Ok, here’s my attempt to answer the question for N colleges and seeing counselors A and B. Counselor A recommends college H and we'd do pretty well listening since A is right most of the time. If B's recommendation differs, we should have a little greater confidence since B is mostly wrong and extra information is helpful. (Think of it like this: B’s answer effectively eliminates one college since B is usually wrong, so A is now selecting mostly correctly from a smaller pool of choices, boosting our confidence.) So my answer is "yes".
There might be an easy way to show it, but I only know Bayes' rule. The probability of H being correct given that A recommends H and B recommends something else is
P(H|A_H, B_~H) = P(A_H, B_~H|H) * P(H) / P(A_H, B_~H),
where
P(A_H, B_~H|H) is prob. that A recommends H and B recommends something else given that H is correct,
P(H) is probability that H is correct college with no other
a priori information, and
P(A_H, B_~H) is the marginal probability of A recommending H and B something else. This last is summed over the two possibilities H is correct and H is wrong:
P(A_H, B_~H)= P(A_H, B_~H|H) * P(H) + P(A_H, B_~H|~H) * P(~H).
We know
P(H) = 1/N;
P(~H) = 1 – P(H) = (N-1) / N
P(A_H |H) = x, and
P(B_H|H) = y or P(B_~H|H) = 1-y. (x and y are 0.8 and 0.1 in the problem)
We can write
P(A_H, B_~H|H) = P(A_H |H) * P(B_~H|H) = x*(1-y)
since the counselors work independently. Likewise the second term is
P(A_H, B_~H|~H) = P(A_H |~H) * P(B_~H|~H). We need values for these.
P(A_H |~H) = ( 1 - P(A_~H |~H) ) / (N-1) = (1-x) / (N-1). [Prob. that H is wrong answer but A chooses it out of N-1 possibilities.]
P(B_~H|~H) = 1 - P(B_H|~H) = 1 – (1-P(B_~H|H)) / (N-1) = (N-2+y) / (N-1). [Prob. that H is wrong answer and B doesn’t give H. Note that B may still be wrong since there are many answers that aren’t H].
The hard part is over (whew!) and the rest is rewriting and simplifying. The answer is
P(H|A_H, B_~H) = x*(1-y)*(N-1) / ( N*(1-xy) + x + y - 2 ).
Sure hope I did that right!
Plugging in x = 0.8 and y = 0.1 for this problem,
P(H|A_H, B_~H) = 0.72 * (N-1) / (0.92*N – 1.1).
This gives probabilities
N P
2 0.973
3 0.868
.
.
8 0.802
9 0.800
10 0.798
inf 0.783
So we definitely do better than A by also listening to B, so long as N<=8.
It’s interesting that we do worse by adding B’s information for N>9. If B were always 100% wrong, then his information would always aid us as can be seen by setting y=0 above.
After that I have no energy left to add counselor C!
Edit: corrected a couple of typos.
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