Sex Partner Math

Via Half Sigma comes a report from "ABC News":

Young British women are more promiscuous than their male counterparts and more likely to be unfaithful, a new survey has revealed.

The study of 2,000 women in the UK, commissioned by More magazine, found that by the age of 21, women have had an average of nine sexual partners; two more than their male partner.

Maybe I'm missing something, but assuming an equal number of men and women, how is it possible for the average man to have more sexual partners than the average woman?

Let's imagine ten men and ten women. Each man sleeps with every woman, therefore each man has had 10 sex partners. But so has every woman.

Let's imagine that each man sleeps with only the first woman. So each man has had only 1 partner, and that one woman has had 10. But on average, the number of sex partners per woman = (1*10 + 9*0)/10 = 1 partner.

The article specifies "by the age of 21." Perhaps British women under 21 are acting on a preference for men older than 21. So by the time they retire from the ring, so-to-speak, men have caught up.

Perhaps the survey restriction to under-22 isn't really necessary in this case? Let's model a situation where younger women have relations with older men thus:

X1   O1

X2 / O2

X3 / O3

X4 / O4

X5 / O5

X6   O6

Pretty crude, but the idea behind each slash mark is that it "partners" the male cohort X below its row with the female cohort O above its row. Thus X3 couples with O1, X4 couples with O2, etc. Let's calculate the average number of sex partners for X1 - X6: (0 + 0 + 1 + 1 + 1 + 1)/6 = 0.67. And also for for O1 - O6: (1 + 1 + 1 + 1 + 0 + 0)/6 = 0.67. Same number of sex partners. But let's kill off the oldest cohorts (5 and 6) and bring two new cohorts (1' and 2') into the game. Everybody couples again (presumably with a different member of the cohort):

X1'    O1'

X2' /  O2'

X1  /  O1

X2  /  O2

X3  /  O3

X4     O4

Now X1 couples with O1', X2 couples with O2', X3 couples with O1, X4 couples with O2. Remember that now X4 and X3 have had sex twice, X2 and X1 have had sex once, and X2' and X1' have never had sex. But O4 and O3 have had sex once, O2 and O1 have had sex twice, and O2' and O1' have had sex once. Averaging for X = (0 + 0 + 1 + 1 + 2 + 2)/6 = 1. Average for O = (1 + 1 + 2 + 2 + 1 + 1) = 1.33.

Run the iteration again and discover that the X average remains at 1 while the O average climbs to (1 + 1 + 2 + 2 + 2 + 2)/6 = 1.67. Both the X's and the O's will all have had two partners by the time they die, but this fact will not be reflected in a survey of the entire population at any snapshot in time.

We should be able to use this model, to calculate the typical age difference in male/female couplings based on the data given in the article (male average = 7, female average = 9) and some rate of sex partner change. I will explore this in a subsequent post.

The article also specifies that each women has had "two more [partners] than their male partner." Doesn't this assumes that all respondents have a partner? Perhaps the survey was taken only among those "sexually active". Perhaps there are a greater percentage of non-active women than there are of men, and that the women who are active are doing extra duty.

So to speak.