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