Are two heads better than one?

This is an interesting question!

But no, the python output is correct (although I do round the values). It's counterintuitive but these are two different questions:

    1. What are the odds that both players lie? (4%)
    2. Given that both players say tails, what are the odds that the coin is heads (~6%)

Trivially, the answer for question (1) is 0.2 * 0.2 = 4%

The answer for question (2) is 0.02 / 0.34 = 6%

One way of expressing this is Bayes Rule: we want P(both say tails | coin is heads):

    * we can compute this as (P(coin is heads | both say tails) * P(coin is heads)) / P(both say tails)
    * P(coin is heads | both say tails) = 0.04 (both must lie)
    * P(coin is heads) = 0.5
    * P(both say tails) = 0.04 * 0.5 + 0.64 * 0.5 = 0.34

This gives us (0.04 * 0.5) / 0.34 = 0.02 / 0.34 ~= 6%

I think that might not be convincing to you, so we can also just look at the results for a hypothetical simulation with 2000 flips:

    * of those 2000 flips, 1000 are tails
    * 640 times both players tell the truth
    * 40 times both players lie
    * 680 times (640 + 40) both players *agree*
    * 320 times the players disagree

We're talking about "the number of times they lie divided by the number of times that they agree"

40 / 680 ~= 6%

We go from 4% to 6% because the denominator changes. For the "how often do they both lie" case, our denominator is "all of our coin flips." For the "given that they both said tails, what are the odds that the coin is heads" case, our denominator is "all of the cases where they agreed" - a substantially smaller denominator!

The three players example is just me rounding 89.6% to 90% to make the output shorter (all examples are rounded to two digits, otherwise I found that the output was too large to fit on many screens without horizontal scrolling).

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).

I am serious, mechanical watches are fascinating but they are much less accurate that a cheap quartz watch.

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.