Answers

1. Yes. Elimination sequences s2, S1 leads to equilibrium (S2, s1).

2. No. Nash equilibrium: (S2; s3).

3. No.

4.

a.

Player B
Player A		s	c
	S	5;5	0;10
	C	10;0	-10;-10

b. No.

c. No.

d. (S, c) and (C, s) are the two Nash equilibria.

5.

a.

Player B
Player A		g	-g
	G	-1;-1	1;0
	-G	0;1	0;0

b. Yes: (G, -g) and (-G, g).

6.

a.

p=(-2-5)/[(-2-5)+(2-6] = 7/11

q= (0+5)/[(0+5)+(4+1)] = 1/2

Hence, (Pr(S1)=7/11, Pr(S2)=4/11; Pr(s1)=Pr(s2)=1/2) is a Nash equilibrium.

b.

The most common outcomes are (S1, s1) and (S1,s2), both with probability 7/22.

c.

EP(S1)=EP(S2)= -1/2. Since Pr(S1)=7/11 and Pr(S2)=4/11, the expected payoff for A’s mixed strategy is -1/2(7/11)- 1/2(4/11) = -1/2. Note that, in general, the expected payoff of a Nash equilibrium mixed strategy is the same as that of each of its component strategies.

EP(s1)=EP(s2)=14/11+20/11=34/11. Hence, the expected payoff for B is 34/11.

7.

1.

EP(s1)=2 x 1/2 + 1 x ¾=5/4

2.

EP(s2)=0 x ¼ + 3 x ¾ = 9/4

3.

EP(S) against s1= 1 x ¼ +2 x ¾ = 7/4

4.

EP(s) against s2= 3 x ¼ + 1 x ¾ = 3/2

5.

EP(S) against s= [1 x 1/3 + 3 x 2/3] x ¼ + [2 x 1/3 + 1 x 2/3] x ¾ = 7/3 x ¼ + 4/3 x ¾ = 19/12

6.

EP(s) against S= 5/4 x 1/3 + 9/4 x 2/3 = 23/12.