Whether gladiator matchup order matters for Bob

Alice and Bob each field four gladiators: Alice's have strengths $1,2,3,4$ and Bob's have strengths $4,5,9,12$. Bouts are one-on-one to the death, with no draws; a gladiator of strength $x$ beats one of strength $y$ with probability $\tfrac{x}{x+y}$, and a survivor keeps the same strength. Before each bout Alice names one of her remaining gladiators, then Bob names one of his to face it. Whoever still has a gladiator when the other side runs out wins. Is there a strategy for Bob that gives him a strictly higher winning probability than every other strategy? Answer $1$ for yes, $0$ for no.

$0$ (no). Model a gladiator of strength $x$ as a lamp with an exponential lifetime of mean $x$; a bout is lost by whichever lamp dies first, giving exactly $\tfrac{x}{x+y}$, and memorylessness preserves strength. The winner is then simply the side with the larger total lamp lifetime, which does not depend on matchup order, so every strategy gives Bob the same chance.