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Fudenberg and Tirole (1992) develop a game of stag-hunting based on an observation originally made by Rousseau. The two players in the game may either cooperate in catching a stag or each may set out on his own to catch a hare. The payoff matrix for this game is given by a Describe the Nash equilibria in this game. b. Suppose \(B\) believes that \(A\) will use a mixed strategy in choosing how to hunt. How will 5 's optimal strategy choice depend on the probability \(A\) will play stag? c. Suppose this game is expanded to \(n\) players (the game Rousseau had in mind) and that all \(n\) must cooperate in order for a stag to be caught. Assuming that the payoffs for one specific player, say \(B\), remain the same and that all the other \(n-1\) players will opt for mixed strategies, how will \(B\) 's optimal strategy depend on the probabilities with which each of the other players plays stag? Explain why cooperation seems less likely in this larger game.

Short Answer

Expert verified
Answer: Player B's optimal strategy choice depends on the expected payoffs of playing Stag versus Hare. If the expected payoff for choosing Stag (2p) is greater than the expected payoff for choosing Hare (1), it is optimal for B to play Stag. If the expected payoff for choosing Stag (2p) is lower than the expected payoff for choosing Hare (1), it is optimal for B to play Hare. If the expected payoffs of both strategies are equal, B is indifferent between playing Stag or Hare.

Step by step solution

01

Part a: Identifying the Nash Equilibria

First, let's identify the Nash equilibria in this game. A Nash equilibrium occurs when no player has an incentive to change their strategy, given that the other players maintain their current strategies. In this case, the payoff matrix is as follows: A Stag Hare Stag (2,2) (0,1) B Hare (1,0) (1,1) There are two pure-strategy Nash equilibria: (Stag, Stag) and (Hare, Hare). In both equilibria, neither player has an incentive to change their strategy, since doing so would result in a lower payoff.
02

Part b: B's Optimal Strategy Choice if A Plays a Mixed Strategy

Suppose Player A adopts a mixed strategy, playing Stag with probability \(p\) and Hare with probability \(1-p\). Given A's strategy, B can choose to play Stag or Hare according to the following expected payoffs: Expected payoff for B if choosing Stag: \(2p + 0(1-p) = 2p\) Expected payoff for B if choosing Hare: \(1p + 1(1-p) = p + 1 - p = 1\) B's optimal strategy choice depends on the expected payoff of playing Stag versus Hare. If \(2p > 1\), it is optimal for B to play Stag. If \(2p < 1\), it is optimal for B to play Hare. If, however, \(2p=1\), B is indifferent between playing Stag or Hare since the expected payoffs of both strategies are equal.
03

Part c: B's Optimal Strategy Choice in an n-Player Game

Now, let's generalize the game to \(n\) players, where all \(n\) players need to cooperate for a stag to be caught. We can assume that the payoffs for B remain the same and all the other \(n-1\) players will opt for mixed strategies. Let \({q_i}\) represent the probability that player \(i\) (different from B) plays Stag. In this case, B's expected payoffs for each strategy are as follows: Expected payoff for B if choosing Stag: \(2(\Pi_{i=1}^{n-1} q_i)\) Expected payoff for B if choosing Hare: \(\displaystyle 1\) It is important to note that the expected payoff for B if they choose Hare is constant and does not depend on the other players' strategies. Comparing the expected payoffs, we find that B will choose Stag if their expected payoff for choosing Stag is greater than the expected payoff for choosing Hare, or when \(2(\Pi_{i=1}^{n-1} q_i) > 1\). As \(n\) gets larger, the likelihood that all the other players play Stag (\((\Pi_{i=1}^{n-1} q_i)\)) becomes smaller. This is because each player's decision to play Stag is multiplied together. As a result, it becomes increasingly difficult for the product to be greater than \(\frac{1}{2}\), meaning that cooperation is less likely in a larger game.

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