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Math Help - probability

  1. #1
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    probability

    can anyone help!!
    We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?
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  2. #2
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    Quote Originally Posted by sonia1 View Post
    can anyone help!!
    We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?
    Read this: Birthday problem - Wikipedia, the free encyclopedia
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  3. #3
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    i've looked at that link but I could'nt still work out an answer for two pairs
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  4. #4
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    Birthday problem

    Hello sonia1
    Quote Originally Posted by sonia1 View Post
    can anyone help!!
    We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?
    Let's assume that the two pairs involve 4 different people; in other words, we are not allowing 3 people, A, B and C all to have the same birthday, and then forming pairs like (A, B), (A, C) and (B, C).

    Then (as is well known - see the Wikipedia article in Mr F's post) the probability p(n) that in a group of n people, at least two people share a birthday is given by:

    p(n) = 1 - \frac{^{365}P_n}{365^n}, where ^{365}P_n is the number of permutations of n chosen from 365, = \frac{365!}{(365-n)!}

    Suppose now that in a room of n people, a pair is found to have a common birthday. Then in the remaining (n-2) people, the probability that there's at least one other pair with the same birthday is p(n-2).

    So the probability that both of these events occur is p(n)\times p(n-2).

    I have put the numbers into an Excel Spreadsheet - see attachment - using the PERMUT function to work out ^{365}P_n, and the answer comes out that with 31 people the probability is just under 0.5, and with 32 it's about 0.532.

    I don't think it's any more complicated than that. Can anyone see a flaw in my reasoning?

    Grandad
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  5. #5
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    thankyou

    thankyou for your help grandad.
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