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11.3: Reduction by Dominance

  • Page ID
    37941
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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.