In the Nash equilibrium, what is Player 2′s belief if he gets to move? Also, the link you gave is for solving simultaneous move games, and the game I presented is a sequential move game.
You are implicitly using circular reasoning. Not picking A is only irrational for some but not all possible beliefs that Player 2 could have if Player 1 does not pick A. And even if we grant your assumption, what should Player 2 do if he gets to move, and if your answer allows for the possibility that he picks Y how can you be sure that Player 1 is irrational?
I’m not using circular reasoning. The choice for player one between A and either B or C is a choice between a certain payoff of 3 and an as of yet uncertain payoff. If player A already chose to play either B or C, the game transforms into a game with a simple 2x2 payoff matrix. Writing down the matrix we see that there is no pure dominant strategy for either player. We know though that there is a mixed strategy equilibrium as there always is one. Player one assumes that player two will play such that player one’s choice does not matter and equalises his expected payoff to 2. Player two again assumes that player one plays in such a way that their choice does not matter and equalises his expected payoff to 2⁄3. As the expected payoff for player one in the second game is lower than in the first game, at least one of the following assumptions has to be false about player one:
Player one maximises expected utility in terms of game payoff
Player one cares only about his own utility, not the utility of player two
Player one assumes player two to not act according to similar principles
“If player [1] already chose to play either B or C, the game transforms into a game with a simple 2x2 payoff matrix.”
No because Player 2 knows you did not pick A and this might give him insight into what you did pick. So even after Player 1 picks B or C the existence of strategy A might still effect the game because of uncertainty.
Distuingish between the reasoning and the outcome. Game theoretic reasoning is memory-less, the exact choice of action of one player does not matter to the other one in the hypothetical. As the rules are known, both players come to the same conclusion and can predict how the game will play out. If in practice this model is violated by one player the other player immediately knows that the first player is irrational.
Sequential move games are essentially a subset of simultaneous move games. If two players both write source code for programs that will play a sequential move game, then the writing of the code is a simultaneous move game.
Things don’t feel so simple to me. (A, X) is a Nash equilibrium (and the only pure strategy NE for this game), but is nonetheless unsatisfactory to me; if player 1 compares that pure strategy against the mixed strategy proposed by Wei_Dai, they’ll choose to play Wei_Dai’s strategy instead. Nash equilibrium doesn’t seem to be a strong enough requirement (“solution concept”) to force a plausible-looking solution. [Edit: oops, disregard this paragraph. I misinterpreted Wei_Dai’s solution so switching to it from the NE pure equilibrium won’t actually get player A a better payoff.]
(I also tried computing the mixed strategy NE by finding the player 1 move probabilities that maximized their expected return, but obtained a contradiction! Maybe I screwed up the maths.)
I think you can just compute the Nash Equilibria. For example, use this site: http://banach.lse.ac.uk/
The answer appears to be “always pick A”. Player 2 will never get to move.
In the Nash equilibrium, what is Player 2′s belief if he gets to move? Also, the link you gave is for solving simultaneous move games, and the game I presented is a sequential move game.
That player one is not rational and he should abandon classical game theory.
You are implicitly using circular reasoning. Not picking A is only irrational for some but not all possible beliefs that Player 2 could have if Player 1 does not pick A. And even if we grant your assumption, what should Player 2 do if he gets to move, and if your answer allows for the possibility that he picks Y how can you be sure that Player 1 is irrational?
I’m not using circular reasoning. The choice for player one between A and either B or C is a choice between a certain payoff of 3 and an as of yet uncertain payoff. If player A already chose to play either B or C, the game transforms into a game with a simple 2x2 payoff matrix. Writing down the matrix we see that there is no pure dominant strategy for either player. We know though that there is a mixed strategy equilibrium as there always is one. Player one assumes that player two will play such that player one’s choice does not matter and equalises his expected payoff to 2. Player two again assumes that player one plays in such a way that their choice does not matter and equalises his expected payoff to 2⁄3. As the expected payoff for player one in the second game is lower than in the first game, at least one of the following assumptions has to be false about player one:
Player one maximises expected utility in terms of game payoff
Player one cares only about his own utility, not the utility of player two
Player one assumes player two to not act according to similar principles
“If player [1] already chose to play either B or C, the game transforms into a game with a simple 2x2 payoff matrix.”
No because Player 2 knows you did not pick A and this might give him insight into what you did pick. So even after Player 1 picks B or C the existence of strategy A might still effect the game because of uncertainty.
Distuingish between the reasoning and the outcome. Game theoretic reasoning is memory-less, the exact choice of action of one player does not matter to the other one in the hypothetical. As the rules are known, both players come to the same conclusion and can predict how the game will play out. If in practice this model is violated by one player the other player immediately knows that the first player is irrational.
“Game theoretic reasoning is memory-less”
No. Consider the tit-for-tat strategy in the infinitely repeated prisoners’ dilemma game.
Why is it irrational for Player 1 to not pick A? Your answer must include beliefs that Player 2 would have if he gets to move.
Sequential move games are essentially a subset of simultaneous move games. If two players both write source code for programs that will play a sequential move game, then the writing of the code is a simultaneous move game.
I don’t think your game is sequential, if Player 2 doesn’t know Player 1′s move.
You really have two games:
Game 1: Sequential game of Player 1 chooses A or B/C, and determines whether game 2 occurs.
Game 2: Simultaneous game of Player 2 maybe choosing X or Y, against Player 1′s unknown selection of B/C.
edit: And the equilibrium case for Player 1 in the second game is an expected payout of 2, so he should always choose A.
Things don’t feel so simple to me. (A, X) is a Nash equilibrium (and the only pure strategy NE for this game), but is nonetheless unsatisfactory to me; if player 1 compares that pure strategy against the mixed strategy proposed by Wei_Dai, they’ll choose to play Wei_Dai’s strategy instead. Nash equilibrium doesn’t seem to be a strong enough requirement (“solution concept”) to force a plausible-looking solution. [Edit: oops, disregard this paragraph. I misinterpreted Wei_Dai’s solution so switching to it from the NE pure equilibrium won’t actually get player A a better payoff.]
(I also tried computing the mixed strategy NE by finding the player 1 move probabilities that maximized their expected return, but obtained a contradiction! Maybe I screwed up the maths.)