Method
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algebraic method Steps (Rule)
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Step-1:
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A play's (p_1, p_2)
p_1=(d - c)/((a + d) - (b + c)) and p_2 = 1 - p_1
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Step-2:
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B play's (q_1, q_2)
q_1=(d - b)/((a + d) - (b + c)) and q_2 = 1 - q_1
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Step-3:
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Value of the game V
V=(a * d - b * c)/((a + d) - (b + c))
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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:
1. Saddle point testing
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 | | Row
Minimum |
| Player A | A_1 | | 1 | 7 | | 1 |
| A_2 | | (6) | [2] | | [2] |
| Column
Maximum | | (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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