# Range for a Strictly determinable game

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?

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 $$-2\le p\le3$$.