detector confidence the situation is as follows: Say, I am going to answer a multiple choice question based on suggestion of my two friends. One gives correct answer 60% percent of the times, the other gives correct answer 40% of the times. Is there any way I can answer a question correctly more than 60% of the time? (in a long run average leaving aside natural variability)
 A: Intuitively, I figured that person B's didn't help, since there was no correlation.  However, I went back to Bays to solve it more rigorously and found an (at least interesting to me) result.
assume N choices, all of which are equally probable
P[ans is 1 | A says 1 AND B says 1] = P[ A says 1 AND B says 1 | ans is 1 ] P[ ans is 1 ] / P[ A says 1 and B says 1 ]
PP[ans is 1 | A says 1 AND B says 1] = (3/5) (2/5) (1/N) / ( (3/5) (2/5) (1/N) + (N-1) (1/N) (1/(N-1)) (2/5) (3/5) ), explanation below
PP[ ans is 1 | A says 1 AND B says 1] = 1 / 2
The other 1/2 chance is spread over all other (the incorrect) answers, but if there are only two choices (i.e. true/false), then in the event that A and B agree, you can guess either, it does not matter.  You have less confidence if B agrees.
*explanation of P[]
since A is right 3/5 and B is right 2/5,
P[ A says 1 AND B says 1 | ans is 1 ] = P[ A 1 | ans 1 ] P[ B 1 | ans 1 ] = (3/5) (2/5)
since the a priori probability of answer 1 is 1/N, 
P[ A says 1 and B says 1 ] = P[ A says 1 AND B says 1 | ans is 1 ] P[ ans is 1 ] = (3/5) (2/5) (1/N)
P[ A says 1 and B says 1 ] = P[ A 1 and B 1 | ans 1 ] P[ ans 1 ] + P[ A 1 and B 1 | ans NOT 1 ] P[ ans NOT 1 ]
since there are N-1 ways for the answer to not be 1, each with a priori probability 1/N, but only a (1/(N-1)) chance to be wrong in such a way to suggest the answer is 1, and the probabilities of being wrong are A 2/5 and B 3/5,
P[ A says 1 and B says 1 ] = (3/5) (2/5) (1/N)  + (N-1) (1/N) (1/(N-1)) (2/5) (3/5)
