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What is the probability that on a specific day (e.g. December 24) two people have birthday in a room with 23 people?

The probability, given a specific day, one person has birthday is in my opinion $\frac{23}{365}$. So the probability for two birthdays is $(\frac{23}{365})^2$.

I believe that my calculation is not correct. If it is wrong, what is calculated by $(\frac{23}{365})^2$ in reality?

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  • $\begingroup$ Welcome to math stack exchange. You need the binomial distributed random variable $X$ with parameters $n=23$ and $p=\frac{1}{365}$ (if we omit the $29$ th February). Then, $P(X=2)$ is the probability for exactly two people with the given birthday. Note that this is not the birthday paradox. $\endgroup$
    – Peter
    Commented Jan 2, 2021 at 12:02
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    $\begingroup$ Exactly two people or at least two people? $\endgroup$
    – Math Lover
    Commented Jan 2, 2021 at 12:05
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    $\begingroup$ Why do you think $\frac{23}{365}$? And what if there are not $23$ but e.g. $400$ persons? A probability will not exceed $1$. $\endgroup$
    – drhab
    Commented Jan 2, 2021 at 12:10
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    $\begingroup$ $(\frac{23}{365})^2$ represents two people both having a birthday on one of $23$ specified days $\endgroup$
    – true blue anil
    Commented Jan 2, 2021 at 13:02
  • $\begingroup$ @Peter I got $0.0018$ that can be considered rare. Indeed in my 40 years teaching in high schools where the number of students per class was around 23, I remember just two cases over about 150-200 classes. Both cases were twins :) $\endgroup$
    – Raffaele
    Commented Jan 2, 2021 at 13:28

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If you mean "exactly two people" the probability is $$\binom {23}2\left (\frac {1}{365}\right)^{2}\left (\frac {364}{365}\right)^{21}.$$

If you mean "at least two people" the probability is $$1-\left (\frac {364}{365}\right)^{23}-\binom {23}1\left (\frac {1}{365}\right)^{1}\left (\frac {364}{365}\right)^{22}.$$

Above no correction for leap years and deviations from the uniform distribution are taken into account.

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