# Pure and Mixed Nash Equilibrium algorithm gives different results

I have a game represented by following table:

It is clear that there is a pure Nash equilibrium at 4,2 (both players do not cooperate, player 1 awarded 4 points and player 2 awarded 2 points).

Now if I used mixed strategy Nash equlibrium algorithm I get a = 1 and b = -1 as the probability of player 1 plays Cooperate and player 2 plays Cooperate respectively. To my understanding, a and b should be equal to 0 so that both players should play the above pure equilibrium strategy.

Why is there such a difference in pure and mixed strategy?

Edit 1: mixed strategy Nash equlibrium algorithm:

• The expected ultility for player 2 playing cooperate is: a*2 + (1-a)*1 (This means that some of the time player 1 plays Cooperate, so the probability is a).
• The expected ultility for player 2 playing non-cooperate is: a*2 + (1-a)*2.
• By equating the two formuala I have (1-a) = 2*(1-a) so a = 1.
• Simmilarly, the expected ultility for player 1 playing cooperate is: 6*b + 3*(1 - b).
• The expected ultility for player 1 playing non-cooperate is: 8*b + 4*(1 - b).
• By equating the two formuala I have 0 = 2*b + (1-b) so b = -1.
• You should probably go into detail about your "mixed strategy Nash equlibrium algorithm". E.g. what do the starting equations mean, etc. – Giskard Jan 3 at 20:35
• @Giskard added to the post. Apologise for not making the question clear. – Leo Long Vu Jan 3 at 21:12

Mixed strategy Nash equilibrium cannot involve strictly dominated strategies. In particular, Cooperate is strictly dominated for player 1 ($$6<8$$ and $$3<4$$). Therefore, no $$b\in[0,1]$$ can make player 1 indifferent between Cooperate and Non-cooperate. You made a mistake by trying to solve for $$b$$ by equating player 1's expected payoffs from his two pure strategies.