“We have to split up… in case somebody better comes along!”
Either from our own real life or from popular culture and the media, we’ve all come across a statement like that. It’s rarely quite so brazen: instead, it’s sometimes concealed behind another reason, whether tactful or simply false. But it still reeks of a lack of commitment and an unwillingness to “give it a try.”
However, it turns out that there’s actually a solid mathematical basis for it. Let’s assume for a moment that you:
- Engage exclusively in monogamous relationships. To each their own, I suppose.
- Are seeking for a relationship that will last indefinitely (e.g. traditional monogamous marriage, “’til death do us part,” and all that jazz).
- Can’t or won’t date your exes.
- Can rate all of your relationships relative to one another (i.e. rank them all, from best to worst)?
- Can reasonably estimate the number of partners that you will have the opportunity to assess over the course of your life. You can work this out by speculating on how long you’ll live (and be dating!) for, and multiplying, though of course there are several factors that will introduce error. When making this assumption, you should assume that you break up from any monogamous relationship that you’re currently in, and that no future monogamous relationship is allowed to last long enough that it may prevent you from exploring the next one, until you find “the one” – the lucky winner you’re hoping to spend the rest of your life with.
Assuming that all of the above is true, what strategy should you employ in order to maximise your chance of getting yourself the best possible lover (for you)?
It turns out that clever (and probably single) mathematicians have already solved this puzzle for you. They call it the Secretary Problem, because they’d rather think about it as being a human resources exercise, rather than a reminder of their own tragic loneliness.
A Mathematical Strategy for Monogamy
Here’s what you do:
- Take the number of people you expect to be able to date over the course of your lifetime, assuming that you never “settle down” and stop dating others. For example’s sake, let’s pick 20.
- Divide that number by e – about 2.71828. You won’t get a round number, so round down. In our example, we get 7.
- Date that many people – maybe you already have. Leave them all. This is important: these first few (7, in our example) aren’t “keepers”: the only reason you date them is to give you a basis for comparison against which you rate all of your future lovers.
- Keep dating: only stop when you find somebody who is better than everybody you’ve dated so far.
And there you have it! Mathematically-speaking, this strategy gives you a 37% chance of ending up with the person who – of all the people you’d have had the chance to date – is the best. 37% doesn’t sound like much, but from a mathematical standpoint, it’s the best you can do with monogamy unless you permit yourself to date exes, or to cheat.
Or to conveniently see your current partner as being better than you would have objectively rated them otherwise. That’s what love will do for you, but that’s harder to model mathematically.
Of course, if everybody used this technique (or even if enough people used it that you might be reasonably expected to date somebody who did, at some point in your life), then the problem drifts into the domain of game theory. And by that point, you’d do better to set up a dating agency, collect everybody’s details, and use a Stable Marriage problem solution to pair everybody up.
This has been a lesson in why mathematicians shouldn’t date.
2 replies to Monogamy and Mathematics
i easily understand the explanation and fomulas how get it really great. amazing… ok that’s all.
Dan Q mentioned this article on danq.me.