NATS1595, Assignment 3 Due : March 12 by 11:59am 6. [6 marks] Consider the Hawk-Dove game from lecture. Suppose, in this scenario, the Benefit is (B = 2) and the Cost is (C' = 4). What is the mixed strategy nash equilibrium for this game? Is the mixed strategy Nash equilibrium evolutionary stable? Hawk and Dove Player 2 Hawk Dove Suppose two individuals are after a common resource B- C B -C Hawk 2 2 B, O with payoff 'B'. Suppose the Player 1 cost to conflict is 'C' Dove O, B 2 Contacts What is the probability of encountering a hawk or a dove? It will depend on how many of each exist in the population. 100% probability of contacting a "hawk" 50% probability of contacting a "hawk" B= 2 C = 1 The Benefit Outweighs the Cost:(B>C Player 2 Hawk Dove B - C 0 Hawk B -CB -C 2 (B, O) 2 2 - Player 1 (0, B) B 2- Dove BY (2'2 Hawk is the dominant strategy, and the Nash Equilibrium corresponds to Player 2 both players playing Hawk. Hawk Dove Hawk NF ( 2, 0 ) This looks like the Prisoner's Player 1 Dilemma. Dove 10, 2) The Benefit Outweighs the Cost: B>C Player 2 What happens as the percentage of "Hawks" Hawk Dove in the population increases? Hawk (1/2,1/2) (2,0) p=0.5 -> ECH)= (0.5)(2) + (1-5) (2)= 0.25+1= 1,25 Player 1 pagott of playing Hark in a 50/50 population of Dove (0,2) (1,1) Doves and Hawks is 1.25 p=0.9 ECH ) = (0.9) (t ) + (o. ) ( 2 ) = 0.65 payoff of playing Hook in a population of 80%. Hawks is 0.65. the more tanks the worse the p= 1 - everyone is a Houck. E ( H) = P(+ ) + (12,# (2) - 2 payoff . Inever playing Dare is never best