Comment by krackers
24 days ago
no in fact that's the proof that adding Bob doesn't help. If Alice & Bob disagree, then since both are correct with same probability it doesn't matter if you pick alice or bob. So WLOG you choose to trust Alice. But now when Alice and Bob agree, that means that you also trust Alice's output [as in you're only right if Alice is right (which is same as when Bob's right since Bob and Alice match)]. So in both cases you are right when Alice is right, i.e. you trust Alice and that means you don't even care about Bob's output.
I think why it feels odd is that most people intuitively answer a different question. If you had to bet on an outcome then Alice and bob agreeing gives you more information. But here you're not dealing with that question, you're either right and wrong; and whether or not Alice & Bob agree, you're effectively "wagering the same" in both cases (where your wager is 0.8, the probability [or expectation] that one is correct).
Although revisiting this, you have to be a bit careful about the argument.
Basically what you're doing is breaking down p(correct) = p(correct & agree) + p(correct & disagree) where former is 0.8*0.8 and latter is 0.8*0.2. Explicitly computing the conditional probability however makes calculating more difficult: p(correct | agree)*p(agree) + p(correct | disagree)*p(disagree). This is something like (16/17) * (0.8*0.8 + 0.2*0.2) + 0.5 * (0.8*0.2*2) which is not easy to arrive at intuitively unless you grind through the calculation.
So _conditioned_ on them agreeing you are right ~94% while conditioned on them disagreeing it's a coin-toss (because when they disagree exactly one is right, and it's equally likely to be alice or bob). Interesting case where the unconditional probability is actually more intuitive and easier than the conditional.