Consider a game with player P1 having choices A, B, and C, and player P2 having choices X, Y, and Z, and the associated 3x3 payoff matrix.
I understand how to find the pure Nash equilibria, if they exist (any cell where neither player can get a better payoff by changing their choice).
I understand how to find a mixed Nash equilibrium in the case where all choices on both sides have nonzero probability - by assuming that one player is indifferent between their choices (same utility for all of them), that gives you the equations for the probabilities of the other player's choices, if a solution exists with all probabilities between 0 and 1.
And I understand that you can find more mixed Nash equilibria by assuming that for each player, one of their three choices - now the utility of that choice doesn't have to be equal to the utility of the other two choices, it can be less - and see if you get a new solution that satisfies those constraints too. Correct so far?
What I don't understand is how to determine which possibilities you have to check for which choices can have probability zero. P1 and P2 each have 3 choices for an option that they could play with probability zero; do I have to check all 9 combinations? Or is there a way to narrow them down?
And if each player has 4 or more possible choices instead of just 3, would the number of "zero-probability" combinations grow exponentially, because you don't just have to check the case of playing each individual choice with probability zero, you also have to check the possibility of playing each subset of choices with probability zero?