# 11.3: Reduction by Dominance

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Sometimes an $$m \times n$$ game matrix can be reduced to a $$2 \times 2$$ matrix by deleting certain rows and columns. A row can be deleted if there exists another row that will produce a payoff of an equal or better value. Similarly, a column can be deleted if there is another column that will produce a payoff of an equal or better value for the column player. The row or column that produces a better payoff for its corresponding player is said to dominate the row or column with the lesser payoff.

## Example $$\PageIndex{1}$$

For the following game, determine the optimal strategy for both the row player and the column player, and find the value of the game.

$G=\left[\begin{array}{ccc} -2 & 6 & 4 \\ -1 & -2 & -3 \\ 1 & 2 & -2 \end{array}\right] \nonumber$

Solution

We first look for a saddle point and determine that none exist. Next, we try to reduce the matrix to a $$2 \times 2$$ matrix by eliminating the dominated row.

Since every entry in row 3 is larger than the corresponding entry in row 2, row 3 dominates row 2. Therefore, a rational row player will never play row 2, and we eliminate row 2. We get

$\left[\begin{array}{ccc} -2 & 6 & 4 \\ 1 & 2 & -2 \end{array}\right] \nonumber$

Now we try to eliminate a column. Remember that the game matrix represents the payoffs for the row player and not the column player; therefore, the larger the number in the column, the smaller the payoff for the column player.

The column player will never play column 2, because it is dominated by both column 1 and column 3. Therefore, we eliminate column 2 and get the modified matrix, M, below.

$\mathrm{M}=\left[\begin{array}{cc} -2 & 4 \\ 1 & -2 \end{array}\right] \nonumber$

To find the optimal strategy for both the row player and the column player, we use the method learned in Section 11.2.

Let the row player's strategy be $$\mathrm{R}=\left[\begin{array}{ll} \mathrm{r} & 1-\mathrm{r} \end{array}\right]$$, and the column player's be strategy be $$C=\left[\begin{array}{c} c \\ 1-c \end{array}\right]$$

To find the optimal strategy for the row player, we, first, find the product RM as below.

$\left[\begin{array}{lll} \mathrm{r} & 1-\mathrm{r} \end{array}\right]\left[\begin{array}{cc} -2 & 4 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} -3 r+1 & 6 r-2 \end{array}\right] \nonumber$

By setting the entries equal, we get

$-3r + 1 = 6r - 2 \nonumber$

or

$r = 1/3 \nonumber. \nonumber$

Therefore, the optimal strategy for the row player is $$\left[\begin{array}{ll} 1/ 3 & 2 / 3 \end{array}\right]$$, but relative to the original game matrix it is $$\left[\begin{array}{lll} 1 / 3 & 0 & 2/3 \end{array}\right]$$.

To find the optimal strategy for the column player we, first, find the following product.

$\left[\begin{array}{cc} -2 & 4 \\ 1 & -2 \end{array}\right]\left[\begin{array}{c} c \\ 1-c \end{array}\right]=\left[\begin{array}{c} -6 c+4 \\ 3 c-2 \end{array}\right] \nonumber$

We set the entries in the product matrix equal to each other, and we get,

$-6c + 4 = 3c - 2 \nonumber$

or

$c = 2/3 \nonumber. \nonumber$

Therefore, the optimal strategy for the column player is $$\left[\begin{array}{l} 2 / 3 \\ 1 / 3 \end{array}\right]$$, but relative to the original game matrix, the strategy for the column player is $$\left[\begin{array}{c} 2 / 3 \\ 0 \\ 1 / 3 \end{array}\right]$$.

To find the expected value, V, of the game, we have two choices: either to find the product of matrices R, M and C, or multiply the optimal strategies relative to the original matrix to the original matrix. We choose the first, and get

$\begin{array}{l} \mathrm{V}=\left[\begin{array}{lll} 1 / 3 & 2 / 3 \end{array}\right]\left[\begin{array}{cc} -2 & 4 \\ 1 & -2 \end{array}\right]\left[\begin{array}{l} 2 / 3 \\ 1 / 3 \end{array}\right] \\ \mathrm{V}=\left[\begin{array}{l} 0 \end{array}\right] \end{array} \nonumber$

Therefore, if both players play their optimal strategy, the value of the game is zero.

We summarize as follows:

## Reduction by Dominance

1. Sometimes an $$m \times n$$ game matrix can be reduced to a $$2 \times 2$$ matrix by deleting dominated rows and columns.
2. A row is called a dominated row if there exists another row that will produce a payoff of an equal or better value. That happens when there exists a row whose every entry is larger than the corresponding entry of the dominated row.
3. A column is called a dominated column if there exists another column that will produce a payoff of an equal or better value. This happens when there exists a column whose every entry is smaller than the corresponding entry of the dominated row.

This page titled 11.3: Reduction by Dominance is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Rupinder Sekhon and Roberta Bloom via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.