Opponent throws rock more than he wants to => You shouldn't throw scissors at all (assumption)
Let your optimal throw be according to probability { R: 1-x, P: x, S: 0 }
Opponent doesn't want to throw R at all in this case; he should respond by matching optimal probabilities on what he can choose: { R: 1/2, P: (1-x)/2, S: x/2 }
You continue to assume that you shouldn't throw any scissors; you respond by matching optimal probabilities on R and P: { R: x/(x+1), P: 1/(x+1), S: 0 }
Since your throw probabilities are optimal 1-x = x/(x+1) and x = 1/(x+1)
Guess what? x = 1/(x+1) is the definition of the golden ratio conjugate! (x = 0.618034) And it works in the first equation as well
Therefore your optimal throw is according to probabilities { R: 0.381966, P: 0.618034, S: 0 }
Therefore your opponent throws according to probabilities { R: .5, P: 0.190983, S: 0.309017 }
You will win (2x-x^2)/2 = 0.427051 of the time, lose (1-2x+2x^2)/2 = 0.263932 of the time, and draw (1-x^2)/2 = 0.309017 of the time. When draws are dropped you win 0.618034 (x) of the time.
I believe this is the optimal solution (but I still have to confirm the above assumption)