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othercriteria comments on binomial variance problem - Less Wrong Discussion
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Comments (14)
Cute problem. And you can probably go a bit further in assessing how good your best guess is by inferring that the class size is at least 20 and lower bounding your variances.
[Or you can be dickish/clever and claim that the problem is underspecified because you're only given the overall boy/girl percentages for the two programs, and not their distribution. E.g., if each class has either exactly 65% or exactly 45% boys, then your observation is consistent with neither of the classes.]
[Actually you can't be dickish/clever that way: The problem isn't underspecified as the goal is to do the best you can with the information you've got. You've got no information/evidence regarding the distribution between classes so your best bet is to treat it as random. From there you can use Bayes theorem, blah blah, etc. etc....]
Or just change the 45% and 65% to 11% and 99%. That makes the correct answer pretty obvious without changing anything important.
Oops, you're right. The variant of the problem I mentioned above got rid of the assumption of binomially distributed boys (equivalently, girls).
The following setup should work, though:
In words, this says that to generate the i-th class, you flip a coin to tell whether it's in program A or program B, conditioned on the program, the proportion of boys is drawn from a program-specific beta distribution, and then the number of boys is drawn from the corresponding binomial distribution. Under the constraints that%20=%200.65)
and %20=%200.45)
, the average proportion of boys matches up with the problem.
However, by taking
or 
small (where 
and 
are adjusted accordingly to maintain the constraint), you can play with the variance so that the observed 55% boys class is more likely under either of the programs. If you had available repeated trials, you might be able to learn 
and 
. In a single trial, you can't be sure that your strategy will do worse than chance.