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Topics Covered

Two-person Zero-sum Pure Strategy Mixed Strategy Maxmin / Minimax Saddle Point 2×2 Game Graphical Method Dominance Property
On this page
  1. 1. Introduction to Game Theory
  2. 2. Two-Person Zero-Sum Games
  3. 3. Maxmin and Minimax Principles
  4. 4. Saddle Point and Pure-Strategy Solution
  5. 5. Mixed Strategies and Games Without Saddle Point
  6. 6. 2 × 2 Games Without Saddle Point
  7. 7. Dominance Property
  8. 8. Graphical Method for 2 × n and m × 2 Games
  9. 9. Relation to Linear Programming
  10. Summary Workflow
  11. Key Take-aways

1. Introduction to Game Theory

DEFINITION

Game theory is the study of decision-making in situations of competition or conflict among two or more rational agents (called players), where each player's outcome depends not only on their own actions but also on the actions of the others. It was formalised by John von Neumann and Oskar Morgenstern (1944) and developed further by John Nash.

Key Terminology

2. Two-Person Zero-Sum Games

DEFINITION

A two-person game has exactly two players (A and B). It is zero-sum if the gain of one player is exactly the loss of the other — sum of payoffs = 0.

Payoff Matrix Convention

Without loss of generality, the payoff matrix shows the gain of player A (the maximiser). Player B is the minimiser — chooses columns to minimise A's gain.

If A has \(m\) strategies and B has \(n\) strategies, the matrix is \(m \times n\):

B₁B₂…Bₙ
A₁a₁₁a₁₂…a₁ₙ
A₂a₂₁a₂₂…a₂ₙ
⋮⋮⋮⋮
Aₘaₘ₁aₘ₂…aₘₙ

\(a_{ij}\) is what A receives (and B loses) when A plays Aᵢ and B plays Bⱼ.

3. Maxmin and Minimax Principles

Maxmin (Player A)

Player A is conservative — for each strategy, A considers the worst (smallest) possible payoff and then chooses the strategy that maximises this worst-case payoff:

\[ v_L \;=\; \max_{i}\, \min_{j}\, a_{ij}. \]

\(v_L\) is the lower value (maxmin value).

Minimax (Player B)

Player B is conservative — for each strategy, B considers the worst (largest) possible payoff to A, then chooses the strategy minimising this worst-case:

\[ v_U \;=\; \min_{j}\, \max_{i}\, a_{ij}. \]

\(v_U\) is the upper value (minimax value).

In every game: \(v_L \le v_U\). If \(v_L = v_U\), the game has a saddle point and a value \(v = v_L = v_U\).

4. Saddle Point and Pure-Strategy Solution

DEFINITION

A saddle point is an entry \(a_{i^*j^*}\) of the payoff matrix that is simultaneously the minimum in its row and the maximum in its column.

If a saddle point exists, both players have an optimum pure strategy — A plays \(i^*\), B plays \(j^*\), and the value of the game is \(v = a_{i^*j^*}\).

How to Find a Saddle Point

  1. Mark the minimum of each row (in the right margin).
  2. Mark the maximum of each column (at the bottom).
  3. If the largest row-minimum equals the smallest column-maximum, that common entry is the saddle point.
EXAMPLE 1 — Game with Saddle Point

Payoff matrix:

B₁B₂B₃Row min
A₁3543
A₂4622
A₃1731
Col max474

Maxmin = max(3, 2, 1) = 3 (row A₁). Minimax = min(4, 7, 4) = 4 (col B₁ or B₃). Since Maxmin = 3 ≠ 4 = Minimax, this matrix has no saddle point — a saddle point requires a cell that is simultaneously the minimum of its row and the maximum of its column, i.e. Maxmin = Minimax.

Now consider a matrix that does have a saddle point:

B₁B₂B₃Row min
A₁-215-2-2
A₂-5-6-4-6
A₃-520-8-8
Col max-220-2

Maxmin = max(-2, -6, -8) = -2 at row A₁. Minimax = min(-2, 20, -2) = -2 at col B₁ or B₃. Maxmin = Minimax = -2 ⇒ saddle point. Optimum: A plays A₁; B plays B₁ (or B₃); value v = -2.

EXAMPLE 2 — Find the saddle point
B₁B₂B₃B₄Row min
A₁86282
A₂89454
A₃75353
Col max8948

Maxmin = max(2, 4, 3) = 4 at A₂. Minimax = min(8, 9, 4, 8) = 4 at B₃. Both = 4 ⇒ saddle point at (A₂, B₃) with value v = 4.

5. Mixed Strategies and Games Without Saddle Point

If a game has no saddle point (\(v_L < v_U\)), neither player can commit to a single pure strategy. Instead they should use mixed strategies: choose each strategy with a certain probability.

Mixed-Strategy Formulation

Let A play strategy Aᵢ with probability \(p_i\) (with \(\sum p_i = 1\)) and B play Bⱼ with probability \(q_j\) (with \(\sum q_j = 1\)). The expected payoff to A is:

\[ E(p, q) \;=\; \sum_{i=1}^{m}\sum_{j=1}^{n} p_i q_j\, a_{ij}. \]

A chooses \(p\) to maximise \(E\); B chooses \(q\) to minimise \(E\). Von Neumann's Minimax Theorem guarantees the optimal value exists and is the same for both players.

6. 2 × 2 Games Without Saddle Point

For a 2 × 2 game without a saddle point:

B₁B₂
A₁ab
A₂cd

Let A play A₁ with probability \(p\) and A₂ with probability \(1 - p\); B plays B₁ with probability \(q\) and B₂ with probability \(1 - q\).

Optimum Strategies

\[ p \;=\; \dfrac{d - c}{(a + d) - (b + c)}, \qquad q \;=\; \dfrac{d - b}{(a + d) - (b + c)}, \] \[ \text{Value of the game } v \;=\; \dfrac{a d - b c}{(a + d) - (b + c)}. \]

(Provided the denominator is non-zero. Sign convention: A is the maximiser.)

EXAMPLE 1

Payoff:

B₁B₂
A₁51
A₂34

Row mins: 1, 3. Col maxes: 5, 4. Maxmin = 3; Minimax = 4 ⇒ no saddle point.

\(a = 5, b = 1, c = 3, d = 4\). Denominator = (5+4) − (1+3) = 5.

\(p = (4 − 3)/5 = 0.2;\;\; q = (4 − 1)/5 = 0.6;\;\; v = (20 − 3)/5 = 17/5 = 3.4\).

A plays A₁ 20 % of the time, A₂ 80 %; B plays B₁ 60 %, B₂ 40 %; expected value = 3.4.

EXAMPLE 2

Payoff matrix:

B₁B₂
A₁24
A₂63

No saddle point. \(p = (3 − 6)/((2+3) − (4+6)) = -3/-5 = 0.6;\;\; q = (3 − 4)/-5 = 0.2;\;\; v = (2·3 − 4·6)/-5 = (-18)/-5 = 3.6\).

A: 60 % A₁, 40 % A₂; B: 20 % B₁, 80 % B₂; v = 3.6.

7. Dominance Property

DEFINITION

One row (or column) dominates another if its entries are uniformly at least as good for the player owning the matrix. Dominated rows / columns can be safely deleted, reducing the matrix size.

Dominance Rules

  1. Row dominance: if every entry of row \(i\) is ≥ the corresponding entry of row \(k\), then A will never play row \(k\) (it's dominated by row \(i\)). Delete row \(k\).
  2. Column dominance: if every entry of column \(j\) is ≤ the corresponding entry of column \(l\), then B will never play column \(l\) (dominated). Delete column \(l\).
  3. Convex (mixed) dominance: if some convex combination of two rows dominates a third, the third can be deleted. Same idea for columns.

Apply dominance repeatedly. The reduced matrix may have a saddle point or become 2 × 2 (solvable by formula) or 2 × n / m × 2 (graphical method).

EXAMPLE 1 — Dominance reduces the matrix

Original matrix:

B₁B₂B₃
A₁324
A₂-142
A₃226

Compare row A₁ and A₃: (3, 2, 4) vs (2, 2, 6). A₁ is not uniformly better than A₃ (A₃ better in col B₃). No row dominance here.

Compare columns B₂ and B₃: (2, 4, 2) vs (4, 2, 6). For B (minimiser), smaller is better. B₂ is not uniformly smaller than B₃. But B₂ has values 2, 4, 2 — and B₃ has 4, 2, 6. Mixed.

Look at B₁ vs B₃: (3, -1, 2) vs (4, 2, 6). Every entry of B₁ ≤ corresponding B₃ entry. So B₃ is dominated by B₁ — delete B₃.

Reduced:

B₁B₂
A₁32
A₂-14
A₃22

Now A₁ vs A₃: (3, 2) vs (2, 2). Every entry of A₁ ≥ A₃ ⇒ A₃ dominated by A₁, delete A₃. Now 2×2:

B₁B₂
A₁32
A₂-14

Row mins: 2, -1. Col max: 3, 4. Maxmin = 2; Minimax = 3 ⇒ no saddle point. Apply 2×2 formula: \(p = (4 − (-1))/((3+4) − (2 + (-1))) = 5/6\); \(q = (4 − 2)/6 = 2/6 = 1/3\); \(v = (3·4 − 2·(-1))/6 = (12 + 2)/6 = 14/6 = 7/3 ≈ 2.33\).

EXAMPLE 2 — Mixed dominance

Consider:

B₁B₂B₃
A₁172
A₂627
A₃516

Take the average of A₁ and A₂: (3.5, 4.5, 4.5). This dominates A₃ (5, 1, 6)? No: 3.5 < 5. So mixed dominance does not apply here.

Direct: B₁ vs B₃: (1, 6, 5) vs (2, 7, 6). B₁ ≤ B₃ everywhere ⇒ B₃ dominated, delete.

8. Graphical Method for 2 × n and m × 2 Games

2 × n Game (A has 2 strategies)

A plays A₁ with probability \(p\) and A₂ with probability \(1 - p\). For each column \(j\), A's expected payoff against pure strategy \(B_j\) is:

\[ E_j(p) \;=\; p \cdot a_{1j} + (1 - p)\, a_{2j}. \]

This is a linear function of \(p\) on \([0, 1]\). Plot all \(n\) lines on the same diagram.

For each \(p\), B will choose the column that minimises A's payoff — the lower envelope of the lines. A wants to maximise this lower envelope. The optimum \(p^*\) is at the highest point of the lower envelope.

Procedure

  1. Draw axis: x from 0 to 1 representing \(p\); y = expected payoff.
  2. For each column \(j\), draw the line \(E_j(p)\) from (0, \(a_{2j}\)) to (1, \(a_{1j}\)).
  3. Sketch the lower envelope (the minimum across all lines).
  4. Find the highest point of the envelope — this gives \(p^*\) and \(v\).
  5. Two lines pass through that point — those two columns are B's active strategies. Solve the 2×2 sub-game.

m × 2 Game (B has 2 strategies)

Symmetric: plot \(E_i(q)\) as a function of \(q\) for each row, find the upper envelope, and locate its lowest point (B is minimiser).

EXAMPLE 1 — 2 × 4 Game

Matrix:

B₁B₂B₃B₄
A₁210-2
A₂1032

Lines:

Lower envelope at various p: at p = 0, min = 0 (E₂); at p = 1, min = -2 (E₄). Plot lines and find the highest point of the lower envelope. Suppose it occurs at intersection of E₂ and E₄.

Solve \(p = 2 - 4p\) ⇒ \(p^* = 2/5\). Reduced 2×2:

B₂B₄
A₁1-2
A₂02

Apply 2×2 formula: \(p = (2-0)/((1+2)-(-2+0)) = 2/5\) ✓; \(q = (2-(-2))/5 = 4/5\); \(v = (1·2 − (-2)·0)/5 = 2/5 = 0.4\).

Graphical solution — 2 × 4 game 321 0−1−2 A's expected payoff E₁E₂ E₃E₄ p* = 2/5, v = 2/5 00.41 p = P(A plays A₁) →
Fig 4.1 — Each line \(E_j(p)\) is A's expected payoff against B's column \(j\). B (the minimiser) picks the lower envelope; A (the maximiser) picks the value of \(p\) at its highest point. That peak — the intersection of \(E_2\) and \(E_4\) — gives \(p^* = 2/5\) and game value \(v = 2/5\), and identifies B₂, B₄ as the 2×2 sub-game to solve.
EXAMPLE 2 — m × 2 Game (4 × 2)

Matrix:

B₁B₂
A₁24
A₂23
A₃32
A₄-26

Lines \(E_i(q) = q \cdot a_{i1} + (1-q) a_{i2}\):

Plot from q = 0 to q = 1. Take the upper envelope and find its lowest point. Find intersection of dominant lines.

Note first that E₂ = 3 − q lies entirely below E₁ = 4 − 2q on \([0,1]\) (their difference E₁ − E₂ = 1 − q ≥ 0), so E₂ never reaches the upper envelope and A₂ is not active. The upper envelope is formed by E₄ (for small \(q\)), then E₁, then E₃; its lowest point is at the intersection of E₁ and E₃:

\[ 4 - 2q \;=\; 2 + q \;\;\Rightarrow\;\; q^* = \tfrac{2}{3}, \qquad v = 2 + \tfrac{2}{3} = \tfrac{8}{3} \approx 2.67 . \]

Active rows: A₁ and A₃ — solve the 2×2 game over these:

B₁B₂
A₁24
A₃32

Here \(a = 2, b = 4, c = 3, d = 2\), denominator \((a+d) - (b+c) = 4 - 7 = -3\). Then \(p = (d-c)/\text{den} = (2-3)/(-3) = 1/3\) (A plays A₁ with prob 1/3, A₃ with 2/3), \(q = (d-b)/\text{den} = (2-4)/(-3) = 2/3\) ✓, and \(v = (ad-bc)/\text{den} = (4-12)/(-3) = 8/3 \approx 2.67\). A guarantees exactly 8/3 against either column, confirming the value.

9. Relation to Linear Programming

Every two-person zero-sum game can be formulated as a linear program. For an \(m \times n\) game with all positive entries (if not, add a constant to make them positive), A's LP is:

\[ \text{Maximise } v \] \[ \text{s.t. } \sum_i p_i a_{ij} \ge v, \quad j = 1, \ldots, n, \] \[ \sum_i p_i = 1, \quad p_i \ge 0. \]

B's LP is the dual. By LP duality, both LPs have the same optimum value — the value of the game.

Summary Workflow

  1. Identify A as maximiser, B as minimiser. Write payoff matrix for A.
  2. Check for saddle point: compute row minima and column maxima.
  3. If saddle point exists ⇒ pure strategies; value = saddle entry.
  4. If not, apply dominance to reduce.
  5. If reduced to 2×2, use the closed-form formula.
  6. If 2×n or m×2 after reduction, use graphical method.
  7. For larger matrices, set up as LP.

Key Take-aways