# 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

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.