This isn't for homework, just a thought.
Let's say there are $n$ people where $n \leq 365$ (I'm not entirely sure how to approach the problem if $n > 365$ and inquire about that down below). What is the probability that there are exactly two pairs with the same birthday, and each of the pairs have a different birthday (e.g. Alice and Bob share a birthday on June 8th, Charlotte and Dylan share a birthday on December 1st.), assuming that birthdays are evenly distributed and we pick a completely random sample of $n$ people.
My attempt to solve is the following:
The sample space is represented by the $365^n$ different combinations of birthdays for the $n$ people.
The number of ways that exactly two pairs share a birthday, and the birthday each pair shares is different, is given by $n \choose 2$$(365)$$n - 2 \choose 2$$(364)(\frac{363!}{(363-(n-4))!})$
The explanation is that we first pick two individuals to have the same birthday, then another two individuals to have a different same birthday from the remaining 364 days, and then finally have rest of the $n-4$ individuals all have different birthdays.
The solution is thus given by the numerator divided by the sample space (having trouble using latex format with the chooses, I apologize).
A couple of questions:
- Is the above solution correct for $n < 365$?
- If I try to think about if $n > 365$ the combinatorics I use in the last part $\frac{363!}{(363-(n-4))!}$ doesn't make sense anymore, so I was thinking about how I could use a different approach. Someone hinted that I could try to calculate the conditional probability that there is exactly one pair that shares a birthday given that I remove a pair that shares a birthday. I'm envisioning the distribution as a bell curve of some sort, but am getting overwhelmed in the details of how to account for all the cases.
I would appreciate any hints or pointers!