Economics 3012 Strategic Behavior Andy McLennan August 25, 2006 Lecture 5 Topics • Problem Set 4 • Examples of Mixed Nash Equilibrium • Finding All Equilibria 1 Problem Set 4 Exercise 118.2 Problem Description: • A voter receives: – 2 utils if her favorite candidate wins, – 1 util in the event of a tie, – 0 utils if her favorite candidate loses. • The cost of voting is c where 0 < c < 1. • Candidate A has k supporters. • Candidate B has m ≥ k supporters. We are looking for a mixed strategy Nash equilibrium in which: • every supporter of Candidate A votes with probability p; • k supporters of Candidate B vote with certainty; • the remaining m − k supporters of Candidate B abstain. 2 Analysis: • In general a voter is pivotal if her vote affects the outcome, either: – between a loss and and a tie or – between a tie and a win. • The two different ways of being pivotal each have value one. • Let p be the probability that a supporter of A votes. • The probability that a supporter of Candidate A is pivotal is just the probability pk−1 that the other k − 1 all are voting. • A supporter of candidate A is indifferent between voting and abstaining if the probability pk−1 of being pivotal is equal to the cost c of voting, so 1 p = ck−1 . 3 • For a supporters of Candidate B who is expected to vote the probability of being pivotal is the sum of: k – the probability pk = ck−1 that all k supporters of Candidate A vote; – the probability kpk−1(1 − p) that exactly k − 1 of them vote. • This is greater than the probability pk−1 = c that a particular group of k − 1 of supporters of A. – Thus all the supporters of Candidate B who are expected to vote strictly prefer to do so. • A supporter of Candidate B who is not expected to vote is pivotal with probability k 1 pk = ck−1 = c · ck−1 < c. – Such a person prefers not to vote. • The expected number of voters 1 pk + k = (ck−1 + 1)k increases with c. 4 Exercise 118.3 Problem Statement: • General A, with three divisions, and General B, with two divisions, each have to allocate their forces to two passes. • For General A a pure strategy is given by the number d ∈ {0, 1, 2, 3} of divisions 1 allocated to the first pass. • A pure strategy for General B is given by the number d ∈ {0, 1, 2} of divisions 2 allocated to the first pass. • General A wins if and only if d ≥ d and 1 2 3 − d ≥ 2 − d . 1 2 5 Analysis: • For General A: – d = 0 is weakly dominated by d = 1. 1 1 – d = 3 is weakly dominated by d = 2. 1 1 • There cannot be a Nash equilibrium in which General B assigns positive probability to d = 1: 2 – General A would never choose d = 0, 3, 1 so d = 1 would lose for sure. 2 – But General B can always get a positive expected payoff. • In any equilibrium General B assigns probability 1/2 to each of d = 0 and 2 d = 2, because if General B assigned 2 unequal probabilities: – General A would assign two divisions to whichever pass General B most frequently assigned two divisions to. – The best response to this would be for General B to assign both divisions to the other pass. 6 • In order for General B to be indifferent between d = 0 and d = 2, the sum of 2 2 probabilities assigned to d = 0 and d = 1 1 1 must be equal to the sum of probabilities assigned to d = 2 and d = 3. 1 1 • In order for General B to not prefer d = 1 2 is must be the case that the sum of the probabilities assigned to d = 0 and d = 3 1 1 must not be greater than 1/2. 7 Exercise 128.1 Problem Statement: • Two consumers simultaneously choose which store to go to. – Each trades with probability 1/2 if they both go to the same store. – Trading at price p results in a utility of 1 − p. – Not trading gives utility 0. • By symmetry it suffices to consider prices p , p with p ≤ p , so we can call the first 1 2 1 2 seller the Discount House and the second seller the Boutique. • We have the following payoff matrix: D B 2 2 D (1−p1 , 1−p1 ) (1 − p , 1 − p ) 1 1 2 2 2 . B (1 − p , 1 − p ) (1−p2 , 1−p2 ) 1 2 1 ! 2 2 8 Analysis: • It is certainly better to go to the Discount House if the other buyer goes to the Boutique. – Going to the Discount House will be a dominant strategy, and (D , D ) will be 1 2 the unique Nash equilibrium, if 1−p1 ≥ 1 − p , i.e., 2 2 p ≤ 2p − 1. 1 2 9 • If 1−p1 < 1 − p , then the game is a Battle 2 2 of the Sexes with: – two pure equilibria (D , B ) and 1 2 (B , D ) 1 2 – a mixed equilibrium. • Let q be the probability that Buyer 2 goes to the Discount House in the mixed equilibrium. Indifference for Buyer 1 implies that q 1−p1 +(1−q)(1−p ) = q(1−p )+(1−q)1−p2 . 1 2 2 2 • Solving this gives 1 − 2p + p 1 2 q = . 2 − p − p 1 2 • Since the game is symmetric with respect to interchange of the two buyers, this is also the probability that Buyer 1 goes to the Discount House. 10