Algebra for Adulterers

Algebra for Adulterers

Algebra for Adulterers

A mathematician's guide to the news.
Aug. 17 2001 3:00 AM

Algebra for Adulterers

What every philandering politician and traveling salesman should know about the odds of getting caught. 

Last month, the Washington Post reported on the April conversation between Susan Levy and her gardener in which the two stumbled onto what seemed to be an amazing coincidence: Both Modesto, Calif., parents had daughters who had had affairs with Rep. Gary A. Condit.


Even though the Levy gardener, Otis Thomas, recanted the allegation about his daughter's affair, the story still prompts this question: How many young women does a congressman have to sleep with in his district to generate a 5 percent chance—or whatever number you choose—that the parents will end up having this sort of intimate conversation?

I can work out an answer, but I have no idea whether it's right. Spurious precision is one of the cardinal mathematical sins, and to dance around it, I begin with the following disclaimer: My calculations are meant as an indication of the methods by which one might answer such a question. I'll try to be honest about the assumptions I make along the way, some of which are unwarranted and others of which I'm pretty sure are seriously false.

Let's start by simplifying the problem, assuming 1) fathers don't know about—or don't speak about—their daughters' sex lives, while mothers know everything; and 2) that each young woman has one mother and that each mother is attached to one young woman. Let's assume too that no one moves in or out of the district and that each mother has a finite set of intimate friends among the other mothers, all of whom she tells absolutely everything every day, so we can ignore all issues of time. We can now rephrase the question:

What is the chance that two of a congressman's young lovers have intimate mothers?

Much depends on the overall atmosphere of intimacy in the congressional district. Let's call the number of people in the district N, and let's say that a mother has, on average, n intimate friends. Then if two mothers are chosen at random from the district, the chance that they are intimate is approximately n/N. (If I have 10 friends, the chance that a random person from a population of 1,000 is my friend is 10/1,000 or 1 percent.) So the chance the two mothers are not intimate is (1 - n/N).

Let g be the number of girlfriends the congressman has; then the number of pairs of mothers he has to worry about is about (1/2)g2. How did I determine that? To specify a pair of mothers, we have to specify Mom 1 and Mom 2. There are g choices for Mom 1. For each of these choices, there are g - 1 (which is approximately g) choices for Mom 2. So, there are g2 possible pairings of mothers in all. But because we've counted both (Mom 1, Mom 2) and (Mom 2, Mom 1) as pairings, the number of different pairs is about (1/2)g2. To be precise, the exact number is (1/2)g(g - 1). Five congressional girlfriends, for example, would produce 10 unique pairs of mothers. Count them yourself, if you are so moved! If you think fathers are as talkative as mothers, you should multiply our figure by 4 to get 2g2 pairs of parents. We'll stick with the smaller number.

Now comes what we call a BFA, or "big false assumption": that intimacies between moms are independent events.

We say two events are "independent" if the occurrence of one doesn't affect the probability of the other. So "It is raining today" and "The next person I meet has blue eyes" are independent events. But "It is raining today" and "The next person I meet is holding an umbrella" are decidedly not independent. The occurrence of the first makes the second much more likely.

If two events are independent, it's easy to calculate the probability that both will take place. Suppose it rains 25 percent of days, and 20 percent of people have blue eyes; then the chance of "It is raining today" and "The next person I meet has blue eyes" is the product of the individual probabilities: 25 percent x 20 percent, or 5 percent. Without the independence assumption, this "product rule" is simply wrong. (See this Howard University law professor's discussion of the product rule as it pertained to DNA evidence in the O.J. Simpson trial.)