Let the three games in a series be called $G_1, G_2$ and $G_3$ respectively, and the probability of winning the game $x$ be denoted as $P(x)$
You can win the series if and only if: You win $G_2$ and you win atleast one of $\{G_1, G_3\}$.
$\large\boxed{P\left ( \substack{\text{winning}\\\text{the series}\\ \\{G_{1}G_{2}G_{3}}} \right ) = P(G_2) \times \Bigl ( P(G_1) + P(G_3) - P(G_1) \times P(G_3) \Bigr )\;}$
Let the probability of winning against player $A$ be $a$ and the probability of winning against player $B$ be $b$.
Then, $a < b.(A \text{ is a stronger player than} B,\text{ so probability of winning against } A$
$ \text{ is smaller compared to}\;B)$
Let $P(xyz)$ be the probability of winning the series in which the games played are against $x, y$ and $z$ in order.
- $P(AAB) = a(a+b-ab) = a^2 + ab - a^2b$
- $P(ABA) = b(a+a-aa) = 2ab - a^2b$
- $P(BAB) = a(b+b-b^2) = 2ab - ab^2$
- $P(BAA) = a(b+a-ba) = a^2 + ab - a^2b$
We can see that not all probabilities are equal, so option E is not correct.
We can also see that options A and D result in the same value, so they are not correct either.
Comparing option B and option C.
Since $a < b$ and $a,b\geq 0$, we have that $2ab - a^{2}b > 2ab - ab^{2}$
Hence, option B is the correct answer.