Optimal Strategy for Rock Paper Scissors with different rewards Imagine Rock Paper Scissors, but where winning with a different hand gives a different reward.

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*If you win with Rock, you get \$9. Your opponent loses the \$9.


*If you win with Paper, you get \$3. Your opponent loses the \$3.


*If you win with Scissors, you get \$5. Your opponent loses the \$5.


*If you tie, you get $0
My first intuition would be that you should play Rock with a probability of 9/(9+3+5), Paper with 3/(9+3+5) and Scissors with 5/(9+3+5) however this seems wrong, as it doesn't take into consideration the risk you expose yourself to (if you play Paper, you have an upside of \$3 but a downside of \$5).
So I put the question to you, in such a game -- what is the ideal strategy.
Edit: By "ideal" strategy, I mean playing against an adversarial player who knows your strategy.

If "optimal" means Nash equilibrium (i.e. a state that is stable wrt. small perturbations of strategies), than it can be computed. If you assume that $x_1$ is the probability of first player to play Rock, $x_2$ his probability to play Scissors and $1-x_1-x_2$ his probability to play Paper, and similarly for $y_i$, then the Payoff of the first player is 
$$f(x_1, x_2, y_1, y_2) = x_1 (9y_2 - 3 (1-y_2-y_3)) + x_2 (-9 y_1 + 5(1-y_2-y_3)) + (1-x_1-x_2)(3y_1-5y_2)$$ 
or something like that. The condition on Nash is that all partial derivatives vanish; you can probably easily compute the probabilities and check, whether you guessed the right solutions (the solution should be unique in this case with $x_i$ and $y_i$ nonzero).
However, in different circumstances, optimal may meen different things; if they are good friends and know that it's a zero sum game, they can also play both Rock all the time.
