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3. Algebraic method example ( Enter your problem )
 Method & Example-1Example-2 Other related methods

1. Method & Example-1

Method
 algebraic method Steps (Rule) Step-1: A play’s (p_1, p_2) p_1=(d - c)/((a + d) - (b + c)) and p_2 = 1 - p_1 Step-2: B play’s (q_1, q_2) q_1=(d - b)/((a + d) - (b + c)) and q_2 = 1 - q_1 Step-3: Value of the game V V=(a * d - b * c)/((a + d) - (b + c))

Example-1
1. Find Solution of game theory problem using algebraic method
 Player A\Player B B1 B2 A1 1 7 A2 6 2

Solution:
Players
 Player B B_1 B_2 Player A A_1 1 7 A_2 6 2

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

 Player B B_1 B_2 RowMinimum Player A A_1 1 7 1 A_2 (6) [2] [2] ColumnMaximum (6) 7

Select minimum from the maximum of columns
Column MiniMax = (6)

Select maximum from the minimum of rows
Row MaxiMin = [2]

Here, Column MiniMax != Row MaxiMin

:. This game has no saddle point.

Matrix size is 2xx2, so dominance rule is not required.

Solution using algebraic method
Here a=1,b=7,c=6,d=2

p_1=(d - c)/((a + d) - (b + c))=(2 -6)/((1 +2) - (7 +6))=(-4)/(3 -13)=2/5

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

q_1=(d - b)/((a + d) - (b + c))=(2 -7)/((1 +2) - (7 +6))=(-5)/(3 -13)=1/2

q_2=1-q_1=1-1/2=1/2

V=(a * d - b * c)/((a + d) - (b + c))=((1 xx 2) - (7 xx 6))/((1 +2) - (7 +6))=(2 -42)/(3 -13)=4

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