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I don't think this is the classic birthday problem, but if it is, I'd love it restated in a way that makes it sound more like the others.

Basically, I think it breaks down to the probability that two given sets of people (of size 5 in my case) have $n$ common birthdays between them. Members of my wife's and my nuclear families (when we were young) happen to share three birthdays.

When queried as "What is the probability that I would meet someone who's family and mine share three birthdays?" (that is in a personal "my" context, with predefined beginning birthdays to match), it would seem to reduce it to a simpler probability (1 in 365 x 365 x 365 or 1 in ~48M?), though I imagine it would depend on the size of the sets, and thus be a permutation of some sort(?).

I'm guessing you start with a set of 5 dates (for each of my family members), and then take a total population (perhaps of the US, where I live), but how would one proceed from there to find another family with $n$ common birthdays? You might bring statistics in to make sure you're only looking at other families with at least $n$ members, but I can't shake the feeling that since I don't see this asked anywhere, it's either incredibly simple or still reduces to the classic birthday problem. What is this probability?

TL;DR:

I'll try to state my query as clearly as possible:

What is the probability that I, as a person in a nuclear family of size $m$ will meet someone in a family (of any size $>n$) that has $n$ common birthday dates with my family?

We can leave the probability of falling in love and staying together with this person to a separate discussion.

Thanks for your consideration!

Note: this is what I thought I was submitting the first time, but somehow a previous cached version is what got submitted - apologies for the badly formatted and incomplete question

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  • $\begingroup$ If you want a really accurate answer you cannot assume all birth dates are equally likely; data suggests they are not. $\endgroup$
    – Gary.
    Commented Aug 7, 2015 at 15:42
  • $\begingroup$ I'm looking from my phone and it seems some of my question got cut off (?) - apologies for the resulting incomplete question. I'll try to fix shortly $\endgroup$
    – Code Jockey
    Commented Aug 7, 2015 at 17:15
  • $\begingroup$ @Gary - I think I generally understand that there are going to be many ways to refine it to avoid pitfalls in accuracy; I'm looking for a rough order of magnitude and basic methodology. Assuming a frictionless vacuum, there's no-one to blow out the candles, or for that matter air for the candles to burn, but I'd be just fine with something even a full order of magnitude off. $\endgroup$
    – Code Jockey
    Commented Aug 7, 2015 at 17:58

1 Answer 1

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You need to state the question clearly to have a clear answer. If you have two families of five, with each family having distinct birthdays, it makes sense to ask the probability that there are exactly three matches. You might as well fix the birthdays of the first family.There are ${365 \choose 5}$ sets of birthdays for the second family to have. We have to assume all $365$ birthdays are equally likely to get a reasonably simple answer. There are ${5 \choose 3}{360 \choose 2}$ ways to select the birthdays to get three matches, so the chance is $$\frac{{5 \choose 3}{360 \choose 2}}{{365 \choose 5}}=\frac {646200}{52521291823}\approx 0.00001230358$$ per Wolfram Alpha

Added: for the restated problem, assume your family's $m$ birthdays are distinct and that the other family has $k$ distinct birthdays. The chance of exactly $n$ matches is $$\frac {{k \choose n}{365-k \choose m-n}}{365 \choose m}=\frac{k!(365-k)!m!(365-m)!}{n!(k-n)!(m-n)!(365+n-m-k)!365!}$$ where the symmetry between $k$ and $m$ is evident.

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  • $\begingroup$ I think some of my question got lost at some point - I'll try to fix shortly. Thanks for your answer so far! $\endgroup$
    – Code Jockey
    Commented Aug 7, 2015 at 17:16
  • $\begingroup$ fixed - not a huge difference from what actually got submitted, but a little clearer, I think - not sure what went wrong (?) $\endgroup$
    – Code Jockey
    Commented Aug 7, 2015 at 18:02

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