# Mixed Strategies — Does this lead to Nash equilibrium?

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.

• Simple: check whether each player is indifferent between all of their pure strategies. See oyc.yale.edu/sites/default/files/… – afreelunch Jan 14 at 0:43
• Just as in the case with pure strategy NE, you assume the row/column player sticks to playing all four actions with probability 1/4, and see if the other player is incentivized to play something other than randomizing over the four actions with equal probability. – Herr K. Jan 14 at 15:28

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. 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.