Game Theory: 2 x n Games
Games where one player has only two courses of action while the other has more than two, are called 2 X n or n X 2 games.
If these games do not have a saddle point or are reducible by the dominance method, then before solving these games we write all 2 X 2 sub-games and determine the value of each 2 X 2 sub-game.
This method is illustrated by the following example.
Example: 2 x n Games
Determine the solution of game for the pay-off matrix given below:
| Player B | ||||
|---|---|---|---|---|
| Player A | I | II | III | |
| I | -3 | -1 | 7 | |
| II | 4 | 1 | -2 | |
Solution.
Obviously, there is no saddle point and also no course of action dominates the other. Therefore, we consider each 2 X 2 sub-game and obtain their values.
(a)
| Player B | |||
|---|---|---|---|
| Player A | I | II | |
| I | -3 | -1 | |
| II | 4 | 1 | |
The saddle point is 1. So the value of game, V1 is 1.
(b)
| Player B | |||
|---|---|---|---|
| Player A | I | II | |
| I | -3 | 7 | |
| II | 4 | -2 | |
This game has no saddle point, so we use the algebraic method.
| Value of game, V2 = | (-3) X (-2) - (7 X 4) ------------------------- (-3 - 2) - (7 + 4) |
= | 11 --- 8 |
(c)
| Player B | |||
|---|---|---|---|
| Player A | II | III | |
| I | -1 | 7 | |
| II | 1 | -2 | |
This game has no saddle point, so we use the algebraic method.
| Value of game, V3 = | (-1) X (-2) - (7 X 1) ----------------------- (-1 - 2) - (7 + 1) |
= | 5 --- 11 |
The 2 X 2 sub-game with the lowest value is (c) and hence the solution to this game provides the solution to the larger game.
Using algebraic method:
A plays ( 3/11, 8/11)
B plays (0, 9/11, 2/11)
Value of game is 5/11.
