Probability of birthdays shared in a group

  • #1
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Main Question or Discussion Point

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!
 
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Answers and Replies

  • #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:
 
  • #3
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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)?
 
  • #4
tiny-tim
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Yes, the probabilities are the same, but the events are different. :wink:
 
  • #5
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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...
 
  • #6
tiny-tim
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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]
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:
 
  • #7
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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]
 
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