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.

Example: 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

Solution.

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

On small screens, use horizontal scrollbar to view full calculation

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.

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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