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In my daughter's class of $23$, three students and the teacher all share the same birthday. Of course, there are $365$ days in the year, and the first case of the shared birthday is not counted in the probability. But is the likelihood of this just $\frac{1}{365}\times 3 = 0.0082?$ For my curiosity, leap years can be omitted.

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    The proper resop$n$se to any post-hoc question like this is "the likelihood of what?" In this case many interpretations are possible: Pr(exactly four people share some birthday), Pr(three students share the teacher's birthday), Pr(at least four people share some birthday), and Pr(I notice anything at all "interesting" about shared birthdays) are all good candidates. The last is the most reasonable interpretation but it has no unique answer. Add to this the fact that this class is not a random sample of people and all bets are off concerning what a valid answer might be.2010-09-20

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No, $1/365^3$ is (roughly) the probability that, given a date and $3$ people, those $3$ people have a birthday on that date. This is much smaller than the probability that, given a date and $23$ people, $3$ of those people have their birthday on that date. It seems now that you want, given $24$ people (and no date fixed beforehand), the probability that $4$ of those people have their birthday on the same date. A generalization of this question can be found at https://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people.