Find Solution of game theory problem using graphical method
| Player A\Player B | B1 | B2 | B3 | B4 |
| A1 | 2 | 2 | 3 | -2 |
| A2 | 4 | 3 | 2 | 6 |
Solution:
1. Saddle point testing
Players
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | `B_4` | | |
| Player `A` | `A_1` | | 2 | 2 | 3 | -2 | |
| `A_2` | | 4 | 3 | 2 | 6 | |
We apply the maximin (minimax) principle to analyze the game.
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | `B_4` | | Row
Minimum |
| Player `A` | `A_1` | | 2 | 2 | 3 | -2 | | `-2` |
| `A_2` | | 4 | (3) | [2] | 6 | | `[2]` |
| Column
Maximum | | `4` | `(3)` | `3` | `6` | | |
Select minimum from the maximum of columns
Column MiniMax = (3)
Select maximum from the minimum of rows
Row MaxiMin = [2]
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` | `B_3` | `B_4` | | |
| Player `A` | `A_1` | | 2 | 2 | 3 | -2 | |
| `A_2` | | 4 | 3 | 2 | 6 | |
column-1 `>=` column-2, so remove column-1
| | | Player `B` | | |
| | | `B_2` | `B_3` | `B_4` | | |
| Player `A` | `A_1` | | 2 | 3 | -2 | |
| `A_2` | | 3 | 2 | 6 | |
Solution using graphical method
First, we draw two parallel lines 1 unit distance apart and mark a scale on each.
The two parallel lines represent strategies of player `A`.
If player `B` selects strategy `B_1`, player `A` can win 2 or 3 units depending on A's selection of strategies.
The value 2 is plotted along the vertical axis under strategy `A_1` and the value 3 is plotted along the vertical axis under strategy `A_2`.
A straight line joining the two points is then drawn.
Similarly, we can plot strategies `B_2,B_3` also. The problem is graphed in the following figure.
The highest point `V` in the shaded region indicates the value of game. From the above figure, the value of the game is `2.44` units.
1. The point of optimal solution occurs at the intersection of two lines
`E_2 = 3p_1 +2p_2`
`E_3 = -2p_1 +6p_2`
Comparing the above two equations, we have
`3p_1 +2p_2 = -2p_1 +6p_2`
Substituting `p_2 = 1 - p_1`
`3p_1 +2(1 - p_1) = -2p_1 +6(1 - p_1)`
Solving `p_1=4/9`
`p_2=1-p_1=1-4/9=5/9`
Substituting the values of `p_1` and `p_2` in equation `E_2`
`V = 3(4/9) +2(5/9)=22/9`
2. The point of optimal solution occurs at the intersection of two lines
`L_2 = 3q_1 -2q_2`
`L_3 = 2q_1 +6q_2`
Comparing the above two equations, we have
`3q_1 -2q_2 = 2q_1 +6q_2`
Substituting `q_2 = 1 - q_1`
`3q_1 -2(1 - q_1) = 2q_1 +6(1 - q_1)`
Solving `q_1=8/9`
`q_2=1-q_1=1-8/9=1/9`
Substituting the values of `q_1` and `q_2` in equation `L_2`
`V = 3(8/9) -2(1/9)=22/9`
This material is intended as a summary. Use your textbook for detail explanation.
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