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Lets say I have following problem:

Zero sum game. Payoff matrix for player one:

-1  4  4

 2  6 -2

I start by writing equations for each strategy.

E1 = -q1 + 4q2 + 4(1 - q1 – q2)
E2 = 2q1 + 6q2 - 2(1 - q1 – q2)

E1 = E2
9q1  + 8q2 = 6

q1 = (6- 8q2)/9
q2 = (6-9q1)/8

But where do I go from here?

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The middle column is strictly dominated by the left column, since $-1 < 4$ and $2 < 6$. Having eliminated the middle column, one is left with the $2 \times 2$ game:

$$ \left[ \begin{array}{ c c } -1 & 4 \\ 2 & -2 \end{array} \right] $$

Given the cyclic preference structure $$A(2,1) = 2 > A(1,1) = -1 < A(1,2) 4 > A(2,2) = -2 < 2 = A(2,1),$$ there is a unique completely mixed equilibrium:

$[Pr(Top),Pr(Bottom)] = [4/9,5/9]$; $[Pr(Left),Pr(Right)] = [2/3,1/3]$.

The value of the game (payoff to player 1) is $-2/3$.

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