I can split the game by adding an extra node without changing the structure of the game
That actually does change the structure of the game, if we assume that physical mistakes can happen with some probability, which seems realistic. (Think about playing this game in real life as player 2. If you do get to move, you must think there’s some non-zero probability that it was because Player 1 just accidentally pressed the wrong button, right?) With the extra node, you get the occasional “crap, I just accidentally pressed not-A, now I have to decide which of B or C to choose” which has no analogy in the original game where you never get to choose between B or C without A as an option.
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.
That actually does change the structure of the game, if we assume that physical mistakes can happen with some probability, which seems realistic. (Think about playing this game in real life as player 2. If you do get to move, you must think there’s some non-zero probability that it was because Player 1 just accidentally pressed the wrong button, right?) With the extra node, you get the occasional “crap, I just accidentally pressed not-A, now I have to decide which of B or C to choose” which has no analogy in the original game where you never get to choose between B or C without A as an option.
Okay, I agree. But what do you think about the extensive-form game in the image below? Is the structure changed there?
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.