There are many descriptions of the "birthday problem" on this site — the problem of finding the probability that in a group of $n$ people there will be any (= at least 2) sharing a birthday.
I am wondering how to find instead the expected number of people sharing a birthday in a group of $n$ people. I remember that expectation means the weighted sum of the probabilities of each outcome:
$$E[X]=\sum_{i=0}^{n-1}x_ip_i$$
And here $x$ must mean the number of collisions involving $i+1$ people, which is $n\choose i$. All $n$ people born on different days means no collisions, $i=0$; two people born on the same day means $n$ collisions, $i=1$; all $n$ people born on the same day means $n$ collisions, $i=n-1$.
Since the probabilities of three or more people with the same birthday are vanishingly small compared to two people with the same birthday, and decreases faster than $x$ increases, is it correct to say that this expectation can be approximated by
$$E[X]\approx {n\choose 0}p_{no\ collisions}+{n\choose 1}p_{one\ collision}$$
This doesn't look right to me and I'd appreciate some guidance.
Sorry – edited to change ${n\choose 1}$ to ${n\choose 0}$ in second equation. Sloppy of me.
Best Answer
The probability person $B$ shares person $A$'s birthday is $1/N$, where $N$ is the number of equally possible birthdays,
so the probability $B$ does not share person $A$'s birthday is $1-1/N$,
so the probability $n-1$ other people do not share $A$'s birthday is $(1-1/N)^{n-1}$,
so the expected number of people who do not have others sharing their birthday is $n(1-1/N)^{n-1}$,
so the expected number of people who share birthdays with somebody is $n\left(1-(1-1/N)^{n-1}\right)$.