7.08a–7.08d - Zero-sum games and pure strategies

You already know

Subtracting the same constant from every number preserves their order. For example, $8 > 6$ and $8 - 5 > 6 - 5$. Multiplying an inequality by $-1$ reverses its direction.

Zero-sum game

In a two-person zero-sum game, the two players' pay-offs add to zero for every pair of choices. A pay-off of $a$ for A means a pay-off of $-a$ for B. A pay-off matrix records one player's pay-offs, with the players' choices labelling its rows and columns. For example, A's pay-off $-2$ means B gains $2$, not that both players lose $2$.

Converting a constant-sum game

If the original pay-offs satisfy $a + b = K$ for every outcome, define

The transformed game has the same best choices as the original game. Recover an original pay-off by adding $K/2$.

Not in the formula booklet: learn it.

Worked example 1: sharing ten points

Question. Convert the opening game to zero-sum form. Interpret the transformed outcome when A plays Q and B plays Y.

Plan. Subtract half the total from each player's score and display A's transformed matrix.

Why this step? B's transformed score is $(10 - a) - 5 = -(a - 5)$, so the transformed scores are opposites.

Answer. The displayed matrix represents the equivalent zero-sum game. At $(\mathrm{Q}, \mathrm{Y})$, both transformed pay-offs are $0$; both original scores are $5$.

Check. At $(\mathrm{P}, \mathrm{Y})$, A's transformed score is $-1$ and B's is $1$. Adding $5$ gives the original scores $4$ and $6$, whose sum is $10$.

Worked example 2: changing the total prize

Question. A's original pay-off matrix is still $\begin{pmatrix}8 & 4 \\ 6 & 5\end{pmatrix}$, but the players now share twelve points. Convert the game to zero-sum form and interpret outcome $(\mathrm{Q}, \mathrm{Y})$.

Plan. Subtract $6$ from both original scores, rather than reusing the previous subtraction of $5$.

Why this step? Zero-sum conversion must use the constant total for this game. A's score of $5$ is now below an equal share.

Answer. At $(\mathrm{Q}, \mathrm{Y})$, the transformed pay-offs are $-1$ for A and $1$ for B. The original scores are $5$ and $7$.

Check. $-1 + 1 = 0$, while $5 + 7 = 12$. A still prefers Q to P when B chooses Y, because $-1 > -2$.

Watch out

Wrong: “Every two-person game can be made zero-sum by subtracting one constant.” This works for a constant total. If the totals vary between outcomes, subtracting one fixed amount from both players cannot make every total zero.

Dominance

One strategy dominates another if it gives its player at least as good a pay-off against every opposing strategy, and a strictly better pay-off against at least one. If every comparison is strict, it is strict dominance; otherwise it is weak dominance. In an A pay-off matrix, a dominating row has entries at least as large, while a dominating column has entries at most as large. Identical rows or columns are equivalent choices rather than strict improvements.

Worked example 3: rows and columns favour different directions

Question. Reduce the following A pay-off matrix as far as possible by dominance. A chooses P, Q or R; B chooses X, Y or Z.

Plan. Look for rows that A can improve in every column, then compare the remaining columns from B's viewpoint.

Delete Q and R. The remaining row is $(4, 1, 6)$, with the original column labels X, Y, Z.

Why this step? B wants A's pay-off to be smaller, so the smallest remaining entry is B's best choice.

Answer. A reduced game consists of row P and column Y, with A pay-off $1$.

Check. If B chooses Y, switching from P to Q or R reduces A's pay-off from $1$ to $0$ or $-1$. If A chooses P, switching from Y increases A's pay-off, so B gains nothing by doing so.

Worked example 4: a tied comparison

Question. Reduce the matrix

using dominance.

Plan. Allow equal pay-offs in a dominance comparison, but require no worse outcome in any column.

Delete Q and R. In the remaining row, $1 < 4$ and $1 < 6$, so B can delete X and Z in favour of Y.

Why this step? Replacing Q by P never costs A anything: X gives a tie, while Y and Z give A an improvement.

Answer. The reduced game is again $(\mathrm{P}, \mathrm{Y})$, with A pay-off $1$.

Check. The equality $4 = 4$ creates no reversal of preference. It would be different if Q's X entry were $5$: then neither P nor Q would dominate the other.

Watch out

Wrong: “Delete the row with the smallest total.” A row total assumes an unstated pattern of opposing choices. For rows $(4, 1)$ and $(2, 3)$, the first is better against X but worse against Y; neither dominates the other.