From my doctoral course
This note comes from the doctoral course “Game Theory and Bargaining”. I rewrote the explanations in my own words and recomputed the numerical examples.
Zero-sum
In a zero-sum game, player 1’s payoff is minus player 2’s, so one matrix is enough (the row player’s payoff). The course example: two higher-education institutions must each launch a new programme to attract students; the entries are “percentage increase in intake share for institution A”. Rows = A’s strategies, columns = B’s strategies.
Maximin and minimax
- The row player is cautious: for each strategy they find the row minimum (worst case) and choose the largest of those minima: maximin.
- The column player likewise finds, for each column, the column maximum and chooses the smallest of them: minimax.
If the two numbers are equal we have a saddle point: the value of the game. Neither wants to change (a Nash equilibrium). In the course example, the row minima for institution A were and the maximin was 13, that is the third strategy.
Mixed strategy
If there is no saddle point, predictability loses the game. The solution: play randomly (choose each strategy with some probability). Von Neumann’s minimax theorem (1928) guarantees that in every finite two-person zero-sum game the value of the game in mixed strategies exists.
A 2×2 example:
| Col 1 | Col 2 | |
|---|---|---|
| Row 1 | 3 | −1 |
| Row 2 | −2 | 4 |
Pure maximin: ; pure minimax: ; they differ → no saddle point. The row player chooses row 1 with probability ; the two columns must become indifferent for them: → → and the value of the game . That is, in the long run the row player wins 1 unit on average, and the opponent can do nothing by anticipating them.
Applications
Problems of direct confrontation over a fixed resource: price competition between two firms for market share, military games, security problems. It is less useful in cooperative relations (most organisational situations are “non-zero-sum”).
Related: Static games: the prisoner’s dilemma and Nash equilibrium.