The structure isn’t changed there, but without the extra node, there is no subgame. That extra node is necessary in order to have a subgame, because only then can Player 2 think “the probabilities I’m facing is the result of Player 1′s choice between just B and C” which allows them to solve that subgame independently of the rest of the game. Also, see this comment and its grandchild for why specifically, given possibility of accidental presses, I don’t think Player 2′s strategy in the overall game should be same as the equilibrium of the 2x2 “reduced game”. In short, in the reduced game, Player 2 has to make Player 1 indifferent between B and C, but in the overall game with accidental presses, Player 2 has to make Player 1 indifferent between A 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.
The structure isn’t changed there, but without the extra node, there is no subgame. That extra node is necessary in order to have a subgame, because only then can Player 2 think “the probabilities I’m facing is the result of Player 1′s choice between just B and C” which allows them to solve that subgame independently of the rest of the game. Also, see this comment and its grandchild for why specifically, given possibility of accidental presses, I don’t think Player 2′s strategy in the overall game should be same as the equilibrium of the 2x2 “reduced game”. In short, in the reduced game, Player 2 has to make Player 1 indifferent between B and C, but in the overall game with accidental presses, Player 2 has to make Player 1 indifferent between A 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.