A teenage boy with a new driver license in pocket and \$1000 saved up from part-time and summer jobs, buys a used station wagon. He figures this will be a rad make-out machine, if a little clunky. Now, he just needs to score a chick...

He has been studying statistics, and conceives a cunning plan to get his new mathematical friend to help him score a chick. The girls in the school are gray and blue; and there is a dance coming up. He will take an anonymous poll to find about dance attendance, color, and likelihood of having sex on the first date. Armed with these statistics, he can maximize chances of getting sexed on the date, by going to the dance or not, asking out a girl of the most likely color.

He finds that, of the 54 girls going to the dance, 33 are gray, 15 of whom have sex on the first date (45%); of the 21 blue girls, 9 have sex on the first date (42%). So, if he goes to the dance, his odds of baptizing the sex machine is better if he goes with a gray girl.

The girls not going to the dance are similarly split: of these 69 girls, 27 are gray, 18 of whom have sex on the first date (66%); of the 42 blue girls, 27 have sex on the first date (64%). His chances in this group are better overall, and he is still more likely to get sexed with a gray girl.

The dance is cancelled a week ahead of time, and he still hasn’t hooked up, so he decides to firm his resolve by rechecking his numbers. His original figures are correct, but now it doesn’t matter if a girl had been going to the dance or not, so he calculates on the combined dance and non-dance groups. Of the 123 girls, 60 are gray, 33 of whom put out on the first date (55%); 36 of the 63 blue girls do the horizontal bop first time out on the floor (57%). How can it be more advantageous to date gray girls when the groups are split, and more advantageous to date blue girls when the groups are merged?