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DanielLC comments on A simple game that has no solution - Less Wrong Discussion
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I thought the answer was Player One picks B with 1/3+ϵ probability and C with 2/3-ϵ probability. Player Two picks Y.
This gives Player One an expected value of 2(1/3+ϵ) + 6(2/3-ϵ) = 14/3-4ϵ and Player Two an expected value of 2(1/3+ϵ) = 2/3+2ϵ.
If Player Two picked X, he'd have an expected value of 2/3-ϵ, and miss out on 3ϵ.
If Player One picked B with a higher probability, Player Two would still pick C, and Player One wouldn't gain anything. If Player One picked C with a higher probability, Player Two would pick X, and Player One would get nothing. If Player One picked A, he'd only get 3, and miss out on 5/3-4ϵ.
Did I mess up somewhere?
If Player One believes that Player Two is going to pick Y, then Player One will pick C, but of course this isn't an equilibrium since Player Two would regret his strategy. All Player Two ever sees is Player One's move, not the probabilities that Player 1 might have used so if C is played Player Two doesn't know if it was because Player One Played C with probability 1 or probability 2/3-ϵ.
Perhaps Player One figures that Player Two knows enough about him to predict him slightly more than explainable by chance, and vice versa.