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I have a simple payoff matrix defined here: https://prnt.sc/m6rp5m
My question is: If both players play all 4 strategies with 1/4 probability, does that lead to nash equilibrium?
I can't quite figure this out. I know how to check if pure stratigies lead to nash equilibrium. You assume the row/columns player sticks to a strategy and check if the other player is incentivized to play differently.
I'd love a pointer or two in the right direction.
In mixed strategies NE, you choose to act (based on that you know the payoffs of the second player) based on the probability that makes the other player indifferent. Still the strategy is not defined as a typical pure strategies case, rather than you mention that A will randomize with probability x and B will randomize(between his actions) with probability y.
Try to read Felix Munoz Garcia game theory book, it very good with examples and strays from scratch to advanced.
No, it's not an equilibrium.
If the column player uses the uniform mixed strategy $$\sigma_C = (1/4, 1/4, 1/4, 1/4),$$ then the four actions available to the row player have the values $-1/4$, $+1/4$, $-1/4$, and $+1/4$, respectively. These values are not identical, so the uniform strategy $$\sigma_R = (1/4, 1/4, 1/4, 1/4)$$ is not a best response for the row player. Specifically, a strategy $\sigma_R$ can only by a best response to $\sigma_C = (1/4, 1/4, 1/4, 1/4)$ if it assigns probability 0 to the first and third row.
