Two-player zero-sum games, the pay-off matrix, play-safe strategies, saddle points and stable solutions, dominance to reduce a game, and mixed strategies including conversion to linear programming.

What this dot point is asking

AQA wants you to analyse a two-player zero-sum game from its pay-off matrix, find each player's play-safe strategy, identify a saddle point and stable solution when one exists, reduce a game using dominance, and find optimal mixed strategies, including converting a larger game into a linear programming problem.

Pay-off matrix and play-safe

Saddle points and stability

When the maximin equals the minimax, neither player can do better by changing alone, so the game is stable and the solution is a pair of pure strategies.

Dominance and mixed strategies

Before solving, reduce the game using dominance. A row is dominated if another row is at least as good entry by entry (never worse for the row player), and a dominated row can be deleted because the player would never choose it. Similarly a column is dominated if another column is never worse for the column player (that is, has smaller or equal entries). Repeatedly removing dominated rows and columns can shrink a large game to a $2\times 2$ one that is solvable by hand.

If no saddle point exists after reduction, each player uses a mixed strategy, choosing each option with a fixed probability so that the opponent cannot exploit a predictable pattern.

For larger games that do not reduce to $2\times 2$, convert to a linear programming problem: add a constant to every pay-off so that all entries are positive (which does not change the optimal strategies, only shifts the value), then set up variables for the strategy probabilities and maximise the game value subject to the constraints that each pure-column expected pay-off is at least the value.

The reason the constant is needed is that the linear programming formulation treats the game value $V$ as a variable to be maximised, and the standard simplex method assumes a non-negative objective. Shifting every entry by a positive constant $k$ makes the value provably positive, so dividing through by $V$ to scale the probability variables is safe (you cannot divide by zero or by a negative). After solving, subtract $k$ to recover the true value of the original game. The strategy probabilities themselves are unaffected by the shift, because adding a constant to every outcome changes what each player wins by the same amount whatever they do, so it cannot change which mix is best.

Why mixed strategies work

A mixed strategy defends against an opponent who could otherwise exploit a predictable choice. If the row player always picked the row with the largest single entry, the column player would simply respond with the column that punishes that row hardest. By randomising in the right proportions, the row player makes every column the opponent might choose yield the same expected pay-off, so the opponent has no profitable reply and cannot push the outcome below the game value. This is the meaning of the equalising condition $E_1 = E_2$ used in the worked solution: the optimal $p$ is the one that leaves the column player indifferent. The same logic, applied to the columns, gives the column player's optimal mix, and the two values agree at the game value $V$, which is the central guarantee of the minimax theorem for zero-sum games.

A useful check on a $2 \times 2$ solution is that the probabilities lie strictly between $0$ and $1$. If solving for $p$ gives a value outside $[0, 1]$, the game in fact has a saddle point and one of the pure strategies is optimal after all, so you should return to the maximin and minimax comparison rather than forcing a mixed solution.

The standard exam sequence is therefore: write the pay-off matrix from the row player's viewpoint, check for a saddle point by comparing maximin and minimax, reduce by dominance if no saddle point is found, then either solve the resulting $2\times 2$ game by equating expected pay-offs or set up a linear program for a larger one. Keep the constant you add for the linear program in mind so you can subtract it back to recover the true game value at the end.