I have an assignment in my game theory book which goes like this:
Consider the following matrix game: $$A=\begin{pmatrix}3&1&4&0\\1&2&0&5\end{pmatrix}$$ a) Determine all the maximin rows and minimax columns. What can you conclude from this about the value of the game?
b) Consider the six $2\times2$ games that can be obtained by choosing 2 columns of the above game as follows: $$A_1=\begin{pmatrix}3&1\\1&2\end{pmatrix},\quad A_2=\begin{pmatrix}3&4\\1&0\end{pmatrix},\quad A_3=\begin{pmatrix}3&0\\1&5\end{pmatrix},$$ $$A_4=\begin{pmatrix}1&4\\2&0\end{pmatrix} ,\quad A_5=\begin{pmatrix}1&0\\2&5\end{pmatrix},\quad A_6=\begin{pmatrix}4&0\\0&5\end{pmatrix}.$$
Determine all the values of these games. Which one must be equal to the value of $A$?
Now, a) is pretty simple, row $1,2$ are both maximin rows and the only minimax column is $2$. This means for the value of the game $v(A)$, that $1\leq v(A)\leq 2$.
For b) I find the following values: $v(A_1)=1\frac23, v(A_2)=3, v(A_3)=2\frac17, v(A_4)=1\frac35, v(A_5)=2, v(A_6)=2\frac29$. So this means that the values of $A_1, A_4$ and $A_5$ lie in the allowed range.
This is where I got stuck, so I calculated the value of the game via a different method and found that $v(A_4)$ is the correct answer. This got me to think about whether it would just be the minimum of these subgames. Hence, I tried the same for the matrices $\begin{pmatrix}1&1\\2&0\\0&2\end{pmatrix}$ (in which all subgames result in $1$, which is the value of the game) and $\begin{pmatrix}1&4&7\\5&3&0\end{pmatrix}$, but for the latter this strategy failed because the value of the game was lower than the minimum values of each of the subgames. (Which is what I expected to be a possibility.)
So this leaves me with the question: How can I answer question b without using the phrase "$A_4$ leads to the minimum expected value", or how do I prove that $v(A)\geq v(A_4)$ without first calculating $v(A)$?
