How to find the mixed strategy equilibrium in the following game:
P2
L R
L (3,1) (0,1)
P1 M (1,1) (1,1)
R (0,1) (4,1)
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asked Sep 27, 2015 at 19:04
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$\begingroup$ We can eliminate M for P1 but even without eliminating it we should be able to calculate the same mixed strategy equilibrium. How to do it ? $\endgroup$– Sub-OptimalSep 27, 2015 at 19:13
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4$\begingroup$ I'm voting to close this question as off-topic because it is a basic question and there is no effort shown at a solution. Because the site is not aimed at solving homework problems we try to avoid these questions. If you write down what you have tried so far the question may be reopened and you will probably get helpful feedback. $\endgroup$– GiskardSep 27, 2015 at 20:32
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$\begingroup$ I have tried solving this question by eliminating the dominated strategy. For player P1, M is a strictly dominated strategy because if he randomizes between L and M with 0.5 probability associated with each then the expected payoffs for that mixed strategy will be (1.5,1) and (2,1) which are in any case better for P1 than what he gets by playing M. So, M can be eliminated. And then we get an infinite number of Nash's equilibria in the game. However, if I do not want to eliminate M beforehand but solve this game with this given structure only then I am unable to find the solution. Please help. $\endgroup$– Sub-OptimalSep 27, 2015 at 20:39
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1 Answer
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As you have suggested in a comment, $M$ will not be played with positive probability in any Nash equilibrium. This is true for any strictly dominated action.
Given that $M$ has zero probability, the rest of your question is answered in another question of yours in another thread at How infinite Nash equilibria are possible in a game?
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$\begingroup$ How do we conform that M will not be played with positive probability. I am interested in the calculation of that. $\endgroup$ Sep 28, 2015 at 17:43
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$\begingroup$ A strategy profile is a Nash equilibrium if for each player, the strategy specified for that player yields at least as good utility as any other alternative strategy given the rest of the strategies of others. One corollary of this is that every strategy that is played with positive probability must be a best response to others' strategies. M is strictly dominated as you have calculated, meaning that whatever the other players choose, it yields a lower payoff for player 1 (compared to the one you calculated), implying that it is not a best response for any possible strategy profile of others. $\endgroup$– ramazanSep 28, 2015 at 19:20
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$\begingroup$ In the above structure let the prob. of L,M,R be p1,p2,p3 respectively for row player and prob. of L, R for column player be q and q-1. Now if we try to calculate the the probabilities here, what will be the solution. I am trying to find out the mixed equilibrium in this way. Ideally it should come out to be that p2=0, p1 and p2 can be anything from 0 to 1 and q=4/7. As I calculated in the other question. So can we find out the same solution by solving for p1,p2,p3 and q. $\endgroup$ Sep 28, 2015 at 20:07
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$\begingroup$ You cannot proceed further by trying to prove it that way. The guaranteed method here is case analysis. You would say: Case1 : Is there an equilibrium with p1>0, p2>0, p3=0? Case2 : Is there an equilibrium with p1>0, p2=0, p3>0? Case3 : Is there an equilibrium with p1=0, p2>0, p3>0? Case4 : Is there an equilibrium with p1>0, p2>0, p3>0? Case5 : Is there a pure strategy equilibrium? However, recognizing that p2=0 due to strict dominance lets you eliminate a big chunk of your cases, which is good. But if you happen to not realize that shortcut, you are still fine. More work but still will work. $\endgroup$– ramazanSep 29, 2015 at 0:34