Range for a Strictly determinable game

Question

The question was to find the range of $p$ so that the following game is strictly determinable.
I calculated the row minima and column maxima ignoring $p$, since we don't know the value of $p$ and obtained $\text{minimax}=-2$ and $\text{maximin}=3$ (though these are not the original maximin and minimax).

I plugged in three values of $p$ for the cases $p<-2$, $-2\le p\le 3$ and $p>3$ and of them $-2\le p\le 3$ is satisfied.

But how do I find this condition mathematical reasonings rather than any trial and error method?

Answer 1

Let $\operatorname{minimax}(p)$ and $\operatorname{maximin}(p)$ denote the minimax and maximin as functions of $p$. Some things to note:

• These functions are monotonic (since $\min$ and $\max$ are monotonic in both arguments).
• Ignoring $p$ is equivalent to substituting $-\infty$ for it in the minimax and $\infty$ in the maximin, so $ \operatorname{minimax}(-\infty)=-2$ and $\operatorname{maximin}(\infty)=3$.
• The value of these functions is either $p$ or the value you get by ignoring $p$ (since all other values in the matrix are dominated by that value).
• Since $p$ occurs in each row and column, we have $\operatorname{maximin}(p)\le p\le\operatorname{minimax}(p)$.

Together, these facts imply that $\operatorname{minimax}(p)=\max(p,-2)$ and $\operatorname{maximin}(p)=\min(p,3)$, and $\max(p,-2)=\min(p,3)$ holds iff $-2\le p\le3$.