11.3: Reduction by Dominance

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.