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This might be a different variant of the typical birthday problem. Given a room of $n$ people, let $N$ be a random variable representing the number of people who have a birthday common with at least one other person. I will start by writing $$N = P_1 + P_2 + \cdots + P_n$$ where $P_i$ is an indicator random variable which is $1$ when person i's birthday is common with at least one other person in the room.

So we have by symmetry

\begin{align*} E[N] &= n E[P_1] \end{align*}

Now I note that person $1$'s birthday $D_1$ is a uniform random variable taking values in $\{1,2, ..., 365 \}$. Same for other people. Hence, by total expectation theorem

\begin{align*} E[N] &= n E[P_1]\\ &= n \sum_{k = 1}^{365} P(D_1 = k) E[P_1 | D_1 = k]\\ &= n \sum_{k = 1}^{365} \frac{1}{365} E[P_1 | D_1 = k]\\ &= n \sum_{k = 1}^{365} \frac{1}{365} \left(1 - \left(\frac{364}{365}\right)^{n-1} \right)\\ &= n\left(1 - \left(\frac{364}{365}\right)^{n-1} \right) \end{align*}

Is this right? If not, help me correct my argument. Thanks.

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    $\begingroup$ It is not clear if you are asking a question. If you are, then what is your question? If you are not, then what is it that you would like us to do with your post? $\endgroup$
    – Ben Grossmann
    Commented Nov 9, 2020 at 21:09
  • $\begingroup$ Yes just proof verification or correction. Thanks. $\endgroup$
    – rims
    Commented Nov 9, 2020 at 21:10
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    $\begingroup$ Your calculation of $E[P_1]$ is not correct. There is no reason for persons 2 and 3 NOT to share a birthday if we are only concerned with the event $\{P_1=0\}$. $\endgroup$
    – Michael
    Commented Nov 9, 2020 at 21:11
  • $\begingroup$ yes I see. How about now? $\endgroup$
    – rims
    Commented Nov 9, 2020 at 21:26
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    $\begingroup$ Yes, you have fixed the mistake. Now it is correct. $\endgroup$
    – Michael
    Commented Nov 9, 2020 at 21:27

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