Suppose that each of $n$ men at a party throws his hat into the center of the room...

Question

Suppose that each of $n$ men at a party throws his hat into the center of the room. The hats are mixed up and then each man randomly selects a hat. What is the probability that at least one of the men selects his own hat?

Answer 1

It has to do with derangements: $n!-D(n)$.

Look here: https://en.wikipedia.org/wiki/Derangement

Answer 2

First of all it's a uniform probability space therefore:

Let's define the event: $A_0$ = {no man selects his own hat}

Therefore we know that $B_0$ = {at least one man select his own hat} = $S_n-A_0$, where $S_N$ is the set of all permutations.

Using the Inclusion–exclusion principle we find that |$A_0$| = $D_n$ = $\sum_{i=1}^{n} \frac{n!}{i!}(-1)^{i+1}$

Therefore $P(B_0) = \frac{n! - D_n}{n!} = \sum_{i=2}^{n}\frac{ (-1)^{i}}{i!}$

Answer 3

By the Inclusion Exclusion principle, we have $$P(\cup A_i)=\sum P(A_i)-\sum P(A_i\cap A_j)+\ldots+(-1)^{n+1}\sum P(\cap A_i),$$

where $A_i$ is the event that the $i^{th}$ person picks his own hat.

A hint: $$P(A_1\cap A_2\cap\ldots\cap A_k)=\frac{(n-k)!}{n!}$$