猜硬币 [[1,-1],[-1,1]] → 值 0，双方各以 0.5/0.5 混合 没有鞍点的教科书博弈：极大值 −1 < 极小值 1，双方必须随机化。线性规划（矩阵移位 +2）经两次枢轴得 z = y = (0.25, 0.25)，即双方各以 50/50 混合，公平博弈值恰为 0——在对方最优应对下谁都占不到便宜。Zero-sum game — 2 row strategies × 2 column strategies (payoffs to the row player)
Payoff matrix:
[ 1 -1 ]
[ -1 1 ]
Saddle-point test:
Row minima: -1, -1 → maximin = -1 (row 1)
Column maxima: 1, 1 → minimax = 1 (column 1)
maximin < minimax → no saddle point; solve mixed strategies by linear programming.
Mixed strategy via LP (matrix shifted by +2 to keep entries ≥ 1; value shifted back by −2):
LP: max z1 + z2 s.t. Bz ≤ 1, z ≥ 0
pivot 1: z1 enters, s1 leaves → Σz = 0.3333
pivot 2: z2 enters, s2 leaves → Σz = 0.5
Row player mixed strategy x = (0.5, 0.5)
Column player mixed strategy q = (0.5, 0.5)
Game value v = 0
Security checks:
Row guarantee: min over columns of xᵀA = 0 ≥ v = 0 ✓
Column cap: max over rows of A·q = 0 ≤ v = 0 ✓
Minimax equality verified: 1/Σz = 1/Σy = 2 (shifted) → v = 0 ✓
反对称 3×3 矩阵 [[0,1,-2],[-1,0,3],[2,-3,0]] → 值 0（公平博弈） 反对称矩阵（Aᵀ = −A）必为公平博弈：双方共享同一最优混合，且值为 0。本例无鞍点，线性规划移位 +4 后求解，验证过的解保证行方对每一列至少得 0，列方将每一行压至至多 0。Zero-sum game — 3 row strategies × 3 column strategies (payoffs to the row player)
Payoff matrix:
[ 0 1 -2 ]
[ -1 0 3 ]
[ 2 -3 0 ]
Saddle-point test:
Row minima: -2, -1, -3 → maximin = -1 (row 2)
Column maxima: 2, 1, 3 → minimax = 1 (column 2)
maximin < minimax → no saddle point; solve mixed strategies by linear programming.
Mixed strategy via LP (matrix shifted by +4 to keep entries ≥ 1; value shifted back by −4):
LP: max z1 + z2 + z3 s.t. Bz ≤ 1, z ≥ 0
...
Game value v = 0
Security checks:
Row guarantee: min over columns of xᵀA = 0 ≥ v = 0 ✓
Column cap: max over rows of A·q = 0 ≤ v = 0 ✓