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satt comments on A simple game that has no solution - Less Wrong Discussion
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I think the following is the unique proper equilibrium of this game:
Player One plays A with probability 1-ϵ, B with probability 1/3 ϵ, C with probability 2/3 ϵ. Player Two plays X with probability 2/3 and Y with probability 1/3.
JGWeissman had essentially the right idea, but used the wrong terminology.
ETA: I've changed my mind and no longer think the proper equilibrium solution makes sense for this game. See later in this thread as well as this comment for the explanation.
Should that be the other way round?
As written, player 1 expects to score 3(1-ϵ) + 2(ϵ/3) + (2ϵ/3)(6/3) = 3 - 3ϵ + 2ϵ/3 + 4ϵ/3 = 3-ϵ (assuming I haven't made a dumb error there), and so would do better by unilaterally switching to the pure strategy A.
But if player 2 plays Y with probability 2/3 and X with probability 1/3, player 1 can expect to score 3(1-ϵ) + 2(ϵ/3) + (2ϵ/3)(12/3) = 3 - 3ϵ + 2ϵ/3 + 8ϵ/3 = 3 + ϵ/3, which beats the pure strategy A.
Edit: no, ignore me, I forgot that the whole point of proper equilibrium is that ϵ is an arbitrary parameter imposed from outside and assumed nonzero. Player 1 isn't allowed to set it to zero. [Slaps self on forehead.]