Question

8. Use the following game matrix for this question Dolores Left Center A, 21,3 Right D, 3 3,4 4, B Middle 2,0X, Y>4 Down2, C1, 2 The Y>4 is correct for Part A. It may or may not work for Parts B & C a. Solving by elimination of strictly dominated strategies, what values of A, B, C, D, X & Y will lead to a single, Nash in pure strategy of Middle, Center? If not possible explain why. Express your values for A, B, C, D, X &Y as inequalities. If a variable can take any value then say it is greater than zero. What is the minimum payoff, expressed in whole numbers? Solving by elimination of strictly dominated strategies, what values of A, B, C, D, X & Y will lead to a single, Nash in pure strategy of Up, Left? If not possible explain why. Express your values for A, B, C, D, X & Y as inequalities. If a variable can take any value then say it is greater than zero. What is the minimum payoff, expressed in whole numbers? What values of A, B, C, D, X & Y will lead to two Nash in pure strategy with one at Teddy: UP; Dolores: Right and the other at Teddy: Down; Dolores: Left and one Nash in mixed strategy? If not possible explain why. Express your values for A, B, C, D, X & Y as inequalities. If a variable can take any value then say it is greater than zero. What is the minimum payoff, expressed in whole numbers? b. c.

Answer 1

a) Since it is a static game, no one knows what other person would do beforehand. Following the norm, let Teddy(P1) be player 1, and Dolores(P2) represent player 2. For Middle, Center to be the Nash equilibrium, the payoff from Middle (to Teddy) and payoff from Center (to Dolores) need to be the highest.

When P1 chooses Up, the optimal move for P2 is center or right, To ensure that (Up, Right) is not the Nash Equilibrium, we need D<4

When P1 chooses Middle, the only optimal move for P2 is Center as Y>4

When P1 chooses Down, optimal move of P2, would depend on values of C and B. If D<4, when P1 would choose Down if P2 moves Right. To ensure that (Down, Right) is not the Nash Equilibrium, B<C and B<2.

Let's check the reaction moves by P1 if P2 chooses from his available options.

When P2 chooses Left, P1 could choose Up if A>2 otherwise he would be indifferent between Middle and Down. If A=2, he would be indifferent in all the three options. When A$\tiny \leq$ 2, to ensure that P2 doesn't choose Left when P1 moves Down, we want C<2. where B is the payoff if P2 moves center.

Now, we want P1 to move Middle when P2 moves Center. To ensure this, X>1 where 1 is the payoff from Up and Down.

So the required conditions are :

(A>2 & X>1 & D<4 & B<C & C>0) OR (A$\tiny \leq$ 2 & X>1 & D<4 & B<C & C<2)

Minimum payoff in this case would be (2,5) (taking X=2>1 and y=5>4)

b) To begin with, the payoff for P1 from UP should be greater than or at least equal to the payoff from other two moves. Hence A$\tiny \geq$ 2. If A=2, C<2 for P2 to choose Left. Therefore we can eliminate (Middle, Left) and (Down, Left) with A $\tiny \geq$ 2 and C<2. But when P1 moves 'Up', there is no way Left could lead to Nash Equilibrium for P2. Hence no conditions on A,B,C,D,X could lead to (Up, Left) as Nash Equilibrium.

c) For 2 Nash Euilibrium to be possible, firstly, note that P2 always chooses Center or Right when P1 moves Up. For P1 to choose only D when P2 moves Right, D>4. To avoid (Middle,Center) as equilibrium, we need X<1 and Y<4. To achieve (Down., Left) as the equilibrium, C>B and C>2 and A<2 (A<2 ensures P1 chooses Down when P2 moves Left). Hence we need,

C>B and C>2 and A<2 and D>4 and X<1 and Y<4.

Minimum payoff is (2,3) and (5,3)

For a mixed strategy Nash, we can assign zero probabilities to Middle (for P1) and Center (for P2) under the conditions derived on X and Y above (X<1 and Y<4).

Let's say, P1 moves 'Up' with Probability p1 and Down with 1-p1. For P2, Pr(Left)=p2 and Pr(Right)=1-p2.

The payoff table is as follows:

		Dolores(p2,1-p2)
		Left	Center	Right
Ted(p1,1-p1)	Up	A.p1, 2.p2	0,0	D.p1, 3.(1-p2)
	Middle	0,0	0,0	0,0
	Down	2.(1-p1), C.p2	0,0	4.(1-p1),B.(1-p2)

If P2 choose Left,

P1 chooses Up if Ap1>2(1-p1), and when P1 chooses Up, for P2 to choose left, 2.p2>3(1-p2). But since p2<=1, 2.p2<3(1-p2) (always). Therefore, when Ted chooses Up,P2 chooses Right always. Now we want P1 to choose Up in its response to achieve Nash. i.e. Dp1>4(1-p1), D>(4(1-p1)/p1). If P1 chooses Down, P2 would choose Left of Cp2>B(1-p2)