In the 2x2 reduced game, Player One’s strategy is 1⁄3 B, 2⁄3 C; Two’s strategy is 2⁄3 X, 1⁄3 Y.
In the complete game with trembling hands, Player Two’s strategy remains unchanged, as you wrote in the starter of the linked thread, invoking proper equilibrium.
Later on in the linked thread, I realized that the proper equilibrium solution doesn’t make sense. Think about it: why does Player 1 “tremble” so that C is exactly twice the probability of B? Other than pure coincidence, the only way that could happen is if some of the button presses of B and/or C are actually deliberate. Clearly Player 1 would never deliberately press B while A is still an option, so Player 1 must actually be playing a mixed strategy between A and C, while also accidentally pressing B and C with some small probability. But that implies Player 2 must be playing a mixed strategy that makes Player 1 indifferent between A and C, not between B and C.
In the 2x2 reduced game, Player One’s strategy is 1⁄3 B, 2⁄3 C; Two’s strategy is 2⁄3 X, 1⁄3 Y. In the complete game with trembling hands, Player Two’s strategy remains unchanged, as you wrote in the starter of the linked thread, invoking proper equilibrium.
Later on in the linked thread, I realized that the proper equilibrium solution doesn’t make sense. Think about it: why does Player 1 “tremble” so that C is exactly twice the probability of B? Other than pure coincidence, the only way that could happen is if some of the button presses of B and/or C are actually deliberate. Clearly Player 1 would never deliberately press B while A is still an option, so Player 1 must actually be playing a mixed strategy between A and C, while also accidentally pressing B and C with some small probability. But that implies Player 2 must be playing a mixed strategy that makes Player 1 indifferent between A and C, not between B and C.