Are two heads better than one?

I imagined it like:

    F  T (Alice)
  F xx ????????
    xx ????????

  T ?? vvvvvvvv
    ?? vvvvvvvv
  ^ ?? vvvvvvvv
  B ?? vvvvvvvv
  o ?? vvvvvvvv
  b ?? vvvvvvvv
  v ?? vvvvvvvv
    ?? vvvvvvvv

In the check-mark (v) region, you get the right answer regardless; they are both being truthful, and of course you trust them when they agree. Similarly you get the wrong answer regardless in the x region.

In the ? region you are no better than a coin flip, regardless of your strategy. If you unconditionally trust Alice then you win on the right-hand side, and lose on the left-hand side; and whatever Bob says is irrelevant. The situation for unconditionally trusting Bob is symmetrical (of course it is; they both act according to the same rules, on the same information). If you choose any other strategy, you still have a 50-50 chance, since Alice and Bob disagree and there is no reason to choose one over the other.

Since your odds don't change with your strategy in any of those regions of the probability space, they don't change overall.

Bob isn't giving you any actionable information. If Alice and Bob agree, you're more confident than you were before, but you're still going to be trusting Alice. If they disagree you're down to 50% confidence, but you still might as well trust Alice.

It depends on what you are doing with the guess. If it is just a question of how frequently you are right or wrong the second person doesn't help. But if you are, for example, betting on your guess the second person improves your odds of coming out ahead significantly, since you can put down a higher wager when they agree than when they disagree.

I paused and wrote out all the probabilities and saw no way to improve beyond 80% - I scrolled down hoping to be proven wrong!

This kinda reminds me of error correction, and where at some level you can have detectable but not correctable error conditions. Adding Bob is just like adding a parity bit: can give you a good indication someone lied, but won't fix anything. Adding Charlie gives you the crudest ECC form, a repetition code (though for storing one bit, I don't think you can do better?)

One way to think about this is you have a binomial distribution with p=0.8 and n=number of lying friends. Each time you increase n, you shift the probability mass of the distribution "to the right" but if n is even some of that mass has to land on the "tie" condition.

I wrote a quick colab to help visualize this, adds a little intuition for what's happening: https://colab.research.google.com/drive/1EytLeBfAoOAanVNFnWQ...

This gives me an idea of how to implement isEven() and isOdd() probabilistically.

Why not unconditionally trust Bob?