2. Example-2
Find Solution of game theory problem using 2Xn Games
| Player A\Player B | B1 | B2 | | A1 | 1 | -3 | | A2 | 3 | 5 | | A3 | -1 | 6 | | A4 | 4 | 1 |
Solution:
1. Saddle point testing
Players
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_1` | | 1 | -3 | | | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
We apply the maximin (minimax) principle to analyze the game.
| | | Player `B` | | | | | | `B_1` | `B_2` | | Row
Minimum | | Player `A` | `A_1` | | 1 | -3 | | `-3` | | `A_2` | | [3] | 5 | | `[3]` | | `A_3` | | -1 | 6 | | `-1` | | `A_4` | | (4) | 1 | | `1` | | Column
Maximum | | `(4)` | `6` | | |
Select minimum from the maximum of columns
Column MiniMax = (4)
Select maximum from the minimum of rows
Row MaxiMin = [3]
Here, Column MiniMax `!=` Row MaxiMin
`:.` This game has no saddle point.
2. Dominance rule to reduce the size of the payoff matrix
Using dominance property
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_1` | | 1 | -3 | | | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
row-1 `<=` row-4, so remove row-1
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
So, we consider each 2`xx`2 sub-game and obtain their values
Players
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
2`xx`2 sub-game : (1) Row 1, 2
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | |
This game has saddle point and value of the game, `V_1` is 3
2`xx`2 sub-game : (2) Row 1, 3
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_2` | | 3 | 5 | | | `A_4` | | 4 | 1 | |
This game has no saddle point, so we use the algebraic method.
Value of game, `V_2=(a * d - b * c)/((a + d) - (b + c))=((3 xx 1) - (5 xx 4))/((3 +1) - (5 +4))=(3 -20)/(4 -9)=17/5`
2`xx`2 sub-game : (3) Row 2, 3
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
This game has no saddle point, so we use the algebraic method.
Value of game, `V_3=(a * d - b * c)/((a + d) - (b + c))=((-1 xx 1) - (6 xx 4))/((-1 +1) - (6 +4))=(-1 -24)/(0 -10)=5/2`
The 2`xx`2 sub-game with the lowest value is `5/2` in (3) and hence the solution to this game provides the solution to the larger game.
so now solution using algebraic method
1. Saddle point testing
Players
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | |
We apply the maximin (minimax) principle to analyze the game.
| | | Player `B` | | | | | | `B_1` | `B_2` | | Row
Minimum | | Player `A` | `A_3` | | -1 | 6 | | `-1` | | `A_4` | | (4) | [1] | | `[1]` | | Column
Maximum | | `(4)` | `6` | | |
Select minimum from the maximum of columns
Column MiniMax = (4)
Select maximum from the minimum of rows
Row MaxiMin = [1]
Here, Column MiniMax `!=` Row MaxiMin
`:.` This game has no saddle point.
Matrix size is 2`xx`2, so dominance rule is not required.
Solution using algebraic method
Here `a=-1,b=6,c=4,d=1`
`p_1=(d - c)/((a + d) - (b + c))=(1 -4)/((-1 +1) - (6 +4))=(-3)/(0 -10)=3/10`
`p_2=1-p_1=1-3/10=7/10`
`q_1=(d - b)/((a + d) - (b + c))=(1 -6)/((-1 +1) - (6 +4))=(-5)/(0 -10)=1/2`
`q_2=1-q_1=1-1/2=1/2`
`V=(a * d - b * c)/((a + d) - (b + c))=((-1 xx 1) - (6 xx 4))/((-1 +1) - (6 +4))=(-1 -24)/(0 -10)=5/2`
A plays `(0,3/10,7/10)`
B plays `(1/2,1/2)`
Value of game is `5/2`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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