# 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.

Can you inform more about this?

## 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
add comment

## 1 Answer

I think you are taking about this problem:

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 $$\lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{k!}= \frac{1}{e}$$

-
Yes, thank you, now I only wanna grasp what e has to do with it. –  909 Niklas Jan 30 '13 at 7:31
@NickRosencrantz : I edit my answer. Again if you face any problem then tell me. –  A.D Jan 30 '13 at 7:35
add comment