# Looking for Hats

• Mar 17th 2010, 09:33 AM
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?

• Mar 17th 2010, 11:50 AM
Anonymous1
If there are $\displaystyle n$ men and $\displaystyle n$ hats. The probability that $\displaystyle k$ men select there own hat is:

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

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

Hope this helps.
Anonymous
• Mar 17th 2010, 01:35 PM
Anonymous1
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.

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

Booyah-Kasha (Cool).
Anonymous
• Mar 17th 2010, 03:15 PM
Plato
Quote:

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 $\displaystyle ^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, 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 $\displaystyle \sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $\displaystyle ^{10}\mathcal{P}_5$
• Mar 17th 2010, 03:20 PM
Anonymous1
Quote:

Originally Posted by Plato
I do not follow the first response. It may work.
Here is another.
Let $\displaystyle ^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, 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 $\displaystyle \sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $\displaystyle ^{10}\mathcal{P}_5$

I think it would speed up the computation to consider the compliment.
• Mar 17th 2010, 03:33 PM
Plato
Quote:

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.
• Mar 17th 2010, 03:42 PM
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.

Well can you see why it would ~ $\displaystyle Hypergeometric?$
• Mar 17th 2010, 03:57 PM
Plato
Quote:

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.$\displaystyle Hypergeometric?$

With an adjustment in indices, that would be the probability that none gets his own hat.