3x3 Nash Equilibrium? I'm trying to figure out a Nash Equilibrium for a 3x3 zero-sum game, and it's not following normal patterns (or I'm making a huge oversight, in which case I'll feel stupid!). Can anyone help me?
The payoff matrix for P1 is (additive inverses for P2):
0.0 0.0 1.0
1.5 3.0 -0.5
-1.5 -2.0 1.5
As far as I can tell, nothing is dominated for either player. Doing the usual calculations where you find probabilities each player makes each play such that the other player is then indifferent to his plays yields negative probabilities though...not sure what's wrong with what I'm doing.
Thanks in advance!
Edit: Some more thinking has led me to believe that I don't think I'm doing this wrong, and that there's a reason I wasn't explicitly taught to do this. It seems to be equivalent to a LP problem in the 3x3 case (and in the general nxn case) where no strategy is strictly dominated, and where there's no pure strategy equilibrium. My confusion arose from the fact that I know a Nash equilibrium is guaranteed to exist -- I guess I was taking that to mean that I should be able to calculate one easily. :)
 A: The Nash equilibrium I got was:
p[1] == 1/2 && p[2] == 1/2 && p[3] == 0 && q[1] == 3/4 && q[2] == 0 &&
  q[3] == 1/4
where p is the vector of probabilities of player 1 and q is of player 2.
As the previous answer pointed some strategies have zero probabilities. You equate payoffs only of strategies that are played with positive probability.
If you want an automatic way of computing equilibria of games like this, you may want to check this link (the code is in Mathematica):
http://www.mathematica-journal.com/2014/03/using-reduce-to-compute-nash-equilibria/
If you want to use pencil and paper. First guess what strategies have positive prob and which have zero prob. Second compute the expected payoff of all strategies. Check all strategies that have positive prob. have the same payoff and that this payoff is no less than the payoff of a strategy with zero prob. If no results, try another guess. It is a force brute method. For each player you have lots of cases to consider: $2^n-n-1$ where $n$ is the number of strategies.
A: There exists a mixed-strategy Nash equilibrium, but some strategies have $0$ weight in that equilibrium.
