Game Theory problem using Algebraic method Example-2

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2. Example-2

Find Solution of game theory problem using algebraic method
_{Player A}\^{Player B} B1 B2
A1 2 -1
A2 -1 1

Solution:
1. Saddle point testing
Players

		Player `B`
		`B_1`	`B_2`
Player `A`	`A_1`	2	-1
Player `A`	`A_2`	-1	1

We apply the maximin (minimax) principle to analyze the game.

		Player `B`
		`B_1`	`B_2`	Row Minimum
Player `A`	`A_1`	2	[-1]	`[-1]`
Player `A`	`A_2`	-1	(1)	`-1`
	Column Maximum	`2`	`(1)`

Select minimum from the maximum of columns
Column MiniMax = (1)

Select maximum from the minimum of rows
Row MaxiMin = [-1]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.

Solution using algebraic method
Here `a=2,b=-1,c=-1,d=1`

`p_1=(d - c)/((a + d) - (b + c))=(1 +1)/((2 +1) - (-1 -1))=(2)/(3 +2)=2/5`

`p_2=1-p_1=1-2/5=3/5`

`q_1=(d - b)/((a + d) - (b + c))=(1 +1)/((2 +1) - (-1 -1))=(2)/(3 +2)=2/5`

`q_2=1-q_1=1-2/5=3/5`

`V=(a * d - b * c)/((a + d) - (b + c))=((2 xx 1) - (-1 xx -1))/((2 +1) - (-1 -1))=(2 -1)/(3 +2)=1/5`

This material is intended as a summary. Use your textbook for detail explanation.
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