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.
Consider the following pay-off matrix
Player A | Player B | ||
---|---|---|---|
B_{1} | B_{2} | ||
A_{1} | -2 | 4 | |
A_{1} | 8 | 3 | |
A_{1} | 9 | 0 |
Solution.
The game does not have a saddle point as shown in the following table.
Player A | Player B | Minimum | Probability | ||
---|---|---|---|---|---|
B_{1} | B_{2} | ||||
A_{1} | -2 | 4 | -2 | q_{1} | |
A_{2} | 8 | 3 | 3 | q_{2} | |
A_{3} | 9 | 0 | 0 | q_{3} | |
Maximum | 9 | 4 | |||
Probability | p_{1} | p_{1} |
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 B_{1} and
the value 4 is plotted along the vertical axis under strategy B_{2}.
A straight line joining the two points is then drawn.
Similarly, we can plot strategies A_{2} and A_{3} 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 = -2p_{1} + 4p_{2} and
E2 = 8p_{1} + 3p_{2}
Comparing the above two equations, we have
-2p_{1} + 4p_{2} = 8p_{1} + 3p_{2}
Substituting p_{2} = 1 - p_{1
}-2p_{1} + 4(1 - p_{1}) = 8p_{1} + 3(1
- p_{1})
p_{1} = 1/11
p_{2} = 10/11
Substituting the values of p_{1} and p_{2} in equation E1
V = -2 (1/11) + 4 (10/11) = 3.4 units