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?

  • $\begingroup$ Hint: given $10$ people the probability that all the birthdays are different is $\frac {365\times \cdots \times 356}{365^{10}}\approx .8831$ under the usual sort of assumptions. $\endgroup$ – lulu Mar 2 '18 at 1:52
  • $\begingroup$ Note: to get a precise answer I expect you'll need to clarify the problem. I'm assuming (possibly incorrectly) that you essentially want to look at groups of $10$ and are wondering how many such groups you'll need to look at before the probability that you'll see a group containing matching birthdays is at least $95\%$, or some number like that. But you should edit to say what, exactly, you had in mind. $\endgroup$ – lulu Mar 2 '18 at 1:55
  • $\begingroup$ Alternatively, if you imagine a system along the lines of "every X months one person leaves and a new person arrives" then, given that the prior group had no match, then the probability that the next group will have a match is $\frac 9{365}$. But, really, you should clarify what you had in mind. $\endgroup$ – lulu Mar 2 '18 at 1:59
  • $\begingroup$ @lulu I can impose that person leaves the house and is replaced by a new individual every month. Does that clarify the problem? $\endgroup$ – kilojoules Mar 2 '18 at 19:15
  • $\begingroup$ Not really...the problem doesn't really refer to the time involved, just the number of replacements. In my posted solution I made the simplifying assumption that only one person is replaced at a time. Definitely changes things if you allow more complex replacements. $\endgroup$ – lulu Mar 2 '18 at 19:17

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.

  • $\begingroup$ I think this is a good model. Alternatively, we might "expect" two people to share a birthday once the probability passes, say, 50% (since it's at least more likely to have happened, than not). I believe we get the same $n = 23$ as the birthday problem, in this case... $\endgroup$ – pjs36 Mar 2 '18 at 2:30
  • $\begingroup$ @pjs36 Yes, I agree. $\endgroup$ – lulu Mar 2 '18 at 2:31

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.

  • 1
    $\begingroup$ I agree with you that the problem is ambiguous. The way I read it, it differs from the standard Birthday Problem in that you are only concerned with the $10$ people living in the house at one time. That is, it doesn't matter if the new guy matches someone who moved out a while back. Of course, different interpretations of the problem are possible and I do not know which reading was intended. $\endgroup$ – lulu Mar 2 '18 at 2:16
  • $\begingroup$ The wrinkle is that people move out. I only live with ten people at any one time. $\endgroup$ – kilojoules Mar 2 '18 at 19:16

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