= 27d567bc2d48d9f40e9ccc20ea370b15)
twanvl comments on A simple game that has no solution - Less Wrong Discussion
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (123)
Could you explain what you mean? What uncertainty is there?
This game has two equilibria: a bad one at (A,A) and good one at (B,B). The game from this post also has two equilibria, but both involve player one picking A, in which case it doesn't matter what player two does (or in your version, he doesn't get to do anything).
If Player 2 gets to move he is uncertain as to what Player 1 did. He might have a different probability estimate in the game I gave than one in which strategy A did not exist, or one in which he is told what Player 1 did.
I'm not convinced that the game has any equilibrium unless you allow for trembling hands. For A,A to be an equilibrium you have to tell me what belief Player 2 would have if he got to move, or tell me that Player 1's belief about Player 2's belief can't effect the game.
In a classical game all the players move simultaneously. So to repeat, your game is: * player 1 chooses A, B or C * then, player 2 is told whether player 1 chose B or C, and in that case he chooses X or Y * payoffs are (A,-) -> (3,0); (B,X) -> (2,0); (B,Y) -> (2,2); (C,X) -> (0,1); (C,Y) -> (6,0)
The classical game equivalent is * player 1 chooses A, B or C * without being told the choice of player 1, player 2 chooses X or Y * payoffs are as before, with (A,X) -> (3,0); (A,Y) -> (3,0).
I hope you agree that the fact that player 2 gets to make a (useless) move in the case that player 1 chooses A doesn't change the fundamentals of the game.
In this classic game player 2 also has less information before making his move. In particular, player 2 is not told whether or not player 1 choose A. But this information is completely irrelevant for player 2's strategy, since if player 1 chooses A there is nothing that player 2 can do with that information.
If the players choose (A,X), then the payoff is (4,0). Changing his choice to B or C will not improve the payoff for player 1, and switching to Y doesn't improve the payoff for B. Therefore this is a Nash equilibrium. It is not stable, since player 2 can switch to Y without getting a worse payoff.
I'm not sure what you mean by "classical game" but my game is not a simultaneous move game. Many sequential move games do not have equivalent simultaneous move versions.
"I hope you agree that the fact that player 2 gets to make a (useless) move in the case that player 1 chooses A doesn't change the fundamentals of the game."
I do not agree. Consider these payoffs for the same game:
A 3,0 [And Player Two never got to move.]
B,X 2,10000
B,Y 2,2
C,X 0,1
C,Y 4,4
Now although Player 1 will never pick A, its existence is really important to the outcome by convincing Player 2 that if he moves C has been played.
Different payoffs imply a different game. But even in this different game, the simultaneous move version would be equivalent. With regards to choosing between X and Y, the existence of choice A still doesn't matter, because if player 1 chose A X and Y have the same payoff. The only difference is how much player 2 knows about what player 1 did, and therefore how much player 2 knows about the payoff he can expect. But that doesn't affect his strategy or the payoff that he gets in the end.