Question

Consider the following payoff table for an evolutionary game:

X Y

X (0,0) (1,4)

Y (4,1) (0,0)

1. If we consider this table as a normal form game, in the mixed strategy Nash equilibrium the probability of X is?
A. 1/2 B. 1/3 C. 1/4 D. 1/5 E. 1/6 F. 1/7 G. 1/8 H. 1/9

2. The mixed equilibrium you found is

Consider the following payoff table for an evolutionary game:

X Y

X (3,3) (1,3)

Y (3,1) (2,2)

1. Is X an evolutionarily stable strategy? Answer
YesNo

2. Is Y an evolutionarily stable strategy? Answer
YesNo

Consider the following payoff table for an evolutionary game:

X Y

X (4,4) (1,3)

Y (3,1) (2,2)

1. Is X an evolutionarily stable strategy? Answer
YesNo

2. Is Y an evolutionarily stable strategy? Answer
YesNo

Consider the following payoff table for an evolutionary game:

X Y

X (3,3) (2,3)

Y (3,2) (1,1)

1. Is X an evolutionarily stable strategy? Answer
YesNo

2. Is Y an evolutionarily stable strategy? Answer
YesNo

Answer 1

Answer to the first question:

1. Suppose player 1 is the row player and player 2 is the column player. Let player 2 play X with probability p and play Y with probability (1 - p).

So when player 1 plays X, he gets 0 with probability p and 1 with probability (1 - p). When player 1 plays Y, he gets 4 with probability p and 0 with probability (1 - p).

Therefore, ( 0*p)+1(1-p) = (4*p) + 0(1-p)

Or, 0 + 1 - p = 4p + 0

Or, 5p = 1

Or, p = (1/5)

Therefore, (1-p) = (4/5)

Now suppose, player 1 plays X with probability q and plays Y with probability (1-q).

Therefore, if player 2 plays X, it gets 0 with probability q and 1 with probability (1 -q). If player 2 plays Y, it gets 4 with probability q and 0 with probability (1 - q).

Therefore, (0*q) + 1(1-q) = 4q + 0(1-q)

Or, 1 - q = 4q

Or, 5q = 1

Or, q = (1/5)

Therefore, (1 - q) = (4/5)

Answer: The probability of X is (1/5).

2. The mixed strategy Nash equilibrium is (1/5 , 4/5), (1/5 , 4/5).