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### Re: Birthday Chances

 on Mar 05, 2002 at 19:29 UTC ( #149476=note: print w/replies, xml ) Need Help??

in reply to Birthday Chances

It is fun to run simulations of situations like that. I think it is most pleasing because all you have to do is write a little code, and you get a whole bunch of data in return.

If you still care, here is the actual answer to that problem.

P(n,d) is the probability that at least two people in a group of n will share the same birthday out of d possible birthdays.
```P(n,d) = 1 - d!/((d-n)!*d^n)     (boy do we need MathML)
Of course, this probability is always less than one since there is some chance nobody will share birthdays, so how many people you need in the room before you make your bet depends on how much you are willing to risk. If you want a 75% chance of winning the bet, you need at least 32 people. For a 99% chance, you only need 55 people.

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