A mathematician says it is a logical impossibility for heterosexual men to have more partners on average than heterosexual women. I think about adding a Fark-style OBVIOUS tag to the Aardvark.
By way of dramatization, we change the context slightly and will prove what will be called the High School Prom Theorem. We suppose that on the day after the prom, each girl is asked to give the number of boys she danced with. These numbers are then added up giving a number G. The same information is then obtained from the boys, giving a number B.
Theorem: G=B
Proof: Both G and B are equal to C, the number of couples who danced together at the prom. Q.E.D.
The Myth, the Math, the Sex (New York Times)
Isn’t that proof irrelevant to the premise? Say there are 10 girls and 1 boy. Say all 10 girls dance with the 1 boy. So G = (1 * 10) = 10 and B = (10 * 1) = 10, sure, but it’s still certainly true in this case that the boy had 10 partners and all 10 girls had only 1 partner.
It only works if you require that the number of boys and girls needs to be equal, e.g. 5 boys and 5 girls. However, even this illustrates that the mathematicians may be missing the point: say 3 of the boys don’t dance with anyone, and 2 dance with all 5 girls. Then the average number of partners for the girls is obviously (2 * 5) / 5 = 2 and for the boys it’s (5 * 2) / 5 = 2, sure. But while it’s not mathematically possible for an equal population of boys and girls to have a different AVERAGE number of partners, it’s certainly possible for one group to have a higher average number of partners among ACTIVE members of said population– in this example, 5 for the boys and 2 for the girls.
I’m not saying that’s necessarily what’s going on here, just that the math works perfectly well in the context of what the statistics may well be saying: that among sexually active adults, it’s entirely possible for men to average more partners than women. So I guess the question is, are the surveys in question limited to sexually active adults, or are there a bunch of virgins included? Because if it’s the former, the results are perfectly believable from a mathematical standpoint.
And even if it’s the latter, no survey is going to encompass the entire population, so there’s no reason to believe that the group of people surveyed doesn’t randomly contain a concentration of high-partner men and low-partner women. Why would the mathematicians assume that it’s a closed system, i.e. that everyone in the survey can only have had other people in the survey as partners? That’s absurd. Yet that’s what the proof assumes.
I’m just sayin’.
Also, Mitt Romney is still a MOTHERFUCKER.
Or are some guys are so forgettable women can’t remember them even if they try?
Logically and mathematically, yes. But even if survey respondents were honest in their answers, surveys can be screwed easily in this type of study. If you surveyed a town of 1000 people and got 99% compliance, that’s extremely good for a survey. BUT if one of the ten people you missed was Jezebel, one of the three female heterosexual prostitutes in town, you’ve skewed the results considerably. In another example, if you surveyed a college campus, you have to correct for the students who have off-campus sex partners outside of their age range, which is probably more likely for females since they are more prone to go outside their age range with older partners, who are more numerous.
I know, the point of the areticle is that people lie about this. But my inner social statistics geek had to surface.
Hey, where did you go?
I just had my annual attack of Ennui 2.0. Fed up with RSS feeds, I went on an RSS fast. Now attempting to catch up with my “day in history” posts for the past week.