TITLE: eHarmony combinatoric question, probability that I should get at least 1 compatible match.
QUESTION [6 upvotes]: Ok.. (as I type this with a smirk on my face) - in all seriousness I am trying to figure out, given 29 degrees of compatibility and 40 million members if I should be getting at least 1 match a day.  There are of course a lot of variables, so I am trying to simplify things.  I also see this as similar to the "birthday problem" in which one wants to see, out of n people, the probability of not having any birthday's in common.
In this case however, we don't care if others get a match, only that I do (muahahaha!), so I am guessing this is, instead of being in line with $_{n}C_{k}$ this is $_{n}C_{1}$ Where I am the $1$!
This is a bit more advanced in that we are looking at the "percentage of compatibility".  So I am going start by taking the 29 degrees of freedom and looking at things in a binary way i.e. either a match is compatible on a degree of freedom (same answer to a question) or they are not. Now by my calculations there are 29 questions, so as with light switches, there are $2^{29} = 536870912$ ways to answer the questions.  So if half the eH members are women, then there is a 20,000,000/536,870,912 = 3.7% chance that I would answer exactly as a potential match.
It seems there should or could be much more to it in that say I went about the computation from the point of view that the probability of not answering the questions the same.  So I think I would have something such as $$(1-2^{28}/2^{29})(1-2^{27}/2^{29})(1-2^{26}/2^{29})\space...\space(1-2^{0}/2^{29})$$
which seems to reduce to: $$((2^{28})(3 \cdot 2^{27})(7 \cdot 2^{26}) \space ... \space(2^{2}(2^{27}-1))(2^{1}(2^{28}-1)) \cdot (2^{29}-1))/(2^{29})^{29}$$  after multiplying the numerator by  $(2^{29})^{29}$.
I am not sure what this reduces too - hopefully more matches than I am currently getting..
I wonder though if I am on the right track?  I however wonder about the 20,000,000.  If I were to take the possible "state spaces" or the $2^{29}$ possible choices for the 29 degrees of freedom (treated as binary yes/no), as it were the possible 365 days in a year birthday state space, then I come up with:
$$\frac{2^{29}!}{(2^{29})^{20,000,000}\cdot (2^{29} - 20,000,000)!}$$
which just seems insane.
Any thoughts (or dating advice ha!)
Thanks,
Brian

REPLY [2 votes]: The probability of at least 1 match is $$1- \mathsf{\text{probability of no matches}} = 1-(1-p)^n$$ where $p = \frac{1}{2^{29}}$ and $n=20000000$. Assuming the choices are equaly likely, and the candidates choose independently. Approximately 3.66%.