Question

4. Consider the following game: PLAYER 2 (0,3) (2,0) (1,7) PLAYER 1 (2,4) (0,6) (2,0) (1,3) (2,4) (0,3) a) Does this game have any pure-strategy Nash equilibrium? If so, identify it (or them) and explain why this is an equilibrium. b) Find a mixed-strategy Nash equilibrium to this game and explain your calculations. Note: a mixed strategy for player i may be expressed by o; = (P1, P2, 1- P1 - p2). c) Is it possible to find another mixed-strategy Nash equilibrium, assuming that one of the players is adopting a pure strategy? If so, identify it and explain your reasoning.

Please answer 3 Questions, thank you.

Answer 1

	a	b	c
A	(0,3)	(2,0)	(1,7)
B	(2,4)	(0,6)	(2,0)
C	(1,3)	(2,4)	(0,3)

Ans A)

If we check for BEst responses for both functions then we find that

For player 1,

When player 2 chooses "a" best response of player 1 is "B"

When player 2 chooses "b" best response of player 1 is "A" or "C"

When player 2 chooses "c" best response of player 1 is "B"

similarly,

For player 2,

When player 1 chooses "A" best response of player 2 is "c"

When player 1 chooses "B" best response of player 2 is "b"

When player 1 chooses "c" best response of player 2 is "b"

if we compare both responses then we see that (C,b) combination is the only combination
that provides equilibrium

Because both are best responses to each other responses

Ans B)

Lets assume that player 1 chooses A with probability p1, B with probability p2 & C with probability (1-p1-p2)

similarly,
for player 2

He chooses a with probability q1, b with probability q2 and c with probability 1-q1-q2

We need to find values of p1,p2,q1,q2 such that for both players

E(A)=E(B)=E(C) and E(a)=E(b)=E(c)

E(A)=2*q2+1(1-q1-q2)
E(B)=2*q1+2*(1-q1-q2)
E(C)=1*q1+2*q2

solving these equations we get

2q2+1-q1-q2=2q1+2-2q1-2q2
q2+1-q1=2-2q2

3q2-q1=1...Equation 1)

E(B)=E(C)
2q1+2-2q1-2q2=q1+q2
2-2q2=q1+q2

q1+3q2=1

therefore solving these 2 equations we get
6q2=2

q2=1/3 and q1=0 and q3=2/3

Now for player 2

E(a)=3p1+4p2+3(1-p1-p2)
E(b)=6p2+4(1-p1-p2)
E(c)=7p1+3(1-p1-p2)

E(a)=E(b)
3p1+4p2+3-3p1-3p2=6p2+4-4p1-4p2
4p2+3-3p2=4-4p1+2p2
3+p2=4-4p1+2p2
p2=1-4p1+2p2
4p1-p2=1

E(b)=E(c)
6p2+4-4p1-4p2=7p1+3-3p1-3p2
2p2+4-4p1=7p1+3-3p1-3p2
2p2+4-4p1=4p1+3-3p2
5p2+1=8p1
5p2-8p1=-1
3p2=1
p2=1/3 and p1=1/3 and p3=1/3

Mixed strategy Nash equilibrium for both players is (1/3,1/3,1/3) & (0,1/3,2/3)

Answer C)

If player 2 doesnot choose to play "a" then we have for player 1 strategy A weakly dominates strategy B therfore we have resultant matrix as below

b c
B (0,6) (2,0)
C (2,4) (0,3)

We can not find another mixed-strategy Nash equilibrium because we can get Pure Nash equilibrium if we solve above game further.

Hence answer is NO