Suppose that in a Hawk-Dove game similar to the one discussed in your workbook, the payoff to each player is -6 if both play Hawk.If both play Dove, the payoff to each player is 3, and if one plays Hawk and the other plays Dove, the one that plays Hawk gets a payoff of 8 and the one that plays Dove gets 0.In equilibrium, we would expect hawks and doves to do equally well.This happens when the proportion of the total population that plays Hawk is
A).45.
B).23.
C).11.
D).73.
E)1.

Question

Quizplus · Accepted Answer

To find the equilibrium proportion of Hawks, we can use the replicator dynamics equation: p_i(t+1) = p_i(t) * (w_i - ), where p_i(t) is the proportion of population playing the Hawk strategy at time t, w_i is the fitness/payoff of the Hawk strategy for player i, and is the average fitness/payoff of the population. Let's assume there are initially p Hawks and (1-p) Doves in the population. Then, the average fitness is: < w > = p * F1S1 + (1-p) * 3, since when both play Hawk, each player gets F1S1, and when both play Dove, each player gets 3. For the Hawk strategy, the fitness of a Hawk player is: w_H = p * (F1S1 - F106) + (1-p)*8, since a Hawk gets F1S1 (which is higher than 3) when facing a Dove, and gets F1S1-F106 when facing another Hawk. For the Dove strategy, the fitness of a Dove player is: w_D = p * F1S1 + (1-p) * 3, since a Dove gets F1S1 when facing a Hawk, and gets 3 when facing another Dove. Now, we need to set up the replicator dynamics equation for both strategies: p_H(t+1) = p_H(t) * (w_H - ) p_D(t+1) = p_D(t) * (w_D - ) Since the population is fixed, we know that p_H(t) + p_D(t) = 1, and we only need one equation to solve for p_H(t+1). Let's use p_D(t) = 1 - p_H(t): p_H(t+1) = p_H(t) * (w_H - ) + (1 - p_H(t)) * (w_D - ) We can substitute the fitness expressions we derived earlier: p_H(t+1) = p_H(t) * [p * (F1S1 - F106) + (1-p)*8 - ] + (1 - p_H(t)) * [p * F1S1 + (1-p) * 3 - ] Simplifying: p_H(t+1) = p_H(t) * [(p-1)F106 + 5p - 5] + 3 - 4p To find the equilibrium proportion of Hawks, we can set p_H(t+1) = p_H(t) = p: p = p * [(p-1)F106 + 5p - 5] + 3 - 4p p^2 * F106 + (5-4F106) p - 3 = 0 Solving for p (using the quadratic formula): p = 0.45 or p = 1.192... Since p must be between 0 and 1, the only valid solution is p=0.45, which means that in equilibrium, 45% of the population should play the Hawk strategy.

Suppose That in a Hawk-Dove Game Similar to the One