Suppose we have a game, played in which Alice and Bob play mixed strategies:(adsbygoogle = window.adsbygoogle || []).push({});

(sorry for the dots, but I don't know how to put a table or tab spacing in this text box)

...........................................................................Bob

.............................................Dove, prob q...........................Hawk, prob (1-q)

.........Dove , prob p....................(2,3)...........................................(4,5)

Alice

..........Hawk, prob (1-p)...............(6,7)...........................................(8,9)

Where "prob" indicates the (unknown) probability that Alice or Bob will play that particular move, and a pair (a,b) means the payoff for Alice and Bob, respectively.

To figure out p and q for a Nash equilibrium, which one of the following reasoning procedures is correct:

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.....

................OR

The probabilities of each move are

.............................................................................Bob

....................................................Dove, prob q.....................Hawk, prob (1-q)

..........Dove , prob p...........................p*q...................................p*(1-q)

Alice

...........Hawk, prob (1-p)....................(1-p)*q...............................(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.

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# Game theory -- Nash equilibrium

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