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 a_0 or b_0 small (where a_1 and b_1 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 a_0 and b_0. In a single trial, you can’t be sure that your strategy will do worse than chance.
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:
%20\\%0Ap_i%20%7C%20z_i%20=%200%20\sim%20\text{Beta}(a_0,%20b_0)%20\\%0Ap_i%20%7C%20z_i%20=%201%20\sim%20\text{Beta}(a_1,%20b_1)%20\\%0Ax_i%20\sim%20\text{Binomial}(n,%20p_i)%0A)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 a_0 or b_0 small (where a_1 and b_1 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 a_0 and b_0. In a single trial, you can’t be sure that your strategy will do worse than chance.