Example 4 There are four rectangular containers $A, B, C, D$, with their base areas and heights being $a^{2}, a; a^{2}, b; b^{2}, a; b^{2}, b$ (where $a \neq b$). A game for two players is defined with the rule that: each person, in turn, takes two of the four containers, and the one with more water wins. Does the first player have a sure-win strategy? If so, how many are there?
Analysis From the problem, it is easy to know that the volumes of the four containers $A, B, C, D$ are $a^{3}, a^{2} b, a b^{2}, b^{3}$, respectively. Dividing the four containers into two groups only has three cases: $A, B$ and $C, D; A, C$ and $B, D; A, D$ and $B, C$. Therefore, the essence of the problem is to compare the sizes of the sums of the volumes in pairs.
Solution (1) If $A, B$ are taken first, then the other can only take $C, D$,
$$
\left(a^{3}+a^{2} b\right)-\left(a b^{2}+b^{3}\right)=(a-b)(a+b)^{2} .
$$

Obviously, $a+b>0,(a+b)^{2}>0$, but the sizes of $a$ and $b$ are uncertain, so the sign of $(a-b)(a+b)^{2}$ cannot be determined, i.e., the size of $a^{3}+a^{2} b$ and $a b^{2}+b^{3}$ is uncertain. This method does not guarantee a win.
(2) If $A, C$ are taken first, then the other can only take $B, D$,
$$
\left(a^{3}+a b^{2}\right)-\left(a^{2} b+b^{3}\right)=(a-b)\left(a^{2}+b^{2}\right) .
$$

Similar to the analysis in (1), it is known that this method also does not guarantee a win.
(3) If $A, D$ are taken first, then the other can only take $B, C$,
$$
\left(a^{3}+b^{3}\right)-\left(a^{2} b+a b^{2}\right)=(a-b)^{2}(a+b) .
$$

Given $a \neq b$, so $(a-b)^{2}(a+b)>0$, i.e., $a^{3}+b^{3}>a^{2} b+a b^{2}$.
Therefore, taking $A, D$ first is the only guaranteed winning strategy.