Here’s another formulation of a birthday problem: given $n$ people, and $m$ days, how to calculate the expected number of people having a birthday on any single collision day, i.e., a day where two or more people have birthdays?
UPD: to clarify: given a date, which is a birthday of at least two people (= collision date), what is the expected number of people who share this birthdate?
Using results in the related question an earlier question and answer
The expected number of people who share a birthday with somebody else is $n-n\left(1-\frac1m\right)^{n-1}$
The expected number of days where two or more people have birthdays is $m – m \left(1-\frac1m\right)^n – n \left(1-\frac1m\right)^{n-1}$
and so dividing the first by the second may look like
$$\frac{n-n\left(1-\frac1m\right)^{n-1} }{ m – m \left(1-\frac1m\right)^n – n \left(1-\frac1m\right)^{n-1} }$$
though note that this approach gives a greater weight to cases of distributions of people among more birthdays
As an example, with $m=2$ and $n=4$ this gives $\frac{28}{11}$. If you consider the sixteen equally probable distributions of birthdays for four people among two days
Day1 Day2
ABCD -
ABC D
ABD C
AB CD
ACD B
AC BD
AD BC
A BCD
BCD A
BC AD
BD AC
B ACD
CD AB
C ABD
D ABC
- ABCD
there are $2$ cases of four people sharing a birthday, $8$ cases of three people sharing a birthday, $12$ cases of two people sharing (as well as $8$ of one and $2$ of zero) making the average number of people per shared birthday $\frac{2\times 4+8 \times3 +12\times 2}{2+8+12}=\frac{28}{11}$ as predicted.
A different approach could say the average should be calculated as the average of $2$ cases with the average being $4$, $8$ cases with the average being $3$ and $6$ cases with the average being $2$, giving a result of $\frac{2\times 4+8 \times3 +6\times 2}{2+8+6}=\frac{11}{4}$