# Nash Equilibria for zero-sum games (Rock Paper Scissors)

I'm trying to figure out a nash equilibria strategy for rock paper scissors and when the strategy would not be optimal. I know it's a zero sum game and I must use a mixed strategy but the practice question i am trying to solve says 'describe how a nash eq. strategy can be found?'.

I understand the concept of nash equilibrium's zero sum games yet I can't find any examples or practice problems with solutions that may lead me to what I'm supposed to answer. Also how would this strategy not be optimal ?

Would it be not optimal if a pure strategy is used instead?

Also, would a strategy for a zero sum game be unique ? (considering 2 players only, i guess) How would I go about proving that?

Thanks,

Barry

-

The basic approach is to calculate the formula for expected value and maximize it by any tools you know (one usually uses linear programming).

If your adversary would play 'scissors' only, there is very specific pure optimal strategy (the mixed one would not be optimal).

In rock-paper-scissors a pure strategy would be rarely optimal, against unknown enemy the only optimal strategy is mixed $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.

It does not need to be unique, consider a game with payoff matrix being just zero everywhere (I know this is a silly game, still it is a zero-sum game, right?). Any strategy would be optimal there ;-)

Hope it helps ;-)

-
That helps ! Thanks a lot ! – penguinsource Dec 9 '12 at 22:58

A fundamental fact about mixed strategy equilibria is that a player only mixes between two (or more) strategies if she is indifferent between the two (or more) strategies. So if you know which strategies are played with positive probability (the support), you can use this condition to constrain the behavior of the other player. This can be quite complicated. But thinking of matching pennies, it is clear that the opponent has to mix with a probability of $0.5$ and $0.5$ to make a player indifferent. In two player games, this gives you linear equations. For games with more players, you get polynomial equation and things can get very difficult.

Also, if there is a profitable deviation, there is a pure strategy profitable deviation. In principle, you can find all Nash equilibria by checking these conditions for every profitable support and then test the equilibria against pure strategy deviations.

For games like matching pennies or rock-paper-scissors, one can also make ones live easier by looking at symmetric equilibria. Nash has shown in his thesis that every (finite) symmetric game has a symmetric equilibrium, but making precise what symetric means takes some group theory.

Lastly, equilibria may not be unique in zero-sum games. An easy example is given by a game in which a player has a redundant strategy, a strategy that gives exactly the same payoff as another strategy, no matter what the other player does. If such a strategy is played in equilibrium, one can mix between the redundant strategies without changing the payoffs. But even if there is more than one equilibrium in a zero-sum game, all equilibria will have the same expected payoff. This is a consequence of von Neumann's minmax theorem.

-
thanks, this helped toooo !! – penguinsource Dec 9 '12 at 23:06
So how do I keep my winning records in rock-paper-scissors? I gots to know, man! – Asaf Karagila Jun 2 '13 at 21:59