1. ## Looking for Hats

Five men are trying to find their hats in a closet. The closet contains their hats as well as 5 additional hats. (total 10 hats) What is the probability that at least one man retrieves his own hat?

2. If there are $n$ men and $n$ hats. The probability that $k$ men select there own hat is:

$\frac{(n-k)!}{n!}\times P_{n-k} \times {{n}\choose{k}}$ Where, $P_n = \sum^n \frac{(-1)^i}{i!}$ is the probability that no matches occur.

For your case, extend the number of hats to $15,$ and let $k$ to be $0,$ then to be $1.$ Now take $1$ minus the sum of your values for $k=0$ and $k=1.$

Hope this helps.
Anonymous

3. I derived the formula above through recursion. I didn't type it out because I'm lazy. But, I just realized if you wanna be slick about it.

$Pr($k men select there own hat $)$ ~ $Hypergeometric(5, k, 15)$

Booyah-Kasha .
Anonymous

Five men are trying to find their hats in a closet. The closet contains their hats as well as 5 additional hats. (total 10 hats) What is the probability that at least one man retrieves his own hat?
I do not follow the first response. It may work.
Here is another.
Let $^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, the number of permutations of N things taken k at a time.
Use the inclusion/exclusion principle.
The number of ways for at least one man to get his own hat is $\sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $^{10}\mathcal{P}_5$

5. Originally Posted by Plato
I do not follow the first response. It may work.
Here is another.
Let $^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, the number of permutations of N things taken k at a time.
Use the inclusion/exclusion principle.
The number of ways for at least one man to get his own hat is $\sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $^{10}\mathcal{P}_5$
I think it would speed up the computation to consider the compliment.

6. Originally Posted by Anonymous1
I think it would speed up the computation to consider the compliment.
That may well be the case.
However, I still do not understand your solution.

7. That came off wrong. I wasn't knocking your solution, I was saying that instead of considering the sum from 1 to 5, consider 1 minus the sum of 0 and 1.

Well can you see why it would ~ $Hypergeometric?$

8. Originally Posted by Anonymous1
That came off wrong. I wasn't knocking your solution, I was saying that instead of considering the sum from 1 to 5, consider 1 minus the sum of 0 and 1. $Hypergeometric?$
With an adjustment in indices, that would be the probability that none gets his own hat.