Question

Consider the following two-person zero-sum game. Assume the two players have the same three strategy options. The payoff table below shows the gains for Player A.

		Player B
Player A	Strategy b1	Strategy b2	Strategy b3
Strategy  a1	3	2	?4
Strategy  a2	?1	0	2
Strategy  a3	4	5	?3

Is there an optimal pure strategy for this game? If so, what is it? If not, can the mixed-strategy probabilities be found algebraically? What is the value of the game?

Answer 1

		B
		b1	b2	b3	Row minimum
A	a1	3	2	-4	-4
	a2	-1	0	2	-1	Maximin
	a3	4	5	-3	-3
	Column Maximum	4	5	2
				Minimax

The value of the game lies between -1 and 2

As you can see there is no pure strategy for the game we have to find the optimal solution algebraical method or linear programming method

Player A linear program

Maximize Z = v

Subject to

v-3x1+x2-4x3<=0

v-2x1+0x2-5x3<=0

v+4x1-2x2+3x3<=0

x1,x2,x3>=0 , v is unrestricted

For B linear program

Minimize Z= v

Subject to

v-3y1-2y2+4y3>=0

v+y1+0y2-2y3>= 0

c-4y1-5y2+3y3>=0, v is unrestricted

Optimal value of game lies between -1 to 2