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Psy 1010: Problem Set 1 Solutions Psy 1010: Problem Set 1 Solutions 1 Stag Hunt (a) The set of players is N = f1; 2g. The set of strategies for each player is fStag, Hareg. (b) We claim that the Nash equilibria of the game are (Stag, Stag) and (Hare, Hare) . These equilibria are highlighted in the payoff matrix below: Stag Hare Stag (2; 2) (0; 1) Hare (1; 0) (1; 1) To check this, we show that in each such case, neither player can do better by deviating. Indeed, if player 1 is playing Stag, then player 2 prefers to play Stag rather than Hare, since this yields payoff 2 rather than 1. For the same reason, if player 2 plays Stag, then player 1 cannot increase his payoff by deviating to Hare. Therefore, (Stag, Stag) is a Nash equilibrium. Likewise, (Hare, Hare) is also Nash. If player 1 is playing Hare, then player 2 has no incentive to deviate to Stag, since 1 > 0. By symmetry, if player 2 is playing Hare, then player 1 also prefers to play Hare. So we see that the pair (Hare, Hare) is also a Nash equilibrium. The (Stag, Stag) equilibrium is preferred by both players since the payoffs are both higher in this case. This is an example of a coordination game, as introduced in lecture. 2 Hawk-Dove (a) The set of players is N = fPlayer 1, Player 2g. The set of strategies is fHawk, Doveg for each player. In order to write the payoff matrix, we first consider each possible outcome. If both animals are aggressive v (as in the (Hawk, Hawk) strategy pair), then each animal receives payoff 2 −c because it wins the resource half of the time but always pays a cost c. If one animal is aggressive but the other is passive, then the Hawk receives payoff v because it wins the resource, while the Dove receives payoff 0 as it gets nothing. v Finally, if both animals are passive, then each gets payoff 2 because each obtains the resource half of the time. Collecting these into a payoff matrix, we find: 1 Hawk Dove v v Hawk 2 − c; 2 − c (v; 0) v v Dove (0; v) 2 ; 2 This indeed agrees with the table presented in lecture! To find the Nash equilibria, we recall that c > v, v so that the payoff 2 −c is negative. Thus in the (Hawk, Dove) strategy pair, player 1 cannot improve his v v payoff by deviating (since v > 2 ), nor can player 2 improve by deviating (since 2 − c < 0. By symmetry, the opposite strategy pair (Dove, Hawk) is also a Nash equilibrium. Therefore, the two equilibria are (Hawk, Dove) and (Dove, Hawk) , as indicated with highlighting in the table above. (b) In this new game, player 1 wins the resource in a fight with probability q. Therefore, the payoff to player 1 in the (Hawk, Hawk) scenario is qv − c, while the payoff to player 2 is (1 − q) − c. The other entries in the payoff matrix are the same as in part (a). This looks like: Hawk Dove Hawk (qv − c; (1 − q)v − c) (v; 0) v v Dove (0; v) 2 ; 2 We analyze this game as before. If player 1 plays Hawk, then player 2 prefers to play Dove (since 0 > −c). v If player 1 plays Dove, then player 2 prefers to play Hawk (since v > 2 ). On the other hand, if player 2 plays Hawk, then player 1 still prefers to play Dove, as 0 > qv − c. Finally, if player 2 plays Dove, then v player 1 will play Hawk, because v > 2 . Combining these results, we see that the two Nash equilibria are (Hawk, Dove) and (Dove, Hawk) . The equilibria of the game are unaffected by the likelihood of winning! c (c) We consider the specific example mentioned in the problem, where one animal pays cost − 8 for fighting 7c and the other animal pays cost − 8 . For concreteness, suppose that player 1 pays the lower cost. The payoff matrix becomes: Hawk Dove v c v 7c Hawk 2 − 8 ; 2 − 8 (v; 0) v v Dove (0; v) 2 ; 2 Now we see that the Nash equilibria depend on the particular value of c. The strategy pair (Hawk, Dove) is still always an equilibrium. If c ≥ 4v, then the pair (Dove, Hawk) is also an equilibrium. However, if v < c < 4v, there is only one Nash equilibrium. The Nash equilibria are therefore: ( (Hawk, Dove) j if c < 4v Nash equilibria = (Hawk, Dove) and (Dove, Hawk) j if c ≥ 4v (d) The payoff matrix will now have a third row and column representing the Bourgeois strategy, but the remaining submatrix will be the same as in part (a). We calculate the payoffs for the remaining cases. Say player 1 is playing Bourgeois and player 2 is playing Hawk. Half of the time, player 1 arrives first v v and plays Hawk, leading to payoffs of ( 2 − c; 2 − c). The other half of the time, player 1 arrives second 2 and plays Dove, yielding payoffs (0; v). Each of these occurs with equal probability, so the expected v c 3v c payoff is simply the average. We arrive at a payoff of ( 4 − 2 ; 4 − 2 ) in this case. Now suppose player 1 plays Bourgeois and player 2 plays Dove. We reason as above - half of the time v v player 1 plays Hawk, giving (v; 0), and the other half of the time he plays Dove, giving ( 2 ; 2 ). In 3v v expectation, we obtain payoffs ( 4 ; 4 ). Finally, say player 1 and player 2 each play Bourgeois. If player 1 arrives first, then player 1 plays Hawk and player 2 plays Dove, giving payoffs (v; 0). If player 2 arrives second, we instead get (0; v). Thus the v v expectation in this case is ( 2 ; 2 ). The other possible payoffs can be determined by symmetry, exchanging the roles of players 1 and 2 above. Collecting these into a matrix, Hawk Dove Bourgeois v v 3v c v c Hawk 2 − c; 2 − c (v; 0) ( 4 − 2 ; 4 − 2 ) v v v 3v Dove (0; v) 2 ; 2 ( 4 ; 4 ) v c 3v c 3v v v v Bourgeois ( 4 − 2 ; 4 − 2 ) ( 4 ; 4 ) ( 2 ; 2 ) The Nash equilibria in this case are (Hawk, Dove), (Dove, Hawk), and (Bourgeois, Bourgeois) , as high- lighted above. 3 War of Attrition v (a) Suppose player 1 is devoting t1 = 0 and player 2 is devoting t2 = c ; we will argue that neither player v can do better by unilaterally deviating. First hold player 2's strategy fixed at t2 = c . By playing t1 = 0, 0 0 v player 1 receives payoff 0. If player 1 deviated to playing t1 with 0 < t1 < c , then player 1 would receive 0 0 v v payoff −ct1, which is strictly worse than 0. If player 1 played t1 = c , then he would receive payoff − 2 , 0 v which is again worse than 0. Finally, if player 1 deviated to t1 > 2 , he would receive v − ct2 = 0, which is also not an improvement. Therefore, player 1 has no incentive to deviate. v Now we consider player 2's strategy, holding player 1's strategy fixed at t1 = 0. By playing t2 = c , 0 player 2 earns payoff v. If player 2 deviates to any other time t2 6= 0, we see that his payoff is still v. If 0 v player 2 deviates to t2 = 0, then he earns payoff 2 . Therefore player 2 also cannot improve his payoff by deviating. v We conclude that the strategy pair (t1 = 0; t2 = 2 ) is a Nash equilibrium. By symmetry, the pair v (t1 = 2 ; t2 = 0) is also Nash. These equilibria correspond to the two equilibria of the Hawk-Dove game: devoting time 0 (giving up v immediately) is similar to playing Dove, and devoting time c (being willing to fight hard) corresponds to playing Hawk. As in the discrete case, we have two equilibria, and in each case one of the players is passive and the other is aggressive. 3 v (b) Suppose player 1 is devoting a time t1 with 0 < t1 < c . We will show that this scenario is not a Nash equilibrium, regardless of player 2's strategy. To do this, we consider three cases. First suppose that player 2 is devoting a time t2 > t1, and hold this strategy fixed. Player 1's payoff is therefore −ct1. In this case, player 1 can certainly do better by playing t1 = 0 and receiving payoff 0. So in this case, player 1 has incentive to deviate. v Next assume that player 2 is devoting t2 = t1, so that player 1's payoff is 2 − ct2. Then player 1 could do better by playing t1 = v=c, in which case he would receive payoff v − ct2 > v=2 − ct2. So in this case player 1 also has an incentive to deviate. Finally, consider the case where player 2 devotes t2 < t1. In this case, player 2 will have an incentive to deviate: he currently receives payoff −ct2, so player 2 can do better by playing t2 = 0. In each of the three possible cases, one of the players can receive a higher payoff by deviating. We v therefore conclude that there is no Nash equilibria in which one player plays ti with 0 < ti < c . v (c) Consider the strategy pair (t1 = 0; t2 = 0). Here player 1 receives payoff 2 , but he could improve by v playing t1 = c and then receive payoff v.
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