We can think of this problem as trying to allocate correct and incorrect guesses across possible worlds.

For a single person, their expected probability of being correct (conditional on venturing a guess) is always 0.5 - that is, every time we allocate probability mass to them making a correct guess in some world, we must allocate the same total probability mass to them making an incorrect guess in another world.

Because of the asymmetry in our win condition, we can do better than 50/50, because we can have our winning scenarios only use one correct guess while our losing scenarios can take up three (but no more) incorrect guesses. So the best we can hope for is to have a 3:1 ratio of successful outcomes to unsuccessful outcomes - i.e., 75% odds.

If we had more than 75% probability mass on correct worlds, we'd have more than $0.75$ weighted person-correct-guesses, which we'd need to balance out with an equal number of weighted person-incorrect-guesses - but we can only fit those into possible worlds with a density of $3$, and we'd have fewer than $0.25$ measure of possible worlds in which to do that allocation.

Suppose we have some mixed strategy for this hat puzzle. For each of the $8$ possible hat assignments and each of the $3$ people involved, we can ask about the probability that the person guesses correctly conditional on that hat assignment, and the probability that they guess incorrectly. After specifying this, we'll have assigned a probability to every possible (world state, person 1 accuracy, person 2 accuracy, person 3 accuracy) tuple, where the world state can be in one of 8 arrangements, and each person's accuracy can be graded as "correct", "silent", or "incorrect".

For a single person, their expected probability of being correct (conditional on venturing a guess) is always 0.5 - that is, the total probability of all outcomes in which they guess correctly must equal the total probability of all outcomes in which they guess incorrectly.

In order for an outcome to lead to a success, we need at least one correct guess and no incorrect guesses. So the probability of success in the puzzle is at most the sum of P(correct) across all three people.

Meanwhile, an unsuccessful outcome can have at most three incorrect guesses, because there are only three people. So the probability of failure is at least one-third as large as the sum of P(incorrect) across all three people.

But for each person, P(correct) = P(incorrect)! So we know that the success probability is at most three times the failure probability, which bounds our success rate by 0.75.