Return-to-variance gap in a two-marble sibling game

Alice and Bob each hold one red and one blue marble, and each reveals one of theirs chosen uniformly at random. Two blues pay Alice $1, two reds pay Alice $3, and mismatched colours pay Bob $2; all money comes from a third party, so nobody pays anything. Let $A_r$ and $B_r$ be expected payoff divided by payoff variance for Alice and Bob respectively. Compute $B_r-A_r$.

$1/3$. Alice gets $1$, $3$, $0$ with probabilities $\tfrac14,\tfrac14,\tfrac12$: mean $1$, variance $\tfrac32$, so $A_r=\tfrac23$. Bob gets $2$ or $0$ with probability $\tfrac12$ each: mean $1$, variance $1$, so $B_r=1$. Same expected payoff, but Alice bears more risk.