well I found it and I think it's surprsing that it's 1/e: http://books.google.se/books?id=OVkoCcszEZ0C&pg=PA39&dq=hat&redir_esc=y#v=onepage&q=hat&f=false

-
Maybe look at the Wikipedia article on derangements. – André Nicolas Jan 30 '13 at 7:34
Also called by various other names. – Glen_b Jan 30 '13 at 7:48

At a party $n$ men take off their hats. The hats are then mixed up and each man randomly selects one. What is the probability that no man select their own hat. Also show that if $n$ is tends to $\infty$, then the probability will become $1 \over e$.
To solve this problem use Poincare's Theorem. Normally your probability will be $$p=\sum_{k=0}^n\frac{(-1)^k}{k!}$$ and note that $$\lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{k!}= \frac{1}{e}$$
An intuitive reason why $e$ is involved: The probability a person gets their own hat back is $\frac{1}{n}$, so the probability they do not get their own back is $1-\frac{1}{n}$. If these events were independent for different people, the probability all $n$ people would get someone else's hat would be $\left(1-\frac{1}{n}\right)^n$, which converges to $\frac{1}{e}$ for large $n$. In actuality the events are not independent (if I don't get my own hat, it's more likely I got your hat, so less likely you got your own), but computing the actual answer shows the dependence is negligible. – Kevin Costello Jan 28 '15 at 23:25