binomial variance problem

nerfhammer

Found in an old Kahneman & Tversky paper:

There are two programs in a high school. Boys are a majority (65%) in program A, and a minority (45%) in program B. There is an equal number of classes in each of the two programs.

You enter a class at random, and observe that 55% of the students are boys. What is your best guess -- does the class belong to program A or to program B?

Found in an old Kahneman & Tversky paper:

There are two programs in a high school. Boys are a majority (65%) in program A, and a minority (45%) in program B. There is an equal number of classes in each of the two programs.

You enter a class at random, and observe that 55% of the students are boys. What is your best guess -- does the class belong to program A or to program B?

[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....]

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:

$\\\\ z\_i \\sim \\text\{Bernoulli\}\(0\.5\$ %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... (read more)

13faul_sname14y

Or just change the 45% and 65% to 11% and 99%. That makes the correct answer pretty obvious without changing anything important.