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.