Consider the game where I = {1, 2}, S_1 = {1, . . . , 7}, S_2 = {A, . . . , G}.
ID: 1194824 • Letter: C
Question
Consider the game where I = {1, 2}, S_1 = {1, . . . , 7}, S_2 = {A, . . . , G}. The payoffs are given by the following matrix: What is the unique outcome that survives iterative deletion of strictly dominated strategies in ? What are the payoffs associated with this unique strategy profile? Is the unique outcome obtained in (1) a Nash equilibrium? Do you think that it is possible for a Nash equilibrium to be eliminated by the procedure of iterative deletion of strictly strategies? Explain your intuition.Explanation / Answer
1. The unique outcome that survives iterative deletion of strictly dominated strategies is (3,C ). the payoffs associated with this unique strategyprofile is (3,4).
2. the unique outcome obatained is a Nash equlibrium. this is because it survives iterated elimination of dominant strategies. in this the player will never choose a strategy that gets eliminated because the payoffs of the players can be increased from deviating or choosing another strategy. The only outcomes that a player will consider choosing will be the ones that survived elimination. In the above case there was only one outcome that survived the elimination, so the players will only choose this strategy. if the players deviates then this would result in lower payoffs. hence their is no incentive for a player to deviate from the unique payoff that survives elimination.The survived strategy is also the strategy that th player will choose for any strategy other player chooses. so player 1 will always choose 3 and player 2 will choose C. hence this is a Nash equilibrium.
No it is not possible for a nash equilibrium to get eliminated . This is because Nash equilibrium is the optimal outcome. here the players have no incentive to deviate. If it gets eliminated then it will never be chosen by a player. But this not the behaviour of a rational agent as he will always like to choose a bundle that is optimum. so both of this gives contradicting results. Hence Nash Equilibrium cannot be eliminated through this method.