Rock-paper-scissors without rock

Idea. First drop the move that is dominated, then solve the remaining 2 by 2 game by making each player indifferent between their two moves.

Step 1: drop paper. Your opponent plays paper or scissors. Your paper ties against paper and loses to scissors. Your scissors beats paper and ties against scissors. So scissors is always at least as good as paper, and you never need paper.

Step 2: the 2 by 2 game. Your gains, with your move on the left and your opponent's on top: $$\begin{array}{rcc}& \text{paper}& \text{scissors}\\ \text{rock}& -1& +1\\ \text{scissors}& +1& 0\end{array}$$ No pure choice works for either side, so both mix.

Step 3: your mix. Play rock with probability $x$. Against paper you gain $-x+(1-x)=1-2x$; against scissors you gain $x$. Make them equal: $1-2x=x$, so $x=\frac{1}{3}$, and you gain $\frac{1}{3}$ whatever the opponent does.

Step 4: the opponent's mix. If the opponent plays paper with probability $y$, your rock gains $1-2y$ and your scissors gains $y$. With $y=\frac{1}{3}$ both give $\frac{1}{3}$, so you cannot get more than $\frac{1}{3}$.

So the value is $\frac{1}{3}$ dollar per round: play rock with probability $\frac{1}{3}$ and scissors with probability $\frac{2}{3}$. Your opponent plays paper with probability $\frac{1}{3}$ and scissors with probability $\frac{2}{3}$.

Note. You still play rock a third of the time: it is your only move that beats scissors, and without it your opponent would play scissors every time.