Rock-Paper-Scissors Without Rock

Key idea. Discard your dominated move, then make each player indifferent in the $2\times 2$ game that remains.

Your opponent chooses only between paper (P) and scissors (S). Your payoffs are:

Paper is dominated. Against each of the opponent’s moves, scissors pays you exactly $1$ more than paper. So you never play paper, and it remains to solve the game

No saddle point.

• Your row minima are $-1$ and $0$, so the maximin over pure strategies is $0$.
• The column maxima are $1$ and $1$, so the minimax is $1$.

These differ, so the optimal play is mixed.

Your mix. Play rock with probability $a$. Your expected payoff is:

• against paper: $-a+(1-a)=1-2a$;
• against scissors: $a$.

Indifference gives $1-2a=a$, so $a=\tfrac13$. You then guarantee $\tfrac13$ whatever the opponent does.

Opponent’s mix. Let the opponent play paper with probability $b$. Your expected payoff is:

• from rock: $-b+(1-b)=1-2b$;
• from scissors: $b$.

Indifference gives $b=\tfrac13$. The opponent then holds you to $\tfrac13$.

Paper earns you only $0\cdot\tfrac13-1\cdot\tfrac23=-\tfrac23$ against this mix, so it is still not worth playing.

Both players can secure $\tfrac13$, so this is the value of the game. You play rock with probability $\tfrac13$ and scissors with probability $\tfrac23$. The opponent plays paper with probability $\tfrac13$ and scissors with probability $\tfrac23$.