ch07_solutions_solved edit.doc

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16 Apr 2012
Solutions to Chapter 7 Exercises
S1. False. A player’s equilibrium mixture is devised in order to keep her opponent indifferent
among all of her (the opponent’s) possible mixed strategies; thus, a player’s equilibrium mixture
yields the opponent the same expected payoff against each of the player’s pure strategies. Note
that the statement will be true for zero-sum games, because when your opponent is indifferent in
such a game, it must also be true that you are indifferent as well.
S2 (a) The game most resembles an assurance game because the two Nash equiibria
occur when both players play the same move. In an assurance game, both players prefer to make
the same move, but there is also a preferred Nash equilibrium with higher payoffs for both
players. In this game, (Risky, Risky) is the preferred equilibrium because it has higher payoffs,
but there is a chance that the players will play the worse Nash equilibrium with lower payoffs.
Even worse, the players might not play an equilibrium at all. Without convergence of
expectations, these results can occur, and this is characteristic of an assurance game.
(b) The two pure-strategy Nash equilibria for this game are (Risky, Risky) and (Safe,
S3. (a) There is no pure-strategy Nash equilibrium here, hence the search for an
equilibrium in mixed strategies. Row’s p-mix (probability p on Up) must keep Column
indifferent and so must satisfy 16p + 20(1 – p) = 6p + 40(1 – p); this yields p = 2/3 = 0.67 and (1
– p) = 0.33. Similarly, Column’s q-mix (probability q on Left) must keep Row indifferent and so
must satisfy q + 4 (1 – q) = 2q + 3(1 – q); the correct q here is 0.5.
(b) Row’s expected payoff is 2.5. Column’s expected payoff = 17.33.
(c) Joint payoffs are larger when Row plays Down, but the highest possible payoff to
Row occurs when Row plays Up. Thus, in order to have a chance of getting 4, Row must play Up
occasionally. If the players could reach an agreement always to play Down and Right, both would
get higher expected payoffs than in the mixed-strategy equilibrium. This might be possible with
repetition of the game or if guidelines for social conduct were such that players gravitated toward
Solutions to Chapter 7 Solved Exercises 1 of 6
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the outcome that maximized total payoff.
S4. The two pure-strategy Nash equilibria are (Don’t Help, Help) and (Help, Don’t Help).
The mixed-strategy Nash equilibrium has the following equilibrium mixtures:
2p + 2(1 – p) = 3p + 0 p = 2/3
2q + 2(1 – q) = 3q + 0 q = 2/3
That is, each player helps two-thirds of the time and doesn’t help one-third of the time.
S5. (a) Evert will play DL the same as before, because her p-mix depends upon
Navratilova’s payoffs.
(b) 70p + 10(1 – p) = 40p + 80(1 – p) p = 7/10
70q + 40(1 – q)= 10q + 80(1 – q) q = 2/5
So the mixed-strategy Nash equilibrium occurs when Evert plays 7/10(DL) + 3/10(CC) and
Navratilova plays 2/5(DL) + 3/5(CC).
Evert’s expected payoff is 70(2/5) + 40(1 – 2/5) = 52.
(c) Compared with the previous game, Evert plays DL with the same proportion,
whereas Navratilova plays DL less, going from 3/5 to 2/5. Navratilova’s q-mix changes because
her mix is dependent on Evert’s payoffs. The changes in Evert’s payoffs lead to a change in
Navratilova’s q-mix. On the other hand, Evert’s p-mix doesn’t change because Navratilova’s
payoffs have remained unchanged.
S6. (a) 0p – 1(1 – p) = 1p – 10(1 – p) p = 9/10
0q – 1(1 – q) = 1q – 10(1 – q) q = 9/10
In the mixed-strategy Nash equilibrium James plays 9/10(Swerve) + 1/10(Straight), and Dean
plays 9/10 (Swerve) + 1/10(Straight). James and Dean play Straight less often than in the
previous game.
(b) James’ expected payoff = 9/10 – 10(1 – 9/10) = – 1/10
Dean’s expected payoff = 9/10 – 10(1 – 9/10) = – 1/10
(c) If James and Dean collude and play an even number of games where they
alternate between (Swerve, Straight) and (Straight, Swerve), their expected payoffs would be 0.
This is better than the mixed-strategy equilibrium, because their expected payoffs are
– 1/10.
Solutions to Chapter 7 Solved Exercises 2 of 6
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