Does Bob have a uniquely best gladiator ordering

Alice owns four gladiators of strengths $1,2,3,4$; Bob owns four of strengths $4,5,9,12$. Fights are one-on-one: in each round Alice first nominates one of her surviving gladiators, then Bob nominates one of his survivors to face it. A bout between strengths $x$ and $y$ is won by the strength-$x$ fighter with probability $\dfrac{x}{x+y}$ (no draws), the loser dies, and the winner keeps the same strength for future bouts. Whoever still has a gladiator when the other side runs out wins. Is there a strategy for Bob that gives him a strictly higher chance of winning than every other strategy? Answer $1$ for yes, $0$ for no.

$0$ (no). Model each gladiator of strength $x$ as a bulb with an exponential lifetime of mean $x$ that only burns while fighting; memorylessness reproduces the $\tfrac{x}{x+y}$ rule and constant strength. Bob wins exactly when his total burn time outlasts Alice's, which does not depend on the order of matchups, so every strategy gives the same probability.