The Secretary Problem – in a nutshell.

Further and deeper exploration of paradoxes and challenges of intuition and logic can be found in my recently published book, Probability, Choice and Reason.

In ‘The Merchant of Venice’, by William Shakespeare, as we have seen in an earlier chapter, Portia sets her suitors a problem to solve to find who is right for her. In the play, there are just three suitors and they are asked to choose between a gold, a silver and a lead casket, one of which contains a portrait which is the key to her hand in marriage.

Let us base a thought experiment around Portia’s quest for love in which she meets the successive suitors in turn. Her problem is when to stop looking and start choosing.

To make the problem of more general interest, let’s say she has 100 suitors to choose from. Each will be presented to her in random order and she has twenty minutes to decide whether he is the one for her. If she turns someone down there is no going back, but the good news is that she is guaranteed not to be turned down by anyone she selects. If she comes to the end of the line and has still not chosen a partner, she will have to take whomever is left, even if he is the worst of the hundred. All she has to go on in guiding her decision are the relative merits of the pool of suitors.

Let’s say that the first presented to her, whom we shall call No.1, is perfectly charming but she has some doubts. Should she choose him anyway, in case those to follow will be worse? With 99 potential matches left, it seems more than possible that there will be at least one who is a better match than No.1.

The problem facing Portia is that she knows that if she dismisses No. 1, he will be gone forever, to be betrothed to someone else.

She decides to move on. The second suitor turns out to be far worse than the first, as does the third and fourth. She starts to think that she may have made a mistake in not accepting the first. Still, there are potentially 96 more to see. This goes on until she sees No. 20, whom she actually prefers to No. 1. Should she now grasp her opportunity before it is too late? Or should she wait for someone even better?

She is looking for the best of the hundred, and this is the best so far. But there are still 80 suitors left, one of whom might be better than No. 20. Should she take a chance?

What is Portia’s optimal strategy in finding Mr. Right?

This is an example of an ‘Optimal Stopping Problem’, which has come to be known as the ‘Secretary Problem.’  In this variation, you are interviewing for a secretary, with you aim being to maximise your chance of hiring the single best applicant out of the pool of applicants. Your only criterion to measure suitability is their relative merits, i.e. who is better than whom. As with Portia’s Problem, you can offer the post to any of the applicants at any time before seeing any more candidates, but you lose the opportunity to hire that applicant if you decide to move on to the next in line.

This sort of stopping strategy can be extended to anything including the search for a place to live to a place, a place to eat, the choice of a used car, and so on.

In each of these cases, there are two ways you can fail to meet your goal of finding the best option out there. The first is by stopping too early, and the second is by stopping too late.

By stopping too early, you leave the best option out there. By stopping too late, you have waited for a better option that turns out not to exist. So how do you find the right balance?

Let’s consider the intuition. Obviously, the first option is the best yet, and the second option (assuming we are taking the options in a random order) has a 50% chance of being the best yet. Likewise, the tenth option has a 10% chance of being the best to that point.

It follows logically that the chance of any given option being the best to that point declines as the number of options there have been before increases. So the chance of coming across the ‘best yet’ becomes more and more infrequent as we go through the process.

To see how we might best approach the problem, let’s go back to Portia and her suitors and look at her best strategy when faced with different-sized pools of suitors. Can she do better using some strategy other than choosing at some random position in the order of presentation to her?

It can be shown mathematically that she can certainly expect to do better, given that there are more than two to choose from. Let’s return to the original play where she has three potential matches.

We can look at it this way. If she chooses No. 1, she has no information with which to compare the relative merits of her suitors. On the other hand, by the time she reaches No. 3, she must choose him, even if he’s the worst of the three. In this way, she has maximum information but no choice.

In the case of No. 2, she has more information than she did when she saw No. 1, as she can compare the two. She also has more control over her choice than she will if she leaves it until she meets No. 3.

So she turns down No. 1 to give herself more information about the relative merits of those available. But what if she finds that No. 2 is worse than No. 1? What should she do?

It can in fact be shown that she should wait and take the risk of ending up with No. 3, as she must do if she leaves it to the last. On the other hand, if she finds that she prefers No. 2 to No. 1, she should chose him on the spot and forego the chance that No. 3 will be a better match.

It can also be shown that in the three-suitor scenario, she will succeed in finding her best available match exactly half the time by selecting No. 2 if he is better than No. 1. If she chooses No. 1 or No. 3, on the other hand, she will only have met that aim one time in three.

If there are four suitors, Portia should use No. 1 to gain information on what she should be measuring her standards against, and select No. 2 if he is a better choice than No. 1. If he is not, do the same with No. 3. If he is still not better than No. 1, go to No. 4 and hope for the best. The same strategy can be applied to any number of people in the pool.

So, in the case of a hundred suitors, how many should she see to gain information before deciding to choose someone?

It can, in fact, be demonstrated mathematically that her optimal strategy (‘stopping strategy’) before turning looking into leaping is 37.

She should meet with 37 of the suitors, then choose the first of those to come after who is better than the best of the first 37. By following this rule, she will find the best of the princely bunch of a hundred with a probability, strangely enough, of 37 per cent.

By choosing randomly, on the other hand, she has a chance of 1 in 100 (1%) of settling upon the best.

This stopping rule of 37% applies to any similar decision, such as the secretary problem or looking for a house in a fast-moving market. It doesn’t matter how many options are on the table. You should always use the first 37% as your baseline, and then select the first of those coming after that is better than any of the first 37 per cent.

The mathematical proof is based on the mathematical constant, e (sometimes known as Euler’s number) and specifically 1/e, which can be shown to be the stopping point along a range from 0 to 1, after which it is optimal to choose the first option that is better than any of those before. The value of e is approximately equal to 2.71828, so 1/e is about 0.36788 or 36.788%. This has simply been rounded up to 37 per cent in explaining the stopping rule. It can also be shown that the chance that implementing this stopping rule will yield the very best outcome is also equal to 1/e, i.e. about 37 per cent.

If there is a chance that your selection might actually become unavailable, the rule can be adapted to give a different stopping rule, but the principle remains. For example, if there is a 50% chance that your selection might turn out to be unavailable, than the 37% rule is converted into a 25% rule. The rest of the strategy remains the same. By doing this, you will have a 25% chance of finding the best of the options, however, compared to a 37% chance if you always get to make the final choice. This is still a lot better than the 1 per cent chance of selecting the best out of a hundred options if you choose randomly. The lower percentage here (25% compared to 37%) reflects the additional variable (your choice might not be final) which adds uncertainty into the mix. There are other variations on the same theme, where it is possible to go back with a given probability that the option you initially passed over is no longer available. Take the case, for example, where an immediate proposal will certainly be accepted but a belated proposal is accepted half of the time. The cut-off proportion in one such scenario rises to 61% as the possibility of going back becomes real.

There is also a rule-of-thumb which can be derived when the aim is to maximise the chance of selecting a good option, if not the very best. This strategy has the advantage of reducing the chance of ending up with one of the worst options. It is the square root rule, which simply replaces the 37% criterion with the square root of the number of options available. In the case of Portia’s choice, she would meet the first ten of the hundred (instead of 37) and choose the first of the remaining 90 who is better than the best of those ten.

Whatever variation you adopt, the numbers will change but the principle stays the same.

All this assumes that we are lacking in some objective standard about which we can measure each of our options objectively, without needing to compare which option is better than which. For example, Portia might simply be interested in choosing the richest of the suitors and she knows the distribution of wealth of all potential suitors. This ranges evenly from the bankrupt suitors to those worth 100,000 ducats.

This means that the upper percentile of potential suitors in the whole population are worth upwards of 99,000 ducats. The lowest percentile is worth up to 1,000 ducats. The 50^th percentile is worth between 49,000 and 50,000 ducats.

Now Portia is presented with a hundred out of this population of potential suitors, and let’s assume that the suitors presented to her are representative of this population.

Say now that the first to be presented to her is worth 99,500 ducats. Since wealth is her only criterion, and he is in the upper percentile in terms of wealth, her optimal decision is to accept his proposal of marriage. It is possible that one of the next 99 is worth more than 99,500 ducats but that isn’t the way to bet.

On the other hand, say that the first suitor is worth 60,000 ducats. Since there are 99 more to come, it is a good bet that at least one of them will be worth more than this. If she has turned down all suitors, however, until she is being presented with the 99^th, her optimal decision now is to accept him. In other words, Portia’s decision as to whether to accept the proposal comes down to how many potential matches she has left to see. When down to the last two, she should choose him if he is above the 50^th percentile, in this case 50,000 ducats. The more there are to come the higher the percentile of wealth at which she should accept. She can set a higher threshold. She should never accept anyone who is below the average unless she is out of choices. In this version of the stopping problem, the probability that Portia will end up with the wealthiest of the available suitors turns out to be 58 per cent. More information, of course, increases the chance of success. Indeed, any criterion that provides information on where an option is relative to the relevant population as a whole will increase the probability of finding the best choice of those available. As such, it seems that if Portia is only interested in the money, she is more likely to find it than if she is looking for love.

References and Links

Optimal Stopping: How to find the perfect apartment, partner and parking spot. Brian Christian. Medium.com https://medium.com/galleys/optimal-stopping-45c54da6d8d0

Mathematics, marriage and finding somewhere to eat. +Plus magazine. David K. Smith. https://plus.maths.org/content/os/issue3/marriage/index