4 ms·

How to use math to choose a spouse

11 points by _pius 17y ago

schammy 17y ago
I don't understand what they mean by "the next best one" after having discarded X number of candidates. "Best" relative to what?
- leif 17y ago
  In the partially ordered set of partners (ordered by <=), given some ordering {p_k}, choose p_N from {p_(X+1), ..., p_100} s.t. p_i <= p_N for all i in {1, ..., X}.
- gojomo 17y ago
  I believe the assumptions are that there is a reliable fitness function giving an overall score, but only one candidate can be evaluated at a time, and once you reject a candidate, you can never go back. So "the next best one" means the first to reach a new high value after the initial run of X. (And note that the first X candidates have no chance -- no matter how high they score -- because you don't even start with any hint of the overall distribution.)
  - jerf 17y ago
    It should be pointed out that while the article sort of munged this point, it is a particular math problem in game theory that, like other game theory problems, has applications of interest well beyond what the problem is "about", whereas it's actually useless in the real world. (How often does Prisoner's Dilemma actually come up for you?) So, if it sounds like an unreasonable assumption that there is a universal fitness function, well, sure. It's a math problem, not a real description of real life. 100 potential partners isn't terribly realistic either. It's either much lower ("the set of people I have dated") or much higher ("the set of all people I have ever met that might be a suitable candidate", and remember, in the face of "love at first sight" that's a very large set for most people, well into the thousands). I find it sort of annoying when people propagate that error, but meh, whatever.
chrischen 17y ago
>I do not believe they were married. Lol!
chrischen 17y ago
So assuming half the population of the world are female, that would give me 3388940742. let's say 20% of them are single (no clue, arbitrary guess), that would give a pool of 677788148 potential partners. Now probably 50% of those are within my acceptable age range. So to get a 37% chance of finding the "best" one in the world, I'd have to date 125516323 women and then choose the next best one (which would take longer since I already have a pretty large pool of exes).
TheElder 17y ago
I met my wife in linear algebra.
dazmax 17y ago
It should be noted that this method only maximizes your chance of getting the best one, it doesn't take into account that the second best is much better than the worst. 37% of the time you use this method, you're just stuck with the last one you meet. If you want to get as good of a wife as you can, only pass up √n before choosing the next one that is better. http://en.wikipedia.org/wiki/Secretary_problem http://en.wikipedia.org/wiki/Secretary_problem

11 points by _pius 17y ago

I don't understand what they mean by "the next best one" after having discarded X number of candidates. "Best" relative to what?

leif 17y ago
In the partially ordered set of partners (ordered by <=), given some ordering {p_k}, choose p_N from {p_(X+1), ..., p_100} s.t. p_i <= p_N for all i in {1, ..., X}.
gojomo 17y ago
I believe the assumptions are that there is a reliable fitness function giving an overall score, but only one candidate can be evaluated at a time, and once you reject a candidate, you can never go back. So "the next best one" means the first to reach a new high value after the initial run of X. (And note that the first X candidates have no chance -- no matter how high they score -- because you don't even start with any hint of the overall distribution.)
- jerf 17y ago
  It should be pointed out that while the article sort of munged this point, it is a particular math problem in game theory that, like other game theory problems, has applications of interest well beyond what the problem is "about", whereas it's actually useless in the real world. (How often does Prisoner's Dilemma actually come up for you?) So, if it sounds like an unreasonable assumption that there is a universal fitness function, well, sure. It's a math problem, not a real description of real life. 100 potential partners isn't terribly realistic either. It's either much lower ("the set of people I have dated") or much higher ("the set of all people I have ever met that might be a suitable candidate", and remember, in the face of "love at first sight" that's a very large set for most people, well into the thousands). I find it sort of annoying when people propagate that error, but meh, whatever.

chrischen 17y ago

>I do not believe they were married. Lol!

chrischen 17y ago

So assuming half the population of the world are female, that would give me 3388940742. let's say 20% of them are single (no clue, arbitrary guess), that would give a pool of 677788148 potential partners. Now probably 50% of those are within my acceptable age range. So to get a 37% chance of finding the "best" one in the world, I'd have to date 125516323 women and then choose the next best one (which would take longer since I already have a pretty large pool of exes).

TheElder 17y ago

I met my wife in linear algebra.

dazmax 17y ago

It should be noted that this method only maximizes your chance of getting the best one, it doesn't take into account that the second best is much better than the worst. 37% of the time you use this method, you're just stuck with the last one you meet. If you want to get as good of a wife as you can, only pass up √n before choosing the next one that is better. http://en.wikipedia.org/wiki/Secretary_problem http://en.wikipedia.org/wiki/Secretary_problem