Zero sum game row elimination

Question

I have a zero sum game $$\begin{pmatrix} 2&0&2\\1&2&3\\4&1&2\end{pmatrix}$$ where player 1 selects the row and wants to maximize the utility and player selects columns and try to minimize the utility. After noticing that the third column is strictly dominated by the second one I can delete it. $$\begin{pmatrix} 2&0\\1&2\\4&1\end{pmatrix}$$ Now is it true that I can say that the first row is strictly dominated because there exists a convex combination between second and third row that is better than first row (i.e. $\frac12 1+\frac12 4>2; \frac12 2+\frac12 1>0$)?

Answer 1

We can preform Iterated elimination of strictly dominated strategies with pure and mixed strategies. Since there is a pure strategy we do not need to use mixed strategies. In pure strategies the payoff of player 1 choosing row 3 is always greater than row 1, $4 > 2$ and $1 > 0$, so row 1 can be eliminated.

If we were to use a mixed strategy, say player 1 plays row 2 with probability $\frac{1}{2}$ and row 3 with probability $\frac{1}{2}$. This gives an expected payoff, regardless of player 2's choice, that is greater than the payoff of row 1, $\frac{5}{2} > 2$ and $\frac{3}{2} > 0$, so row 1 can be eliminated. Either result leaves the game

$$ \begin{pmatrix} 1 & 2 \\ 4 & 1 \end{pmatrix} $$

and this game has no pure Nash Equilibrium.