SPOILER ALERT: solution presented here. Rot-13 would be immensely painful, so I’ll just present some facts from which an LW reader can piece together a solution. The probability of drawing 11 blue balls followed by 9 green balls from an urn that’s 45% blue is 7.0567033E-7. If the urn is 65% blue it’s 6.8969856E-7. The log-likelihood-ratio is 0.099425348 decibans. So for practical purposes the answer is “my best guess is still 50/50” but the posterior probability is really 0.50572313. If you draw 100 rather than 20 it’s 0.52858571.
I think even if it were a real-life problem I would have correctly guessed, without doing arithmetic, which class had more evidence; but I was sort of spoilered by knowing that the answer has to be the “counterintuitive” one.
ETA: another way to do the sums is that each boy provides 1.5970084 decibans of evidence for program A, and each girl 1.9629465 for program B.
SPOILER ALERT: solution presented here. Rot-13 would be immensely painful, so I’ll just present some facts from which an LW reader can piece together a solution. The probability of drawing 11 blue balls followed by 9 green balls from an urn that’s 45% blue is 7.0567033E-7. If the urn is 65% blue it’s 6.8969856E-7. The log-likelihood-ratio is 0.099425348 decibans. So for practical purposes the answer is “my best guess is still 50/50” but the posterior probability is really 0.50572313. If you draw 100 rather than 20 it’s 0.52858571.
I think even if it were a real-life problem I would have correctly guessed, without doing arithmetic, which class had more evidence; but I was sort of spoilered by knowing that the answer has to be the “counterintuitive” one.
ETA: another way to do the sums is that each boy provides 1.5970084 decibans of evidence for program A, and each girl 1.9629465 for program B.
If you look at girls in addition to boys, it’s no longer quite so counterintuitive.