The problem specifies that Adam and Eve can play whatever they want, and asks for an equilibrium in which they choose probabilities other than 100%.

The fact that they’d both be better off in one situation than another says nothing about which situations are equilibria. Think prisoner’s dilemma.

If they both follow this strategy, suppose Jack flips heads and calls red. If Jill flips heads and calls red, he gets nothing; if Jill flips tails and calls red, he gets $20, so his expected gain is $10. This is better than the 74/13 that he gets under your strategy when he flips heads.

And if Jack flips tails and calls red, then whether Jill flips heads or tails, if she calls red too, Jack gets $20, so his expected gain is $20, also better than the 100/13 that he gets under your strategy if he flips tails.

So both of them are better off if they call red all the time, than if they follow your mixed probability strategy. So they’ll call red 100% of the time if that’s allowed, and 99.99% of the time if it’s not. Wouldn’t they? If not, why not?

You said Ron pointed out that there is “an equilibrium” when they both play red, but I think it’s more than that — it’s not just that there’s *an* equilibrium, it’s that that is the *best* equilibrium for both players.

Oops. I’m fixing the title. As for the rest: The rules of the game allow Jack and Jill to play whatever they want.

If the strategy of both players playing red 100% of the time, leaves both Jack and Jill better off then your suggested mixed probability strategy, then both of them will play red 100% of the time. If they are disallowed from playing red 100% of the time, then they’ll both play red 99% of the time, to get as close as possible to the optimum, because the optimum in this case is also the optimum for each of them *individually*. If you don’t think they would do this, why not?

You said, “If Jill plays the strategy I recommended, and Jack calls red 999999 out of 10000000 times, then Jill will switch to playing red all the time, whereupon Jack switches to playing red all the time.” Well yes, that’s what they’ll do if the rules of the game permit them to play red 100% of the time. And if that’s not permitted, then they’ll both keep playing red 99% of the time.

In that case the mixed-strategy equilibrium at P(red|head)=1/5, P(red|tail)=37/65 exists if both players are trying to minimize their scores, but not if they’re trying to maximize them. In that case there are only pure-strategy equilibria.

This is assuredly false. Taking Jill’s strategy as given, Jack’s payoff is independent of his strategy, so choosing this strategy both minimizes and maximizes his payoff.

If we treat the players’ probabilities separately, many of the mixed second derivatives are positive, so a player who deviates from the minimum-score equilibrium gives the other player an incentive to deviate also. (It’s an equilibrium only in the sense of indifference, not in the sense of stability.) If Jill uses those probabilities, then Jack is only indifferent between colors until Jill notices she can do better. If he switches to e.g. always playing red, then he’s no worse off if Jill continues her strategy, and better off if she notices his predictability and exploits it. Only if Jill is trying to minimize her score should she stay at P(red|head)=1/5, P(red|tail)=37/65.

Actually, maybe that’s what you intend? The economese version of the problem isn’t clear about which direction is better, although the translated version seems to obviously intend maximizing.