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army1987 comments on A simple game that has no solution - Less Wrong Discussion
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Assume that each player's hand may tremble with a small non-zero probability p, then take the limit as p approaches zero from above.
... Let's do that!
Simple model: A plays A, B and C with probabilities a, b, and c, with the constraint that each must be above the trembling probability t (=p/3 using the p above). (Two doesn't tremble for simplicity's sake)
Two picks X with probability x and Y with probability (1-x).
So their expected utilities are:
One: 3a + 2b+6c(1-x)
Two: 2b(1-x) + cx = 2*b + (c - 2b) x
It seems pretty clear that One wants b to be as low as possible (either a or c will always be better), so we can set b=t.
So One's utility is (constant) - 3c+6c -6cx
So One wants c to maximize (1-2x)c, and Two wants x to maximize (c-2t)c
The Nash equilibrium is at 1-2x=0 and c-2t=0, so c=2t and x=0.5
So in other words, if One's hand can tremble than he should also sometimes deliberately pick C to make it twice as likely as B, and Two should flip a coin.
(and as t converges towards 0, we do indeed get One always picking A)
Excellent! This does indeed work given the assumption that Player 1 can not set the probability of himself picking B at zero. But if Player 1 can set the probability of him picking B at zero, the game still has no solution.
Good thinking, but I picked the payoffs so this approach wouldn't give an easy solution. Consider an equilibrium where Player 1 intends to pick A, but there is a small but equal chance he will pick B or C by mistake. In this equilibrium Player 2 would pick Y if he got to move, but then Player 1 would always intend to Pick C, effectively pretending he had made a mistake.