Results 1 to 8 of 8

Math Help - Looking for Hats

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    42

    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?

    Thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    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
    Last edited by Anonymous1; March 17th 2010 at 01:15 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by JoAdams5000 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Quote Originally Posted by Plato View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by Anonymous1 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by Anonymous1 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Problem of hats
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: December 23rd 2010, 11:07 AM
  2. Prisoner' Hats
    Posted in the Math Puzzles Forum
    Replies: 2
    Last Post: November 24th 2009, 03:03 AM
  3. Variations on colours and hats
    Posted in the Math Challenge Problems Forum
    Replies: 12
    Last Post: July 8th 2009, 04:38 PM
  4. Hats off to helpful mathaticians
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 30th 2009, 09:41 PM

Search Tags


/mathhelpforum @mathhelpforum