Birthday Probability

Solution 1:

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.