# Birthday problem: expected birthday collision “size”?

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?

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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
A     BCD
BCD   A

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