I live with ten people at a time. How many total housemates should I have gone through to expect that two of my housemates would share a birthday?

Question

I live in a house with ten people. I've had about 50 people live here since I moved in. Nobody has ever shared the same birthday before, but a new housemate shares a birthday with a current housemate. What were the odds of that happening? How many people would have to cycle through a house of ten people for us to expect that two of the housemates share a birthday at some point?

Answer 1

I don't see how this is different from the birthday paradox, see Wikipedia for instance. On a group pf 23 people, there are around 50% chances that two of them share a birthday.

With the cycling of people the problem seems easier, ie. when a new roommate arrives, you can easily compute the probability of him/her sharing a birthday. But then again, if you live with 10 people and expect 13 people to arrive in the next years, you have 50% chances of that happening. If the house accepts no more than 10 people at a time, then you should see the closed formula of the birthday paradox.

Answer 2

I don't think the problem is entirely clear, but perhaps it makes sense to model it this way:

We suppose that the house starts with $10$ people. Then, at intervals, one person at a time leaves and is replaced. How many such replacements do we expect to have before the probability that at one moment the house will contain a birthday match is at least $95\%$ (or whatever threshold you care to specify).

Of course we assume that each birthday from $1$ to $365$ is equally probable and that each birthday in your sample is independent of all the others.

To do it, let's compute the probability that we have seen no match after $N$ replacements.

The probability that there was no match in the starting group is $$\frac {365\times \cdots \times 356}{365^{10}}\approx 0.883051822$$

Given that, the probability that the first replacement does not create a match is $$\frac {365-9}{365}\approx 0.975342466$$

Thus the probability that we have seen no match after $N$ replacements is $$0.883051822\times (0.975342466)^N$$

We set this to $.05$, our confidence threshold, and take logs to solve. With this choice of threshold we get $\boxed {N\approx 115}$ So you need a lot of cycling.

Note: if you have had $50$ roommates in total, that means $N=40$. With that choice, the probability that there will have been a match at some point is about $.67$, so it isn't particularly surprising.