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Math Help - probability of same birthdays

  1. #1
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    probability of same birthdays

    in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

    (365^n -(365!/(365 - n)!))/365^n

    why does this not give the probability as 1 for n = 365
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  2. #2
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    Quote Originally Posted by hmmmm View Post
    in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

    (365^n -(365!/(365 - n)!))/365^n

    why does this not give the probability as 1 for n = 365
    Read this: Birthday problem - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by hmmmm View Post
    in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

    (365^n -(365!/(365 - n)!))/365^n

    why does this not give the probability as 1 for n = 365
    the problem assumes there are 365 days you can be born on (ignores leep years and junk)
    if there are 365 people in a room, it is possible for each of them to have a different birthday since there are 365 days to have one on.
    366 (or more), on the other hand, evaluates to 1.
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