Friday, August 17, 2007

A brief musing on bonking

Chris Dillow introduced me to the blog of Norman Geras, from where I found this:

I bring you 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.

One of the things it shows is that when, in surveys, men report having many more sexual partners than women, something is amiss.

Interesting, I thought. That doesn't allow for homosexuals or, even, for the annoying (when I was a teenager) tendency for heterosexual girls to dance with each other rather than me or my equally spotty and lecherous mates. So I went to the source article.

There I found one answer in Professor Gale's original posing of the theorem but also several more fundamental errors:
But there is just one problem, mathematicians say. It is logically impossible for heterosexual men to have more partners on average than heterosexual women. Those survey results cannot be correct.

“I have heard this question before,” said Cheryl D. Fryar, a health statistician at the National Center for Health Statistics and a lead author of the new federal report, “Drug Use and Sexual Behaviors Reported by Adults: United States, 1999-2002,” which found that men had a median of seven partners and women four.

But when it comes to an explanation, she added, “I have no idea.”

The journalist has mis-represented the Prof and Cheryl, or the journalist, has made the usual mistake of mixing up the different sorts of average: mean, median and mode. Let us take a limiting example.

Take a population of 200, 100 boys and 100 girls, who are all asked the "how many sexual partners" question and all answer it honestly. 2 boys say "none", 98 boys say "3". 98 girls say "1", 2 (very popular) girls say "98". Total number of sexual partners is 294 both ways.

MeanSum of activity divided by number of samples2.942.94
ModeMost common answer31
MedianAnswer half-way through an ordered sample31

And another one for interest - add 2 (faithfully virginal) nuns :

MeanSum of activity divided by number of samples2.942.88
ModeMost common answer31
MedianAnswer half-way through an ordered sample31

So, although the Good Prof is correct and the total must be the same (assuming complete honesty and heterosexuality). However, the mean can be different if the population numbers are not equal - and I believe there are more lasses around than lads? Mode and median can clearly be different between the sexes (and, although I haven't shown it here, different from each other, just trust me :) I believe, from the sociological point, it would be the mode that one would take as showing the characteristic behaviour?

Discuss but let's keep it theoretical. It's been a long week and I apologise for any errors in spelling, logic, statistics or arithmetic :)

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