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

*

*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.

Let $(x_1,x_2,x_3)$ be the first player's strategy (i.e., his probabilities of playing rock, paper, and scissors respectively), and let $(y_1,y_2,y_3)$ be the second player's strategy.  The expected payoff to the first player is
$$
P(x,y)=9(x_1y_3-x_3y_1)+3(x_2y_1-x_1y_2)+5(x_3y_2-x_2y_3).
$$
To constrain the probability sums to be $1$, we take $x_3=1-x_1-x_2$ and $y_3=1-y_1-y_2$.  So
$$
P(x,y)=9\left(x_1(1-y_1-y_2)-(1-x_1-x_2)y_1\right)+3(x_2y_1-x_1y_2)+5\left((1-x_1-x_2)y_2-x_2(1-y_1-y_2)\right) \\
=9(x_1-y_1)+ 17(x_2 y_1 -x_1y_2) + 5(y_2-x_2).
$$
The first derivatives are zero when
$$
\frac{\partial}{\partial x_1}P(x,y)= 9 -17y_2=0 \\
\frac{\partial}{\partial x_2}P(x,y)= 17y_1 -5=0 \\
\frac{\partial}{\partial y_1}P(x,y)=-9+17x_2 = 0\\
\frac{\partial}{\partial y_2}P(x,y)=-17x_1+5=0,
$$
or at $(x_1,x_2,x_3)=(y_1,y_2,y_3)=(5/17, 9/17, 3/17)$.  The Nash equilibrium is to play to beat each move with probability proportional to that move's reward.
To check that this is indeed a Nash equilibrium, suppose $y_1=5/17$ and $y_2=9/17$.  Then 
$$
P(x)=9(x_1-5/17)+17\left(x_2 (5/17)-x_1 (9/17)\right)-5(x_2-9/17)=0;
$$
that is, the first player's expected payoff is zero with any strategy.  So the first player cannot improve his payoff by changing his strategy unilaterally, and by symmetry, neither can the second player; this is the definition of a Nash equilibrium.
