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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.

Henry
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michaelkoss
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  • Why would you multiply by 3? If you replaced 3 by a different number the answer should be smaller, not larger. Also, do you want the probability that 4 people out of 24 share the same birthday or that 3 people share a birthday with the teacher? – Qiaochu Yuan Sep 19 '10 at 18:39
  • I would like the probability of four people sharing a birthday. – michaelkoss Sep 19 '10 at 18:43
  • I multiplied by three because there are three people who share a birthday with another person. – michaelkoss Sep 19 '10 at 18:43
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    @michaelkoss: but if there were fifty people who share a birthday would you multiply by 50? – Qiaochu Yuan Sep 19 '10 at 18:45
  • @Qiaochu: I see your point. It should be exponential, not multiplication. (1/365)^3 would be greater than (1/365)^50. Thanks for pointing out that error. – michaelkoss Sep 19 '10 at 19:01
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    @michaelkoss, that still isn't the correct number. That would be the probability that 4 people out of 4 people have the same birthday, but for 4 out of 24 it is much higher, and more complicated to compute. – Jonas Meyer Sep 19 '10 at 19:05
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    The proper resopnse 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. – whuber Sep 20 '10 at 22:12

1 Answers1

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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.

Community
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Jonas Meyer
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