Unit 04 — Theory of Games
Two competitors, one table of outcomes. How to turn a story into a payoff matrix, find a saddle point, cut a game down by dominance, and solve a 2 × 2 game with mixed strategies.
Game theory appeared in both 2025-26 papers as Q3A (5 marks), and both times the task was the same: read a short story about two competing firms and write the payoff matrix Final QP p. 3 Re-Exam QP p. 2. Neither paper asked you to solve a mixed-strategy game — but Assignment 7 asks you to formulate and read off the result, so learn the saddle-point test too.
- Formulate the 2 × 2 payoff matrix — 2 of the 5 marks are for the table itself, 1 for labelling strategies, 1 for interpretation Key p. 3. Both versions solved →
- Find the saddle point and the value of the game — every Assignment 7 game has one. Assignment 7 solved →
1. What a Game Is — The Vocabulary Core syllabus concept
1Understand the Concept
Game theory studies decisions where the outcome for you depends on what a competitor chooses at the same time. Two firms bidding for one contract, two brands setting prices — each chooses a strategy without knowing the other's choice, and the pair of choices fixes the result.
This course only covers two-person zero-sum games: exactly two players, and whatever one gains the other loses. So a single number in each cell is enough — it is the gain of the row player, and automatically the loss of the column player.
2Terms you must be able to define
| Term | Meaning |
|---|---|
| Player | A competitor. The row player is the maximiser (wants the cell values high); the column player is the minimiser (wants them low). |
| Strategy | A complete course of action a player can choose — e.g. Low tariff or Premium tariff. |
| Payoff matrix | Rows = row player's strategies, columns = column player's strategies, cell = row player's gain (negative = loss). |
| Zero-sum | Row player's gain + column player's gain = 0 in every cell. One number per cell. |
| Pure strategy | A player always uses the same single strategy. |
| Mixed strategy | A player chooses among strategies with fixed probabilities (e.g. 60% A, 40% B). |
| Maximin | Row player's guaranteed minimum: the largest of the row minima. |
| Minimax | Column player's guaranteed ceiling: the smallest of the column maxima. |
| Saddle point | A cell that is the minimum of its row and the maximum of its column. Exists when maximin = minimax. |
| Value of the game (V) | The expected payoff when both play optimally. V = 0 ⟹ fair game. |
2. Formulating the Payoff Matrix from a Story Asked in the 2025-26 papers
1The exam shape
The question gives four (2 × 2) or nine (3 × 3) sentences like "If SpeedEx bids conservatively and FastMove bids aggressively, FastMove wins +8 while SpeedEx loses 8." Your job is to turn them into one table. It is mechanical — the only real decision is whose payoff the table shows.
2The four-step method
| Step | What you do |
|---|---|
| 1. Pick the row player | Read the line "the payoff represents X's advantage". X is the row player. If no such line is given, take the first-named firm as rows and say so. |
| 2. Label rows and columns | Rows = X's strategies, columns = the rival's strategies, in the order the question lists them. |
| 3. Convert every sentence to X's gain | X gains k → +k. X loses k → −k. The rival gains k → X's payoff is −k. The rival loses k → +k. "Both equal / random award" → 0. |
| 4. Interpret in one line | Run the saddle-point test (concept 3) and say what each firm should do and whether the game is fair. This is the interpretation mark. |
3Worked Example — the Final Exam Q3A shape
SpeedEx and FastMove each bid Aggressively (A) or Conservatively (C). (C, C) and (A, A): contract awarded randomly, expected profit 0 each. (C, A): FastMove wins +8, SpeedEx loses 8. (A, C): SpeedEx wins +8, FastMove loses 8. Formulate the zero-sum payoff matrix.
- Step 1 — Row player No "advantage of" line is given, so take the first-named firm, SpeedEx, as the row player. Cell value = SpeedEx's profit (FastMove's is the same number with the sign flipped).
- Step 2 — Convert each sentence
Outcome (SpeedEx, FastMove) Sentence SpeedEx's payoff (C, C) random award, 0 each 0 (C, A) SpeedEx loses 8 −8 (A, C) SpeedEx wins +8 +8 (A, A) random award, 0 each 0 - Step 3 — The payoff matrix
SpeedEx ↓ / FastMove → C (Conservative) A (Aggressive) Row min C (Conservative) 0 −8 −8 A (Aggressive) +8 0 0 ← maximin Column max 8 0 ← minimax - Step 4 — Interpret
Maximin = 0 = Minimax ⟹ saddle point at (A, A), value of the game V = 0
Both firms should bid aggressively — for each, A is never worse than C whatever the rival does. With V = 0 the game is fair: neither firm has a built-in advantage. This matches the official key's matrix exactly Key p. 3.
- Mixing up whose payoff is shown. If the stem says "payoff represents Green's advantage", every cell is Green's gain — including the cells where the sentence is written from Solar's side.
- Forgetting the sign flip. "TBL's payoff is 6" in a table of ABL's payoffs is −6, not 6.
- No labels. Write the strategy names on both axes (1 mark), not just a 2 × 2 grid of numbers.
- Stopping at the table. One sentence of interpretation (best strategy for each, fair or not) is the fifth mark.
3. Pure Strategies and the Saddle Point Core syllabus concept
1Understand the Concept
Each player assumes the worst about the other. The row player asks, for every row, "what is the least I could get here?" and picks the row whose least is largest — the maximin. The column player does the mirror image — the minimax. If the two numbers meet in the same cell, neither player can do better by switching alone: that cell is a saddle point, both play pure strategies, and the cell value is the value of the game.
2The method
- Write the row minimum beside every row.
- Write the column maximum under every column.
- Maximin = largest row minimum. Minimax = smallest column maximum.
- Maximin = Minimax ⟹ saddle point. Optimal strategies are that row and that column; V = the common value.
- Maximin < Minimax ⟹ no saddle point. V lies between them; use dominance and/or mixed strategies (concepts 4–5).
3Worked Example — 3 × 3 with a saddle point
Find the optimal strategies and the value of the game (payoffs to player A).
| A ↓ / B → | B1 | B2 | B3 |
|---|---|---|---|
| A1 | 6 | 2 | 4 |
| A2 | 7 | 5 | 6 |
| A3 | 3 | 1 | 8 |
- Step 1 — Row minima and column maxima
A ↓ / B → B1 B2 B3 Row min A1 6 2 4 2 A2 7 5 6 5 A3 3 1 8 1 Col max 7 5 8 - Step 2 — Compare Maximin = max(2, 5, 1) = 5 (row A2) Minimax = min(7, 5, 8) = 5 (column B2) Maximin = Minimax = 5 ⟹ saddle point at (A2, B2)
- Step 3 — State the answer
A always plays A2, B always plays B2, value of the game V = 5 in A's favour — so the game is not fair (V ≠ 0).
Circle the smallest number in each row, box the largest in each column. A number that is both circled and boxed is the saddle point. Every game in Assignment 7 has one — so check this before reaching for any formula.
4. The Dominance Rule Core syllabus concept
1Understand the Concept
A strategy is dominated when another strategy is at least as good against every move of the opponent. A rational player never uses a dominated strategy, so you can delete it and work with a smaller matrix. Dominance is how a 3 × 3 game is cut down to a 2 × 2 that the formula in concept 5 can solve.
2The two rules — note they point in opposite directions
| Player | Delete a strategy when… | Why |
|---|---|---|
| Row player (maximiser) | every entry of row i is ≤ the corresponding entry of some other row | That row never earns more — the row player wants big numbers. |
| Column player (minimiser) | every entry of column j is ≥ the corresponding entry of some other column | That column never concedes less — the column player wants small numbers. |
3Worked Example — reduce 3 × 3 to 2 × 2
| A ↓ / B → | B1 | B2 | B3 |
|---|---|---|---|
| A1 | 3 | −1 | 4 |
| A2 | 1 | 2 | 3 |
| A3 | −2 | 0 | 1 |
Reduce by dominance, then solve.
- Step 1 — Saddle point? Row minima: −1, 1, −2 ⟹ maximin = 1 Column maxima: 3, 2, 4 ⟹ minimax = 2 1 ≠ 2 ⟹ no saddle point; 1 ≤ V ≤ 2
- Step 2 — Row dominance Compare A3 (−2, 0, 1) with A2 (1, 2, 3): every entry of A3 is smaller ⟹ delete row A3
- Step 3 — Column dominance on what is left B3 (4, 3) vs B1 (3, 1): every entry of B3 is larger ⟹ the minimising column player deletes column B3
- Step 4 — The reduced game
A ↓ / B → B1 B2 A1 3 −1 A2 1 2 Still no saddle point (maximin 1, minimax 2), so solve it with mixed strategies — continued in concept 5.
5. Mixed Strategies — the 2 × 2 Formula Core syllabus concept
1Understand the Concept
With no saddle point, any pure strategy can be exploited by the opponent. The answer is to randomise: play each strategy with a probability chosen so the opponent gains nothing by knowing it. For a 2 × 2 game these probabilities have a closed form.
2The formula
3Worked Example — finishing the dominance example
Solve the reduced game from concept 4: A1 = (3, −1), A2 = (1, 2).
- Step 1 — Read a, b, c, d a = 3, b = −1, c = 1, d = 2 D = (a + d) − (b + c) = (3 + 2) − (−1 + 1) = 5
- Step 2 — Player A's mix p = (d − c)/D = (2 − 1)/5 = 0.2 ⟹ A1 with 20%, A2 with 80%
- Step 3 — Player B's mix q = (d − b)/D = (2 + 1)/5 = 0.6 ⟹ B1 with 60%, B2 with 40%
- Step 4 — Value
V = (ad − bc)/D = (3·2 − (−1)(1))/5 = 7/5 = 1.4
Check: 1 ≤ 1.4 ≤ 2 ✓. In the full 3 × 3 game, A plays (0.2, 0.8, 0) and B plays (0.6, 0.4, 0) — the deleted strategies get probability 0.
When dominance leaves a 2 × 3 or 3 × 2 game, plot each of the opponent's strategies as a straight line of expected payoff against p (from 0 to 1). The row player picks the highest point of the lower envelope; the two lines that cross there give the 2 × 2 sub-game, which you then solve with the formula above. Your papers have not asked for it, so know the idea rather than the drawing.
- Using the mixed formula when there is a saddle point. Always test for a saddle point first; the formula gives nonsense probabilities otherwise.
- Reversing the dominance direction for columns. The column player deletes the column with the bigger numbers.
- Leaving out deleted strategies in the final answer. State them with probability 0.
- Calling any V > 0 "fair". Fair means V = 0 exactly.