Complete question:
Consider the game of chicken. Two players drive their cars down the center of the road directly at each other. Each player chooses SWERVE or STAY. Staying wins you the admiration of your peers (a big payoff) only if the other player swerves. Swerving loses face if the other player stays. However, clearly, the worst output is for both players to stay! Specifically, consider the following payouts. Player two Stay swervePlayer one stay -6 -6 2 -2 swerve -2 2 1 1
a) Does either player have a dominant strategy?
b) Suppose that Player B has adopted the strategy of Staying 1/5 of the time and swerving 4/5 of the time. Show that Player A is indifferent between swerving
and staying.
c) If both player A and Player B use this probability mix, what is the chance that they crash?
Explanation:
a. There is no dominant strategy for either player. Suppose two players agree to live. Then the best answer for the player is to swerve(-6 versus -2). Yet if the player turns two, the player will remain one (2 vs 1).
b. Player B must be shown to be indifferent among swerving and staying if it implements a policy (stay= 1⁄4, swerving= 5/4).
When we quantify a predicted award on the stay / swerving of Player A, we get
E(stay)= (1/5)(-6)+ (4/5)(2)= 2/5 E(swerve)= (1/5)(-2)
c. They both remain 1/5 of the time. The risk of a crash (rest, stay) is therefore (1/5)(1/5)= 1/25= 4%
