Suppose we have a game, played in which Alice and Bob play mixed strategies: (Sorry about the spacing, but I don't know how to put a table or tab spacing in this text box.)
Alice plays Dove with probability p and Hawk with probability (1-p)
Bob plays Dove with probability q and Hawk with probability (1-q)
Payoffs are, with Alice being the first coordinate, and Bob the second:
Dove, Dove: (2,3)
Dove, Hawk: (3,4)
Hawk, Dove (5,6)
Hawk, Hawk (7,8)
To figure out p and q for a Nash equilibrium, which one of the following reasoning procedures is correct:
First method:
Alice's payoff, if she plays Dove, is 2q + 4(1-q) = A
Alice's payoff, if she plays Hawk, is 6q + 8(1-q)= B
Bob's payoff, if he plays Dove, is 3p + 5(1-p) = C
Bob's payoff, if he plays Hawk, is 7p + 9(1-p)= D
For a Nash equilibrium, A=B & C=D, so we solve.....
Second method:
The probabilities of each move are
Dove, Dove: p*q
Dove, Hawk: p*(1-q)
Hawk, Dove: (1-p)*q
Hawk, Hawk (1-p)*(1-q)
so that Alice's payoff, if she plays Dove, is 2pq + 4p(1-q) = A
Alice's payoff, if she plays Hawk, is 6(1-p)q + 8(1-p)(1-q)= B
Bob's payoff, if he plays Dove, is 3pq + 5(1-p)q = C
Bob's payoff, if he plays Hawk, is 7p(1-q) + 9(1-p)(1-q)= D
For a Nash equilibrium, A=B & C=D, so we solve.....
Which method (or neither) is correct? Thanks for any indications.
Both players are trying to maximize their payoffs. In this game, switching from Dove to Hawk always increases your payoff. So the only Nash equilibrium is (Hawk,Hawk).
In order to have a $2 \times 2$ game with a mixed strategy equilibrium, Bob must be able to make Alice's expected payoffs equal, and Alice must be able to make Bob's expected payoffs equal. This leads to the first method (but the resulting $p$ and $q$ must be probabilities, therefore in the interval $[0,1]$).
The second method is just completely wrong. $2pq + 4p(1-q)$ is not ``Alice's payoff if she plays Dove'''.