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Probability of birthdays shared in a group

  1. Mar 10, 2010 #1
    In a group of 23 people, the probability that any 2 people share a birthday is:

    1 - [tex]\frac{365!}{342!365^{23}}[/tex]

    Why can't I just do the following?

    [tex](23\mathbf{C}2)(\frac{1}{365})[/tex]

    Thanks!
     
    Last edited: Mar 10, 2010
  2. jcsd
  3. Mar 10, 2010 #2

    tiny-tim

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    Hi IniquiTrance! :smile:

    Do you mean 1 - 23C2/365 ?

    because there are "overlaps" that you aren't subtracting …

    the individual events (of one particular pair not sharing the same birthday) are not independent … eg the pairs Tom and Dick, Tom and Harry, and Dick and Harry, are not independent. :wink:
     
  4. Mar 10, 2010 #3
    Hmm, I kind of see what you're saying...

    But why is say P(Tom and Harry|Dick and Harry) different than P(Tom and Harry)?
     
  5. Mar 10, 2010 #4

    tiny-tim

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    Yes, the probabilities are the same, but the events are different. :wink:
     
  6. Mar 10, 2010 #5
    Thanks for your response. I'm on the verge of uynderstanding it, can you think of any other way to explain it?

    I thought the definition of independence is that [tex]E_{i}[/tex] and [tex]E_{j}[/tex] are independent events so long as [tex]P(E_{i}|E_{j}) = E_{i}[/tex]

    and [tex]P(E_{j}|E_{i})= E_{j}[/tex]

    A bit confused...
     
  7. Mar 10, 2010 #6

    tiny-tim

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    I prefer to write it P(Ei and Ej) = P(Ei)P(Ej).

    (because, that way, you can string more than two together)

    But it's much easier just to use common-sense, and to say that the three events of Tom and Dick, Tom and Harry, and Dick and Harry, sharing (or not sharing) a birthday are obviously not independent. :smile:
     
  8. Mar 11, 2010 #7
    I wrote something about this a while ago in my http://yabm.wordpress.com/2010/02/16/a-bunch-of-people-in-a-room/" [Broken]
     
    Last edited by a moderator: May 4, 2017
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