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In Tamer Basar Noncooperative Game theory pg $33$ there is a $2 \times 3$ game (zero sum game)

$A = \begin{bmatrix} 1 & 3 & 0 \\ 6 & 2 & 7 \end{bmatrix}$

(each element is a cost, player 1 wants to minimize his cost, player 1 is the row player)

They obtained the mixed strategy for player 1 as $(y_1 =2/3, y_2 = 1/3)$

Then, column 3 was eliminated because it yielded a lower average cost for player 2 than the average expected cost

So we have

$A = \begin{bmatrix} 1 & 3 \\ 6 & 2 \end{bmatrix}$

To compute the mixed strategy NE (for player 2) I first obtain equations $z_1 + 3z_2 = 6z_1 + 2z_2$ $\Rightarrow -2z_1 + 3 = 4z_1 + 2$ since $z_2 = 1-z_1$,

so $z_1 = 1/6, $and $ z_2 = 5/6$ (my answer)

But the text says that $z_1 = 1/3, z_2 = 2/3$ is the correct answer

Can someone resolve this discrepancy and clarify how the mixed strategy for the reduced game related to the original game?

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You're right and the text is wrong, as also confirmed by this game solver.

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  • $\begingroup$ So I'm not crazy...thought I was going to pull a nash just about now $\endgroup$ – Carlos - the Mongoose - Danger Oct 16 '15 at 15:37

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