MHB Variation of the shared birthday problem

[Asifbymagic]

I work as part of a team of fourteen. No big challenge to work out the probability of two of us sharing a birthday. It's a well-documented puzzle.

In my team, we've gone one better: we have two dates where two people have birthdays on that day!

Trying to work out the probability has us stumped. Not of 4/14 not having unique birthdays, but of two pairs of shared birthdays.

Can someone work me through a solution?

Thanks! Alex

 

[Ackbach]

I am not an expert in probability, but I've studied it some. Here's my approach.

Let's try counting, and let's start with the $10$ folks with unique birthdays. There'd be $365$ ways to choose the first birthday, $364$ for the second, and so on (ignoring leap year). That is, you'd have $\dfrac{365!}{355!}$ ways to choose these birthdays. Now let's choose the first pair. There are $355$ ways to choose the first pair's birthday, and thus $354$ ways to choose the second pair's birthday. But this is simply reasoning the same way as before, so you have $\dfrac{365!}{353!}$ ways to choose everyone's birthday. However, this number, as is, is too restrictive, because the way I've counted so far assumes which people would be in the pairs. We need to count how many ways we can get the pairs. An easier way of doing this is to count how many ways we can get the ten non-pairs. That's simply $\displaystyle\binom{14}{10}$. There are $365^{14}$ total ways to assign the birthdays randomly. So, our probability would be
$$\frac{\displaystyle\left(\frac{365!}{353!}\right)\binom{14}{10}}{365^{14}}\approx 6.26\times 10^{-3}.$$