Pure strategy- Saddle point in a generic matrix I am studding Game theory. I know that a game can be solved in pure strategy if it admits a saddle point and I know how to find the saddle point, when the matrix has values. Now I am trying to understand how can I show that a matrix 4x2 admits a saddle point for any value of a generic matrix: $\begin{pmatrix} a & b \\\ c & d\\\ a & d\\\ c & b \end{pmatrix} $ Can someone help me on this:

If $b \ge d$, you can eliminate the rows with $d$ in column $2$.

But then, once that's done, column $2$ has equal values, so one of the two remaining rows can be eliminated.

So now just one row, with $2$ columns.

But then one column can be eliminated.

So now just a $1\times 1$ game.

Thus, an optimal pure strategy is given by the row and column, respectively, in the original matrix, corresponding to the remaining cell in the $1\times 1$ matrix.

Since neither player can do better by any change, it must yield a saddle point.

The reasoning for the case $d \ge b$ is analogous.

The key idea is this: By removing a row or column which is weakly or strongly dominated, the new game still has at least one of the original pure-strategy saddle points if there was one originally, and can't add any new pure-strategy saddle points that weren't there originally.