Pay-off matrices and play-safe strategiesAQA A-Level Further Maths: Revision notes

In a two-person zero-sum game two players each choose a strategy at the same time, and one player's gain is exactly the other's loss. The game is shown as a pay-off matrix:

• rows are the strategies of the row player (Rowan), columns are those of the column player (Colin);
• each entry is the pay-off to the row player for that pair of choices;
• a negative entry means Rowan loses, which is a gain for Colin.

Colin's pay-off matrix is the negative of the transpose of Rowan's: his strategies become the rows and every sign is changed. To construct a matrix from a description, work out the pay-off for every combination, row by row, using the stated rule. To interpret an entry, say who gains, by how much, and in what units. For example, if the entry in row 2, column 3 is $-5$ and the pay-off is in pounds, then if Rowan plays his second strategy and Colin his third, Rowan loses £5 to Colin.

A play-safe strategy guards against the worst outcome, assuming the opponent plays to hurt you.

• Rowan (maximiser): find the minimum of each row, then choose the row with the largest of these. This is the maximin.
• Colin (minimiser of Rowan's pay-off): find the maximum of each column, then choose the column with the smallest of these. This is the minimax.

Example, for $\begin{pmatrix} 3 & -1 & 2 \\ 1 & 0 & 4 \\ -2 & 5 & 1 \end{pmatrix}$: row minima $-1, 0, -2$ so maximin $=0$, play row 2. Column maxima $3, 5, 4$ so minimax $=3$, play column 1. Every game has play-safe strategies; they only give a stable outcome in some games. If Colin's matrix is used instead, Colin finds the minimum of each of his rows and chooses the largest, which gives the same result.

A game has a stable solution (a saddle point) when maximin $=$ minimax.

• The entry at that row and column is the value of the game.
• It is stable because neither player can do better by changing strategy alone: the entry is the smallest in its row, so if Rowan stays in that row Colin has no better column, and the largest in its column, so if Colin stays in that column Rowan has no better row.
• At a stable solution, both players use their play-safe strategies.

If maximin $\neq$ minimax, there is no stable solution (it always holds that maximin $\leq$ minimax), and the value of the game lies between them. For the matrix above the value is between $0$ and $3$. Optimal play then needs mixed strategies (covered later).

To prove whether a stable solution exists, show the working in order:

1. Write the row minima and state the maximin and the row it comes from.
2. Write the column maxima and state the minimax and the column it comes from.
3. Compare. Equal means a stable solution exists, at that row and column, with value equal to the common number. Different means no stable solution exists.

Worked example: $\begin{pmatrix} 4 & 2 & 5 \\ 3 & 1 & 6 \\ 1 & 0 & 2 \end{pmatrix}$. Row minima $2, 1, 0$, so maximin $=2$ (row 1). Column maxima $4, 2, 6$, so minimax $=2$ (column 2). They are equal, so the game is stable at (row 1, column 2) with value $2$.

To explain the value in context: when both play safe, the row player gains $2$ units per play and the column player loses $2$.