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Chapter 10: Problem 4

Players \(A\) and \(B\) have found \(\$ 100\) on the sidewalk and are arguing about how it should be split. A passerby suggests the following game: "Each of you state the number of dollars that you wish \(\left(d_{d}, d_{B}\right) .\) If \(d_{A}+d_{B} \wedge 100\) you can keep the figure you name and I'll take the remainder. If \(d_{A}+d_{B}>100,\) I'll keep the \(\$ 100 . "\) Is there a unique Nash equilibrium in this game of continuous strategies?

Short Answer

Expert verified
Answer: The unique Nash equilibrium of the game is (50, 50), where both players A and B choose to take $50 each.

Step by step solution

01

Understand the concepts and constraints

In this game, players A and B need to choose the amounts \((d_A, d_B)\) such that \(d_A + d_B \leq 100\), otherwise the passerby takes the entire amount. The Nash equilibrium will be a pair of amounts \((d_A^*, d_B^*)\) such that no player can be better off by changing their strategy, given the other player's strategy.
02

Setup players' utility functions

The objective of both players is to maximize their utilities. Let's assume that the utility function of player A is \(U_A(d_A, d_B) = d_A\), and the utility function of player B is \(U_B(d_A, d_B) = d_B\). Both players want to maximize their respective utility functions, given the other player's strategy.
03

Find the Best Response Functions

To find the Nash equilibrium, we need to find the best response functions (BRF) for both players. The BRFs are the amounts that each player will choose to maximize their utility, given the other player's amount. For player A, the BRF is given by the following equation: \(BRF_A(d_B) = \max_{d_A} U_A(d_A, d_B) = \max_{d_A} d_A\) Subject to the constraint: \(d_A + d_B \leq 100\) Similarly, for player B, the BRF is given by the following equation: \(BRF_B(d_A) = \max_{d_B} U_B(d_A, d_B) = \max_{d_B} d_B\) Subject to the constraint: \(d_A + d_B \leq 100\)
04

Solve for the Nash equilibrium

The Nash equilibrium can be found by solving the BRFs for both players: \((d_A^*, d_B^*) = (BRF_A(d_B^*), BRF_B(d_A^*))\) Using the constraint \(d_A + d_B \leq 100\), we can rewrite the BRFs as follows: \(BRF_A(d_B^*) = 100 - d_B^*\) \(BRF_B(d_A^*) = 100 - d_A^*\) Now, substituting the BRFs for equilibrium amounts \((d_A^*, d_B^*)\), we get: \((d_A^*, d_B^*) = (100 - d_B^*, 100 - d_A^*)\) Since the game is symmetric, i.e., both players face the same constraints and have the same objective, the Nash equilibrium will have both players choosing the same amounts: \(d_A^* = d_B^* = \frac{100}{2} = 50\) Thus, the unique Nash equilibrium of the game is \((d_A^*, d_B^*) = (50, 50)\).

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Most popular questions from this chapter

Players \(A\) and \(B\) are engaged in a coin-matching game. Each shows a coin as either heads or tails. If the coins match, \(B\) pays A \(\$ 1 .\) If they differ, \(A\) pays \(B \$ 1\) a. Write down the payoff matrix for this game, and show that it does not contain a Nash equilibrium. b. How might the players choose their strategies in this case?

The game of "chicken" is played by two macho teens who speed toward each other on a single-lane road. The first to veer off is branded the chicken, whereas the one who doesn't turn gains peer group esteem. Of course, if neither veers, both die in the resulting crash. Payoffs to the chicken game are provided in the following table. a. Does this game have a Nash equilibrium? b. Is a threat by either not to chicken-out a credible one? c. Would the ability of one player to firmly commit to a not-chicken strategy (by, for example, throwing away the steering wheel) be desirable for that player?

In A Treatise on the Family (Cambridge; Harvard University Press, 1981 ), G. Becker proposes his famous Rotten Kid theorem as a game between a (potentially rotten) child, \(A\), and his or her parent, \(B . A\) moves first and chooses an action, \(r,\) that affects his or her own income \(Y_{A}(r)\) \(\left(Y_{A}^{\prime}>0\right)\) and the income of the parent \(Y_{s}(r)\left(Y_{B}^{\prime}<0\right) .\) In the second stage of the game, the parent leaves a monetary bequest of \(L\) to the child. The child cares only for his or her own utility, \(U_{A}\left(Y_{A}+L\right),\) but the parent maximizes \(U_{B}\left(Y_{B}-L\right)+X U_{A},\) where \(A>0\) reflects the parent's altruism toward the child. Prove that the child will opt for that value of \(r\) that maximizes \(Y_{A}+Y_{B}\) even though he or she has no altruistic intentions. Plint: You must first find the parent's optimal bequest, then solve for the child's optimal strategy, given this subsequent parental behavior.)

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.

The mixed-strategy Nash equilibrium for the Battle of the Sexes game described in Example 10.4 may depend on the numerical values of the payoffs. To generalize this solution, assume that the payoff matrix for the game is given by where \(K \geq 1 .\) Show how the Nash equilibrium in mixed strategies for this game depends on the value of \(K\)
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