|Game Theory and Business Strategy.|
Chapter 5 Problems 4 and 5 (1st Edition)
Chapter 8 Problem 3 (2nd Edition)
There is no pure-strategy Nash equilibrium here, hence the search for an equilibrium in mixed strategies.
If the row player plays "up" with probability p, and the column player plays "left" with probability q, the Nash equilibrium is (p=2/3, q=1/2).
Row's p-mix (probability p on Up) must keep Column indifferent so must satisfy:
16p + 20(1 - p) = 6p + 40(1 - p)
this yields p = 2/3 = 0.67 and (1 - p) = 0.33.
Column's expected payoff = 17.33.
Similarly, Column's q-mix (probability q on Left) must keep Row indifferent so must satisfy
q + 4(1 - q) = 2q + 3(1 - q)
correct q here is 0.5. Row's expected payoff is 2.5
That this is an equilibrium can be confirmed by checking that for each player, the expected payoff from using each of her two strategies is the same.
(b) Note that it is difficult for the players to coordinate on (Down,Right) even though both would be better off. The reason is that, if the row player believed that the column player was likely to play right with high probability, the row player would prefer to play up. However, this cooperation may be possible if the game between these players were to be repeated (opening the way to punishment) or if guidelines for social conduct were such that players gravitated toward the outcome that maximized total payoff.
False. When a game has several Nash equilibria in pure strategies, it may also have a mixed strategy equilibrium.
Unfortunately, a mixed strategy equilibrium not only mixes between the equilibria, but also between all of the non-equilibrium outcomes. Consider the natural monopoly market entry example in class, in which the mixed strategy equilibrium resulted in expected payoffs of 0 for each firm. This is because, by playing the mixed strategy equilibrium, there is good chance that neither firm will enter, and a chance that both firms will enter. Ideally, the firms should flip the same coin, so that they do mix only between the equilibria.