Graphical Method: Game Theory

The method discussed in the previous section is feasible when the value of n is small, because the larger value of n will yield a larger number of 2 X 2 sub-games.

In this section, we discuss Graphical Method for solving 2 X n games.

This method can only be used in games with no saddle point, and having a pay-off matrix of type n X 2 or 2 X n.

exampleExample: Graphical Method for Game Theory

Consider the following pay-off matrix

Player A Player B
  B1 B2
A1 -2 4
A1 8 3
A1 9 0


The game does not have a saddle point as shown in the following table.

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Player A   Player B Minimum Probability
B1 B2
A1 -2 4 -2 q1
A2 8 3 3 q2
A3 9 0 0 q3
Maximum 9 4    
Probability p1 p1    

Maximin = 4, Minimax = 3

First, we draw two parallel lines 1 unit distance apart and mark a scale on each. The two parallel lines represent strategies of player B.
If player A selects strategy A1, player B can win –2 (i.e., loose 2 units) or 4 units depending on B’s selection of strategies. The value -2 is plotted along the vertical axis under strategy B1 and the value 4 is plotted along the vertical axis under strategy B2. A straight line joining the two points is then drawn.
Similarly, we can plot strategies A2 and A3 also. The problem is graphed in the following figure.

Graphical Method Game Theory

The lowest point V in the shaded region indicates the value of game. From the above figure, the value of the game is 3.4 units. Likewise, we can draw a graph for player B.

The point of optimal solution (i.e., maximin point) occurs at the intersection of two lines:

E1 = -2p1 + 4p2 and
E2 = 8p1 + 3p2

Comparing the above two equations, we have

-2p1 + 4p2 = 8p1 + 3p2

Substituting p2 = 1 - p1
-2p1 + 4(1 - p1) = 8p1 + 3(1 - p1)
p1 = 1/11
p2 = 10/11

Substituting the values of p1 and p2 in equation E1

V = -2 (1/11) + 4 (10/11) = 3.4 units

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