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$$ \begin{array}{cccc} & Q & R & S \\ K & [2,0] &[4,5]&[1,1] \\ L &[3,2]&[1,0]&[0,0] \\ M &[1,1]&[1,0]&[0,0] \\ \end{array} $$

What is the mixed strategy Nash equilibrium in this game?

My own calculated mixes doesn't add up. TY in advance.

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  • $\begingroup$ Nobody has a better option than getting [4,5]. So the equilibrium should be: P1 chooses K 100% of the time, P2 chooses R 100% of the time. $\endgroup$
    – genisage
    Oct 16, 2014 at 21:06

1 Answer 1

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First note $M$ for player 1 (row player) is strictly dominated by $K$. Hence $M$ is never played with positive probability. Similarly, $S$ is strictly dominated by a mixture of $Q$ and $R$ for player 2 (column player). Hence $S$ is never played with positive probability.

Now let $\alpha$ be the probability of player $K$ for player 1 and $\beta$ be the probability of $Q$. Then indifference condition yields $$2\beta+4(1-\beta)=3\beta+1(1-\beta)$$ and $$5(1-\alpha)=2\alpha.$$ Then you know what the mixed Nash is.

There are of course two pure Nash $(K,R)$ and $(L,Q)$.

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  • $\begingroup$ How you find the equilibrium using alpha and bita value, describe this please. $\endgroup$
    – alhelal
    Apr 1, 2019 at 10:26

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