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.
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.
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 \(m \times n\):
| B₁ | B₂ | … | Bₙ | |
|---|---|---|---|---|
| A₁ | a₁₁ | a₁₂ | … | a₁ₙ |
| A₂ | a₂₁ | a₂₂ | … | a₂ₙ |
| ⋮ | ⋮ | ⋮ | ⋮ | |
| Aₘ | aₘ₁ | aₘ₂ | … | aₘₙ |
\(a_{ij}\) is what A receives (and B loses) when A plays Aᵢ and B plays Bⱼ.
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).
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 \(v = v_L = v_U\).
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 \(v = a_{i^*j^*}\).
Payoff matrix:
| B₁ | B₂ | B₃ | Row min | |
|---|---|---|---|---|
| A₁ | 3 | 5 | 4 | 3 |
| A₂ | 4 | 6 | 2 | 2 |
| A₃ | 1 | 7 | 3 | 1 |
| Col max | 4 | 7 | 4 |
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:
| B₁ | B₂ | B₃ | Row min | |
|---|---|---|---|---|
| A₁ | -2 | 15 | -2 | -2 |
| A₂ | -5 | -6 | -4 | -6 |
| A₃ | -5 | 20 | -8 | -8 |
| Col max | -2 | 20 | -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.
| B₁ | B₂ | B₃ | B₄ | Row min | |
|---|---|---|---|---|---|
| A₁ | 8 | 6 | 2 | 8 | 2 |
| A₂ | 8 | 9 | 4 | 5 | 4 |
| A₃ | 7 | 5 | 3 | 5 | 3 |
| Col max | 8 | 9 | 4 | 8 |
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.
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.
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:
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.
For a 2 × 2 game without a saddle point:
| B₁ | B₂ | |
|---|---|---|
| A₁ | a | b |
| A₂ | c | d |
Let A play A₁ with probability \(p\) and A₂ with probability \(1 - p\); B plays B₁ with probability \(q\) and B₂ with probability \(1 - q\).
(Provided the denominator is non-zero. Sign convention: A is the maximiser.)
Payoff:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 5 | 1 |
| A₂ | 3 | 4 |
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.
Payoff matrix:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 2 | 4 |
| A₂ | 6 | 3 |
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.
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.
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).
Original matrix:
| B₁ | B₂ | B₃ | |
|---|---|---|---|
| A₁ | 3 | 2 | 4 |
| A₂ | -1 | 4 | 2 |
| A₃ | 2 | 2 | 6 |
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:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 3 | 2 |
| A₂ | -1 | 4 |
| A₃ | 2 | 2 |
Now A₁ vs A₃: (3, 2) vs (2, 2). Every entry of A₁ ≥ A₃ ⇒ A₃ dominated by A₁, delete A₃. Now 2×2:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 3 | 2 |
| A₂ | -1 | 4 |
Row mins: 2, -1. Col max: 3, 4. Maxmin = 2; Minimax = 3 ⇒ no saddle point. Apply 2×2 formula: \(p = (4 − (-1))/((3+4) − (2 + (-1))) = 5/6\); \(q = (4 − 2)/6 = 2/6 = 1/3\); \(v = (3·4 − 2·(-1))/6 = (12 + 2)/6 = 14/6 = 7/3 ≈ 2.33\).
Consider:
| B₁ | B₂ | B₃ | |
|---|---|---|---|
| A₁ | 1 | 7 | 2 |
| A₂ | 6 | 2 | 7 |
| A₃ | 5 | 1 | 6 |
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.
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:
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.
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).
Matrix:
| B₁ | B₂ | B₃ | B₄ | |
|---|---|---|---|---|
| A₁ | 2 | 1 | 0 | -2 |
| A₂ | 1 | 0 | 3 | 2 |
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:
| B₂ | B₄ | |
|---|---|---|
| A₁ | 1 | -2 |
| A₂ | 0 | 2 |
Apply 2×2 formula: \(p = (2-0)/((1+2)-(-2+0)) = 2/5\) ✓; \(q = (2-(-2))/5 = 4/5\); \(v = (1·2 − (-2)·0)/5 = 2/5 = 0.4\).
Matrix:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 2 | 4 |
| A₂ | 2 | 3 |
| A₃ | 3 | 2 |
| A₄ | -2 | 6 |
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:
Active rows: A₁ and A₃ — solve the 2×2 game over these:
| B₁ | B₂ | |
|---|---|---|
| A₁ | 2 | 4 |
| A₃ | 3 | 2 |
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.
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:
B's LP is the dual. By LP duality, both LPs have the same optimum value — the value of the game.