Algebraic Method: Game Theory
In this section, we will talk about the algebraic method used to solve mixed strategy games. Here we have provided formulas and examples of algebraic method.
Consider the zero sum two person game given below:
| Player B | |||
|---|---|---|---|
| Player A | I | II | |
| I | a | b | |
| II | c | d | |
Formulas: Algebraic Method
The solution of the game is:
A play’s (p, 1 - p)
where:
| p = | d - c -------------------- (a + d) - (b + c) |
B play’s (q, 1 - q)
where:
| q = | d - b ------------------- (a + d) - (b + c) |
| Value of the game, V = | ad - bc -------------------- (a + d) - (b + c) |
Algebraic Method Example 1: Game Theory
Consider the game of matching coins. Two players, A & B, put down a coin. If coins match (i.e., both are heads or both are tails) A gets rewarded, otherwise B. However, matching on heads gives a double premium. Obtain the best strategies for both players and the value of the game.
| Player B | |||
|---|---|---|---|
| Player A | I | II | |
| I | 2 | -1 | |
| II | -1 | 1 | |
Solution.
This game has no saddle point.
| p = | 1 - (-1) ----------------------- (2 + 1) - (-1 - 1) |
= | 2 ---- 5 |
1 – p = 3/5
| q = | 1 - (-1) ----------------------- (2 + 1) - (-1 - 1) |
= | 2 ---- 5 |
1 – q = 3/5
| V = | 2 X 1 - (-1) X (-1) -------------------------- (2 + 1) - (-1 - 1) |
= | 1 ---- 5 |
Example 2: Algebraic Method in Game Theory
Solve the game whose payoff matrix is given below:
| Player B | |||
|---|---|---|---|
| Player A | I | II | |
| I | 1 | 7 | |
| II | 6 | 2 | |
Solution.
This game has no saddle point.
| p = | 2 - 6 ----------------------- (1 + 2) - (7 + 6) |
= | 2 ---- 5 |
1 – p = 3/5
| q = | 2 - 7 ----------------------- (1 + 2) - (7 + 6) |
= | 1 ---- 2 |
1 – q = 1/2
| V = | 1 X 2 - (7 X 6) -------------------------- (1 + 2) - (7 + 6) |
= | 4 |
