THE ALGEBRA OF TRUE LOVE

A Settlement Strategy For Marriage

By now, lawyers are quite familiar with attempts to apply mathematical formulas to certain settlement decisions -- the ones that concern when, and for how much, to settle civil cases. But what about the type of settlement decision that really obsesses twenty- and thirty-something lawyers -- the decision when to "settle" on a particular partner, and get married? Could mathematics apply there too?

You might think not. You might think that the decision when to settle on a particular partner, and pop the question, is the type of highly subjective judgment call that is tailor-made for hours of angsting on Ally McBeal-- not the type of cold, rational calculation that is tailor-made for a quick mathematical fix.

There is, indeed, a mathematical solution as to when to settle on a marriage partner. For the marriage-minded, I'll present that solution first. Then, for the math-minded, I'll detail the reasons this solution is objectively correct.

The Mathematical Solution To When To Marry

I'll start by assuming, for simplicity, that you start dating at age 18 and want to marry by 37 (perhaps for biological clock reasons, if you're a woman, or, if you're a guy, because the club kids will be mocking your bald spot by then). That's a 19 year period. Applying the formula set forth in the math section below, you should date for about 7 years -- from the ages of 18 to 25 -- without marrying.

Don't even consider it! Enjoy your single life. Sow your wild oats. Feel no guilt, during this period. Others may call you a cad, a Casanova, part of the Peter Pan Syndrome -- or a flirt, a party girl, a heartbreaker, a femme fatale. Or, if you get really extreme, a sexual predator. But don't give in to the peer pressure, the name-calling, the suicide attempts, the tears. Stay strong. And remain secure in the knowledge that your behavior, callous and hedonistic as it may appear, is actually ordained by immortal mathematical theorems.

If you follow this advice, you should still be single when you hit age 25. But you should also think to yourself, at that point: Who is the best prospect I've met so far? Call that person the Best Ex. Keep a picture of the Best Ex next to your bed. He or she is going to set your standard.

On the day of your 25^th birthday, you should suddenly be in the market for a serious relationship leading to marriage. Yes, this should be an overnight transformation. And, yes, it is just as rational as it is abrupt -- and potentially shocking to your girl/boyfriend, who may have written you off as marriage material based on your past behavior. It's a new kind of Seven Year Itch -- not an irrational itch to cheat, but an entirely rational itch to marry.

Listen up, Mr. or Ms. Twenty-Five: The party's over. For commitment-phobic though you might be, it's time to swallow your fear and get ready to shop for rings when 25 rolls around. Because now fear of marriage just isn't rational anymore -- assuming you do want to marry by 37, that is. For the more commitment-phobic you are, after age 25, the more you hurt your prospects of marrying the best catch you can get and ending up with someone who, by your own lights, is second-best instead.

Of course, even after the magical 25^th birthday, you shouldn't marry just anyone. Instead, marry the first person you meet who blows the Best Ex out of the water. Marry that person. Get down on one knee and do it. And whatever you do, don't look back. You made the optimal choice -- by definition (and the math section below explains why that is).

Look at the case of Oscar winner Gwyneth Paltrow. A few years ago, in her early 20s, she was engaged to the Dashing Brad Pitt (probably her Best Ex). Realizing she was not yet 25 (and obviously displaying a strong command of mathematics), she called it off. Now 26, however, she should be searching for permanent love. Will her off-and-on flame Ben Affleck fill the bill? Or should it be relationship Armageddon? It will be easy for Gwynnie to decide: she need only compare Affleck to Pitt, her best pre-25 catch. If she finds Ben to be Brad's superior, time to head for the chapel. If not, it's back to the bars (or, in her case, the cool Hollywood parties).

The Math Behind The Solution

Part I - The Optimal Stopping Problem

Kudos for this approach are due to the great British mathematician Arthur Cayley -- who first considered this type of settling problem in the late 1800s. Although Cayley never knew Ally McBeal, nor could he watch her show, he still felt her pain. Indeed, Cayley felt the pain of any person who, like Ally, is confronted with a series of options over time (for Ally, the options would be Bachelors #1, #2, #3 and so on until marriage or cancellation). Anyone, that is, who must forgo one option forever in order to move on to the next; anyone who cuts and runs -- holding close the hope that the next relationship will be better but staving off the fear that the last was the best there is.

In short, although he lived in the 1800's, Cayley anticipated the dilemma of the serial monogamist. Better than that, he took steps toward solving it -- thereby furthering the chances that the love thang that turns into a marriage thang will be the very best it can be. And who says mathematicians aren't romantics?

To mathematicians, a dilemma of this sort is known as an Optimal Stopping Problem. It's easiest to illustrate, at least initially, by imagining you're offered a series of bags of cash that vary in amount -- rather than, say, a series of persons of varying degrees of fantastic-ness.

Once you open a bag o' cash (labeled, in the best cartoon fashion, with a large dollar sign), you can choose to either (1) accept it, or (2) reject it and go for a new bag. An important rule: You can never go back to bags that you've already opened and rejected. Another important rule: At some point, the bags o' cash are going to run out, and if you go through all of them and reject them all, you get the last one by default.

You want to maximize your chance of taking the most valuable bag. The question is, when should you stick, and accept a particular bag of money -- rather than rejecting that bag, and moving on to the next one?

Getting a rough, intuitive answer to this question is easy if you have some information about the kinds of amounts you are likely to be offered (what mathematicians call the "distribution function"). For example, if you knew you were going to see 30 bags, and you knew the amount in each was a random number of dollars that is equally likely to be anything from 1 to 1000, then you'd pretty much wait until you saw something up over 900 and jump on it. If your luck was bad and after 26 bags you'd seen no $900+ bags, you'd grab an $800 or maybe $750 if you saw one, just to play it safe. Nothing to it.

But let's add a twist -- one which gets us closer to the marriage scenarios I discussed above.

The twist -- which makes the optimal stopping problem more difficult -- is that you have no idea what kinds of amounts of money to expect. (This wrinkle was added by Merrill Flood in 1950.) Of course, this new feature -- that you don't know in advance the quality of opportunities you can expect to have -- is also characteristic of the search for that one perfect love.

Now you need a new strategy: (1) Watch the money bags for a while, to get a feel for what constitutes a nice chunk of change, and (2) then pick one that looks good, relative to this newly developed definition of what, indeed, a "nice chunk of change" is. But how long should you watch and wait, developing your sense of a "nice chunk of change" before choosing?

In 1954, the statistician Herb Robbins showed, amazingly, that there was a precise, mathematical answer to this very question. Robbins' answer is this: (1) If there are N bags, then you begin by rejecting the first N/e bags, where e is the calculus constant 2.71828.... Since 1/e is about .37, this means you watch 37% of the bags go by before even considering choosing one. But (2) you keep a record of the highest cash valued-bag as the first 37% of bags go by -- call it the Best Cash. Then (3) you stick with the next bag you open with cash of a value greater than the Best Cash you've seen so far.

Perhaps you are underwhelmed by this explanation. Perhaps you are having flashbacks of not understanding the numerical solution to the meaning of life provided in The Hitchhiker's Guide to the Galaxy. Perhaps you were a liberal arts major because calculus made you as sick as cotton candy on a rollercoaster. Take heart. Here's an easy application:

Suppose there are 100 bags. Then N=100, and N/e is about 37. That means you should reject the first 37 bags, but be careful to write down the value of the bag, among the first 37, that contained the most cash. Say the value was $10,000. Then, starting with bag 38, keep opening the rest of the bags, and stick with the first bag of cash with a value in excess of $10,000. If none of the bags has that much cash in it, just take the last one -- you did the rational thing, but you still got screwed.

So now you get it, even if you read Keats and Yeats in the back of Calc One -- rather than, say, acing Calc One but (like many math majors) believing Keats and Yeats were pronounced the same way, yet still not knowing what way that was. But you're still saying, What does this mean for me, Al Franken? Or, more likely, for me, Ally McBeal?

That brings me to the marriage application we started with. If you need to marry between 18 and 37, then N is 19 (taking each year in which you might meet a potential mate, rather than each bag of money you're offered, as the relevant unit). 19/e is approximately 19/2.7 -- which is approximately 7. So, applying Robbins' formula, you wait N/e years -- or 7 years -- that is, until you're 25 -- to even consider marrying. And the 7-year period between 18 and 25 is also the period during which you settle on your Best Ex -- the man or woman to beat when you choose the person you'll marry after the age of 25.

Once you do marry, I assure you that you'll realize that, like Pangloss, you're living in the best of all possible worlds. You'll thank Cayley and Robbins for solving all your problems. And for doing so mathematically.

Feel free to live happily ever after.