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.

  • $\begingroup$ Can you continuously modify your strategy based on your opponent's past plays? $\endgroup$ – Hugh May 21 '14 at 1:44
  • $\begingroup$ Sure. Although, I'm more imagining a circumstance where you are playing a single round against an opponent who knows your exact strategy. $\endgroup$ – Heptic May 21 '14 at 2:06

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.

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  • $\begingroup$ Great answer, and surprising conclusion (at least to me)! Really appreciate the effort you put into explaining it. $\endgroup$ – Heptic May 21 '14 at 3:53
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    $\begingroup$ No, "always play paper" doesn't beat this strategy. If you play paper, you will win with paper when I play rock (payoff +3 with probability $5/17$) and I will win with scissors when I play scissors (payoff -5 with probability $3/17$). Your expected payoff is $0$. $\endgroup$ – mjqxxxx May 21 '14 at 14:07

This table summarizes the possible outcomes in playing this game once: $$ \begin{array}{c|c|c|c|c} \text{Hero Plays} & \text{Villain Plays} & \text{Hero's Earnings} \\ \hline R & R & +\$0 \\ R & P & -\$3 \\ R & S & +\$9 \\ P & R & +\$3 \\ P & P & +\$0 \\ P & S & -\$5 \\ S & R & -\$9 \\ S & P & +\$5 \\ S & S & +\$0 \end{array} $$ Assume that all outcomes are equally likely. Letting $X$ be our earnings we can compute our expected earnings given which choice we make \begin{align*} \Bbb E(X|R) &= \frac{1}{3}\cdot (\$0)+\frac{1}{3}\cdot (-\$3)+\frac{1}{3}\cdot(\$9) \\ &= -\$1+\$3 \\ &= \$2 \\ \Bbb E(X|P) &= \frac{1}{3}\cdot (\$3)+\frac{1}{3}\cdot (\$0)+\frac{1}{3}\cdot(-\$5) \\ &= \$1-\$\frac{5}{3} \\ &= -\$\frac{2}{3} \\ \Bbb E(X|S) &= \frac{1}{3}\cdot (-\$9)+\frac{1}{3}\cdot (\$5)+\frac{1}{3}\cdot(\$0) \\ &= -\$3+\$\frac{5}{3} \\ &= -\$\frac{4}{3} \\ \end{align*} According to these computations, the only winning strategy is to play rock in which case the expected value is $\$2$.

Note: The above assumes that hero makes a choice against a villain that is playing randomly.

Edit: I missed the "Nash-Equilibrium" tag when I first read this question. You're probably looking for something a bit more sophisticated than this! I'll keep the answer posted in case others find these computations useful.

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  • $\begingroup$ @AwalGarg I don't understand your comment. $\endgroup$ – Brian Fitzpatrick May 21 '14 at 2:20
  • $\begingroup$ His comment means that the villain could reason likewise and decide to always play rock, assuming that the hero is playing randomly. "So I can clearly not choose the cup in front of me!" $\endgroup$ – mjqxxxx May 21 '14 at 2:57
  • $\begingroup$ @mjqxxxx I make this completely clear in my answer, conceding that this solution doesn't account for a thinking villain. Was this part left unread? $\endgroup$ – Brian Fitzpatrick May 21 '14 at 2:59
  • $\begingroup$ No, your answer is fine and explains your assumptions clearly. I was just explaining what he meant by "Likewise". $\endgroup$ – mjqxxxx May 21 '14 at 3:38
  • $\begingroup$ @BrianFitzpatrick Likewise what mjqxxxx said... $\endgroup$ – user3459110 May 21 '14 at 8:01

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.

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