One way to look at dating and other life choices is to consider them as decision-time problems. Imagine, for example that have a number of candidates for a job, and all can be expected to say yes. You want a recipe that maximizes your chance to pick the best. This might apply to a fabulously wealthy individual picking a secretary or a husband (Mr Right) in a situation where there are 50 male choices. We’ll assume that you have the ability to recognize who is better than whom, but that your pool has enough ego that you can’t go back to anyone once you’ve rejected the person.
Under the above restrictions, I mentioned in this previous post that you maximize your chance of finding Mr Right by dating without intent to marry 36.8% of the fellows. After that, you marry the first fellow who is better than any of the previous. My previous post had a link to a solution using Riemann integrals, but I will now show how to do it with more prosaic math — a series. One reason for doing this by series is that it allows you to modify your strategy for a situation where you can not be guaranteed a yes, or where you’re OK with number 2, but you don’t like the high odds of the other method, 36.8%, that you’ll marry no one.
I present this, not only for the math interest, but because the above recipe is sometimes presented as good advice for real-life dating, e.g. in a recent Washington Post article. With the series solution, you’re in a position to modify the method for more realistic dating, and for another related situation, options cashing. Let’s assume you have stock options in a volatile stock company, if the options are good for 10 years, how do you pick when to cash in. This problem is similar to the fussy suitor, but the penalty for second best is small.
The solution to all of these problems is to pick a stopping point between the research phase and the decision phase. We will assume you can’t un-cash in an option, or continue dating after marriage. We will optimize for this fractional stopping point between phases, a point we will call x. This is the fraction of guys dated without intent of marriage, or the fraction of years you develop your formula before you look to cash in.
Let’s consider various ways you might find Mr Right given some fractional value X. One way this might work, perhaps the most likely way you’ll find Mr. Right, is if the #2 person is in the first, rejected group, and Mr. Right is in the group after the cut off, x. We’ll call chance of of finding Mr Right through this arrangement C1, where
C1 = x (1-x) = x – x2.
We could used derivatives to solve for the optimum value of x, but there are other ways of finding Mr Right. What if Guy #3 is in the first group and both Guys 1 and 2 are in the second group, and Guy #1 is earlier in the second line-up. You’d still marry Mr Right. We’ll call the chance of finding Mr Right this way C2. The odds of this are
C2 = x (1-x)2/2
= x/2 – x2 + x3/2
There is also a C3 and a C4 etc. Your C3 chance of Mr Right occurs when guy number 4 is in the first group, while #1, 2, and 3 are in the latter group, but guy number one is the first.
C3 = x (1-x)3/4 = x/4 – 3x2/4 + 3x3/4 – x4/4.
I could try to sum the series, but lets say I decide to truncate here. I’ll ignore C4, C5 etc, and I’ll further throw out any term bigger than x^2. Adding all smaller terms together, I get ∑C = C, where
C ~ 1.75 x – 2.75 x2.
To find the optimal x, take the derivative and set it to zero:
dC/dx = 0 ~ 1.75 -5.5 x
x ~ 1.75/5.5 = 31.8%.
That’s not an optimal answer, but it’s close. Based on this, C1 = 21.4%, C2 = 14.8%, C3 =10.2%, and C4= 7.0% C5= 4.8%Your chance of finding Mr Right using this stopping point is at least 33.4%. This may not be ideal, but you’re clearly going to very close to it.
The nice thing about this solution is that it makes it easy to modify your model. Let’s say you decide to add a negative value to not ever getting married. That’s easily done using the series method. Let’s say you choose to optimize your chance for either Mr 1 or 2 on the chance that both will be pretty similar and one of them may say no. You can modify your model for that too. You can also use series methods for the possibility that the house you seek is not at the last exit in Brooklyn. For the dating cases, you will find that it makes sense to stop your test-dating earlier, for the parking problem, you’l find that it’s Ok to wait til you’re less than 1 mile away before you settle on a spot. I’ll talk more about this latter, but wanted to note that the popular press seems overly impressed by math that they don’t understand, and that they have a willingness to accept assumptions that bear only the flimsiest relationship to relaity.
Robert Buxbaum, January 20, 2020
You left something out of your calculation: in order to know that you have achieved a global maximum at a critical point, you first need to know that your critical point is a local maximum and then you have to check the endpoints of the the interval (in this case x is between zero and one).
The reason why your critical point is is local and global maximum is that the curve in this case is a parabola with its nose pointed upwards. (For general curves, you might have to use the second derivative test and there may be more than critical point.)
You might want to check out “sequential analysis” as a reasonable statistical method for reducing sample size needed before making a rational decision based upon real probabilities.