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 $\frac{2\times 4+8 \times3 +6\times 2}{2+8+6}=\frac{11}{4}$