6
$\begingroup$

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

$\endgroup$

closed as too localized by Potato, Amzoti, Lord_Farin, Zev Chonoles, Dan Rust Jun 21 '13 at 23:20

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ My dating advice: don't use internet sites. Just go out there and talk to women. Gain experience by analyzing your rejections and adapting your tactics accordingly. You'll get it right eventually. $\endgroup$ – Raskolnikov May 29 '13 at 18:51
  • $\begingroup$ Does it only count as a match if all 29 degrees match? $\endgroup$ – Austin Mohr May 29 '13 at 18:56
  • $\begingroup$ @Austin, I don't know, I am thinking not exactly as the numbers (although I don't know if I am going about it right), seem to say that there are just to many possibilities, that to have the $2^{29}$ possible combinations of answers exactly matching, I get the 3.7% (and as also noted below by ronno), so there has to be some leeway. $\endgroup$ – Relative0 May 30 '13 at 3:29
  • $\begingroup$ @Raskolnikov. Seems that the complexity of the dating analysis is extra-ordinarily difficult. I am thinking that trying to design an adaptable algorithm for love is uncomputable. My algorithm either doesn't halt and return an answer or the answer is inconsistent over repetitive trials. I suppose the problem lies in that the state space is a subspace of an infinite dimensional vector space. Seems that the difficulty of designing a search algorithm to which minimizes time and money could be more efficient. This is what I am trying to figure out.. $\endgroup$ – Relative0 May 30 '13 at 4:57
  • 1
    $\begingroup$ We don't know how the matching algorithm works, so this question is unanswerable. Also, your computations make too many simplifying assumptions to be useful. For example, people probably do not answer the questions according to a uniform random distribution. $\endgroup$ – Potato Jun 21 '13 at 21:37
2
$\begingroup$

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%.

$\endgroup$
  • $\begingroup$ seems extremely close to my first guess - is this just a coincidence? I don't see offhand how to reduce your solution to mine. I however took the derivative of both (respect to n). For yours, $\frac{d(1-(1-p)^{n})}{dn} = (1-p)^{n}ln(1-p)$, and for mine $\frac{d(n/p)}{dn} = 1/p$ where $p = 1/(2^{29})$. Now I can't seem to find software that will do that computation as a whole, but in parts $1-(1/2^29)^{20000000} = 1$ (of course not exactly 1 but..) and $ln(1-p) = -(1/p)$ the right side is the negative of the derivative of my equation. So are yours and mine equal? $\endgroup$ – Relative0 May 30 '13 at 4:39
  • 1
    $\begingroup$ For small $p$, $(1-p)^n\approx 1-np$. $\endgroup$ – Raskolnikov May 30 '13 at 5:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.