VSE - Introduction to Game Theory Problem set #1 - Due Tuesday, March 16, Suggested Solution

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1 VSE - Introduction to Game Theory Problem set #1 - Due Tuesday, March 16, Suggested Solution Problem 1: Two people enter a bus. Two adjacent cramped seats are free. Each person must decide whether to sit or stand. Sitting alone is more comfortable than sitting next to the other person, which is more comfortable than standing. (a) Suppose that each person cares only about her own comfort. Model the situation as a strategic game. Find its Nash equilibrium/equilibria (b) Suppose that each person is altruistic, ranking the outcomes according to the other person s comfort, and, out of politeness, prefers to stand than to sit if the other person stands. Model the situation as a strategic game. Find its Nash equilibrium/equilibria (c) Compare the people s comfort in the equilibria of the two games. Solution: 1 \ 2 Sit Stand (a) Sit 2,2 3,1 Stand 1,3 1,1 (b) 1 \ 2 Sit Stand Sit 2,2 0,3 Stand 3,0 1,1 Underlined numbers are best responses of both players. In the game (a) the only Nash equilibrium is (Sit,Sit). In the game (b) the only Nash equilibrium is (Stand,Stand). (c) These two people are more comfortable in case (a) where they both sit. In part (b) they are altruistic, care about the comfort of the other person as well and they both end up standing. Problem 2: Find all the mixed strategy Nash equilibria of the strategic games below. Sketch best response functions of both players (similar graphs to those during the third lecture). (a) T 6,0 0,6 (b) T 0,1 0,2 B 3,2 6,0 B 2,2 0,1 Solution: (a) There are no Nash equilibria in pure strategies in this game. To find mixed strategy Nash equilibrium we use the following notation: 1

2 player 1 plays Top with probability p and Bottom with probability 1 p player 2 plays Left with probability q and Right with probability 1 q For player 1 to mix his strategy, the second player has to choose his mixing strategy such that the first player s payoff from playing Top (q 6 + (1 q) 0) is the same as payoff from playing Bottom (q 3 + (1 q) 6). If one of these payoffs was higher, there wouldn t be any reason for mixing: q 6 + (1 q) 0 = q 3 + (1 q) 6 9q = 6 q = 6 9 = 2 3 Similarly, for player 2 to mix his strategy, the payoff from playing Left (p 0 + (1 p) 2) has to be the same as payoff from playing Right (p 6 + (1 p) 0). If one of these payoffs was higher, there wouldn t be any reason for mixing: p 0 + (1 p) 2 = p 6 + (1 p) 0 8p = 2 p = 2 8 = 1 4 Hence, the only mixed strategy Nash equilibrium of this game is: {( 1, ) ( 3 4 4, 2, 1 3 3)} (b) This game has two Nash equilibria in pure strategies (T,R) and (B,L). Now, we look for mixed strategies equilibria: For player 1 to mix his strategy, the payoff from playing Top (q 0 + (1 q) 0) has to be the same as payoff from playing Bottom (q 2 + (1 q) 0). If one of these payoffs was higher, there wouldn t be any reason for mixing: q 0 + (1 q) 0 = q 2 + (1 q) 0 2q = 0 q = 0 So the second player always chooses Right (if he played Left with positive probability, player 1 would never mix, only play Bottom). Now the first player has to mix in such a way that he makes Right more attractive for the second player than Left. 2

3 Therefore, the payoff from playing Left (p 1 + (1 p) 2) has to be smaller or equal than payoff from playing Right (p 2 + (1 p) 1). p 1 + (1 p) 2 p 2 + (1 p) 1 2p 1 p 1 2 Hence, there are infinitely many mixed strategy Nash equilibriua of this game: {(p, 1 p), (0, 1)}, where p 1 2. Problem 3: Consider the two-player game with vnm preferences in which the players preferences over deterministic action profiles are the same as in Hawk-Dove and their preferences over lotteries satisfy the following two conditions. First, each player is indifferent between the outcome (Dove, Dove) and the lottery that assigns probability 1 to (Hawk, 2 Hawk) and probability 1 to the outcome in which she plays Hawk and the other player 2 plays Dove. Second, each player is indifferent between the outcome in which she plays Dove and the other player plays Hawk and the lottery that assigns probability 2 to the 3 outcome (Hawk, Hawk) and probability 1 to the outcome (Dove, Dove). Find payoffs 3 whose expected values represent these preferences (take each players payoff to (Hawk, Hawk) to be 0 and each players payoff to the outcome in which she plays Dove and the other player plays Hawk to be 1). Find the mixed strategy Nash equilibrium of the resulting strategic game. Solution: Denote by u i a payoff function whose expected value represents player i s preferences. The conditions in the problem imply that for player 1 we have: u 1 (Dove, Dove) = 1 2 u 1(Hawk, Hawk) u 1(Hawk, Dove) 3

4 u 2 (Dove, Dove) = 1 2 u 2(Hawk, Hawk) u 2(Dove, Hawk) u 1 (Dove, Hawk) = 2 3 u 1(Hawk, Hawk) u 1(Dove, Dove) u 2 (Hawk, Dove) = 2 3 u 2(Hawk, Hawk) u 2(Dove, Dove) Given that u 1 (H, H) = u 2 (H, H) = 0 and u 1 (D, H) = u 2 (H, D) = 1 we have: u 1 (Dove, Dove) = 1 2 u 1(Hawk, Dove) 1 = 1 3 u 1(Dove, Dove) Therefore: u 1 (Dove, Dove) = 3, u 1 (Hawk, Dove) = 6 u 2 (Dove, Dove) = 3, u 2 (Dove, Hawk) = 6 Therefore, this game is represented by the following payoff table: 1 \ 2 Hawk Dove Hawk 0,0 6,1 Dove 1,6 3,3 There are two Nash equilibria in pure strategies in this game (Hawk,Dove) and (Dove,Hawk). To find mixed strategy Nash equilibrium we use the following notation: player 1 plays Hawk with probability p and Dove with probability 1 p player 2 plays Hawk with probability q and Dove with probability 1 q For player 1 to mix his strategy, the payoff from playing Hawk (q 0 + (1 q) 6) has to be the same as payoff from playing Dove (q 1 + (1 q) 3). If one of these payoffs was higher, there wouldn t be any reason for mixing: q 0 + (1 q) 6 = q 1 + (1 q) 3 4q = 3 q = 3 4 Similarly, for player 2 to mix his strategy, the payoff from playing Hawk (p 0 + (1 p) 6) has to be the same as payoff from playing Dove (p 1 + (1 p) 3). If one of these payoffs was higher, there wouldn t be any reason for mixing: p 0 + (1 p) 6 = p 1 + (1 p) 3 p = 3 4 4

5 Hence, the only mixed strategy Nash equilibrium of this game is: {( 3, ) ( 1 4 4, 3, 1 4 4)} Problem 4: Find strictly dominated strategies using also mixed strategies and use iterative elimination of strictly dominated strategies to simplify the following game as much as possible. 1 \ 2 L C R T 3,1 0,5 8,10 M 8,5 0,2 1,0 B 4,3 2,1 5,5 Solution: We start with best responses and NE in pure strategies. Strategy that is best response to some of opponent s actions can never be eliminated. 1 \ 2 L C R T 3,1 0,5 8,10 M 8,5 0,2 1,0 B 4,3 2,1 5,5 For the firs player, all strategies (T, M, and B) are best responses to some of opponent s action. Only strategy that is never best response to other player s actions is C of the second player. This is our candidate for elimination. C is not strictly dominated by L or R only. So we are looking for such mixing of L and R that this mixed strategy strictly dominates C. If player 1 plays B then any combination of L and R strictly dominates B, because payoffs for L and R are both higher than that of C. So we need to concentrate on the remaining two options of the first player: T and M: pl + (1 p)r > C 5

6 T : p 1 + (1 p) 10 > 5 p < 5 9 M : p 5 + (1 p) 0 > 2 p > 2 5 If we pick a random number p from interval ( 5 9, 2 5) then mixed strategy pl+(1 p)r gives a higher payoff than strategy C. Hence, C is strictly dominated and we can eliminate this strategy. For example, for p = 1, we have: 1L + 1 R gives payoffs: (5.5, 2.5, 4) while C yields payoffs (5,2,1). After the elimination of C our table representing the game is as follows: T 3,1 8,10 M 8,5 1,0 B 4,3 5,5 Now, the only strategy that is never best response to any opponent s actions is B of the first player. No pure strategy strictly dominates B, we are looking for such value of p that pt + (1 p)m > B pt + (1 p)m > B p 3 + (1 p) 8 > 4 p < 4 5 p 8 + (1 p) 1 > 5 p > 4 7 If we pick a random number p from interval ( 4 7, 4 5) then mixed strategy pt +(1 p)m gives a higher payoff than strategy B. Hence, B is strictly dominated and we can eliminate this strategy. For example, for p = 3, we have: 3T + 1 M gives payoffs: (4.25,6.25) while B yields payoffs (4,5). T 3,1 8,10 M 8,5 1,0 Now, each action of each player is sometimes a best response and hence no more strategies can be eliminated. 6