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.
2026-04-12 07:43:07.1775979787
Birthday Probability
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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.