MaxMin: how much does the min "see"? Consider the following quantity:
$$ \max_{a \in \{-1,1\}}\left( \min_{b \in \{-1,0,1\}} ab\right).$$
Since the min is inside, we apply it first, but what value $b$ will be chosen? If the minimum "knew" that there was a maximum waiting for it outside, it would choose $b=0$ because the max gets to pick next and can counteract any nonzero value $b$ takes to make the entire quantity positive. But the min isn't "supposed to see" the outside (or it would be equivalent to taking the min of the max, rather than the max of the min). However, the minimum does "see" the $a$ in the term it's minimizing: how does the min "deal" with this term, without "knowing" the values that $a$ can take?
I apologize for the somewhat lacking tagging and title; I honestly don't know how to refer to this problem.
 A: If you haven't yet got the hang of multiple levels of abstraction, here's a very explicit way to think about it. Remember that $$\max\limits_{a\in\{a_1,a_2,\ldots\}}({\sf whatever})$$ is just the maximum over the set $$\{{\sf whatever}\text{ when }a=a_1, {\sf whatever}\text{ when }a=a_2, {\sf whatever}\text{ when }a=a_3, \ldots\}.$$ So, unpacking your big expression, we have
$$\begin{array}{cccccc}
\max_{\color{red}{a\in\{-1,1\}}}\bigg(\color{blue}{\min_{b\in\{-1,0,1\}}ab}\bigg) & = & \max \bigg\{ & \color{blue}{\min_{b\in\{-1,0,1\}}ab}\text{ when }\color{red}{a=-1}, & \color{blue}{\min_{b\in\{-1,0,1\}}ab}\text{ when }\color{red}{a=1} & \bigg\} \\
& = & \max\bigg\{ & \min_{b\in\{-1,0,1\}}-b, & \min_{b\in\{-1,0,1\}}b & \bigg\} \\
& = & \max\big\{ & \min\{1,0,-1\}, & \min\{-1,0,1\} & \big\} \\
& = & \max\,\{ & -1, & -1 & \} \\
& = & -1.
\end{array}$$
A: Note that $\displaystyle{m(a)=\min_{b \in \{-1,0,1\}} ab}$ is a function of $a$ and the min does not know what is "outside" (the domain of $m$). 
Here, we have $m(a)=a$ when $a<0$, $m(0)=0$ and $m(a)=-a$ when $a>0$, so $m(a)=-|a|$.
Hence, $\displaystyle{\max_{a \in \{-1,1\}}\left( \min_{b \in \{-1,0,1\}} ab\right)=\max_{a \in \{-1,1\}} (-|a|)=-1}$. 
