Oddments

Oddments#

A very efficient method for solving \(2\times 2\) matrix games with mixed strategy solutions (MSS) is using the technique of oddments. We calculate oddments by taking the absolute value differences across the rows and down the colums. From the oddments, we can calculate the proportions needed for \(\vec r\) and \(\vec c\) and then determine the overall game value.

Teaching Example

Given the example game below, first verify that no domination exists for either player. Then, solve for Rose’s and Colin’s optimal strategy mixture (OSM).

\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B&\\ \hline A&7&2\\B&-1&6 \end{array}\end{array}\end{split}\]

To begin, we calculate the row oddments which are the absolute value difference of the entries in the row.

\[\begin{split}\begin{array}{ll}&\hspace{9mm}\text{Colin}\\\text{Rose}&\begin{array}{r|ccccccc}&A&B&&\text{Row Diffs}&\text{Oddments}\\\hline A&7&2&&|7-2|&5&\\B&-1&6&&|-1-6|&7 \end{array}\end{array}\end{split}\]

Next, we calculate the column oddments in a similar way.

\[\begin{split}\begin{array}{ll}&\hspace{9mm}\text{Colin}\\\text{Rose}&\begin{array}{r|ccccccc}&A&B&&\text{Row Diffs}&\text{Oddments}\\\hline A&7&2&&|7-2|&5&\\B&-1&6&&|-1-6|&7\\\text{Column Diffs}&|7-(-1)|&|2-6|\\&8&4 \end{array}\end{array}\end{split}\]

Creating the optimal strategy proportions (OSP) works as follows:

  • OSP for Rose A is equal to $\(\frac{\text{Row B Oddment}}{\text{Sum of Oddments}}\)$

  • OSP for Rose B is equal to $\(\frac{\text{Row A Oddment}}{\text{Sum of Oddments}}\)$

Notice that the numerators swap position compared to the oddments as demonstrated with arrows below.

\[\begin{split}\begin{array}{ll}&\hspace{9mm}\text{Colin}\\\text{Rose}&\begin{array}{r|ccccccc}&A&B&&\text{Row Diffs}&\text{Oddments}&&\text{Rose OSP}\\\hline A&7&2&&|7-2|&5&\searrow&\frac{7}{12}\\B&-1&6&&|-1-6|&7&\nearrow&\frac{5}{12}\\\text{Column Diffs}&|7-(-1)|&|2-6|&&\\\text{Oddments}&8&4&&&\end{array}\end{array}\end{split}\]

Colin’s OSP are calculated in an analagous way as shown below.

\[\begin{split}\begin{array}{ll}&\hspace{9mm}\text{Colin}\\\text{Rose}&\begin{array}{r|ccccccc}&A&B&&\text{Row Diffs}&\text{Oddments}&&\text{Rose OSP}\\\hline A&7&2&&|7-2|&5&\searrow&\frac{7}{12}\\B&-1&6&&|-1-6|&7&\nearrow&\frac{5}{12}\\\text{Column Diffs}&|7-(-1)|&|2-6|&&\\\text{Oddments}&8&4&&&\\&\searrow&\swarrow\\ \text{Colin OSP}&\frac{4}{12}&\frac{8}{12} \end{array}\end{array}\end{split}\]

Thus, we have from the OSP the following optimal strategy sets for Rose and Colin.

\[\begin{split}\vec r = \left[\begin{array}{rr} \frac{7}{12}\\ \frac{5}{12}\end{array}\right]\hspace{5mm}\text{and}\hspace{5mm}\vec c = \left[\begin{array}{rr}\frac{1}{3}\\ \frac{2}{3}\end{array}\right]\end{split}\]

We find the game value by determining Rose’s (or Colin’s) expected value given the opponent plays her or his optimal strategy proprotions:

\[E(\text{Rose }A) = (7) \frac{1}{3} + (2) \frac{2}{3} =\frac{11}{3}\]

If we insert the game solver code, we can evaluate our solution to verify it. Remember that we enter the matrix game values as

example_game = np.array([[7,2],[-1,6]])

We first must initialize python:

Hide code cell source

 
## Do not change this cell, only execute it. 
## This cell initializes Python so that numpy and scipy packages are ready to use.

%matplotlib inline
import matplotlib.pyplot as plots
plots.style.use('fivethirtyeight')
import numpy as np
import scipy as sp
from scipy.optimize import linprog

Next, the game solver code:

Hide code cell source

 
def solver(payoff_matrix):
    payoff_matrix = np.array(payoff_matrix)
    row_counts, col_counts = payoff_matrix.shape
# 1. Shift matrix to ensure all values are positive
# This ensures the game value V > 0
    shift = abs(payoff_matrix.min()) + 1
    A_shifted = payoff_matrix + shift
# 2. Solve for Row Player (Maximizer)
# We minimize 1/V = sum(p_i/V). Let x_i = p_i/V
# Objective: min (1 * x1 + 1 * x2 + ...)
    c_row = np.ones(row_counts)
# Constraints: A_shifted.T * x >= 1  =>  -A_shifted.T * x <= -1
    A_ub_row = -A_shifted.T
    b_ub_row = -np.ones(col_counts)
    res_row = linprog(c_row, A_ub=A_ub_row, b_ub=b_ub_row, method='highs')
# Game Value V = 1 / sum(x_i)
    v_shifted = 1 / res_row.fun
    row_strategy = res_row.x * v_shifted
# 3. Solve for Column Player (Minimizer) using the Dual
# Objective: max (1 * y1 + 1 * y2 + ...) => min (-1 * y1 - 1 * y2)
    c_col = -np.ones(col_counts)
# Constraints: A_shifted * y <= 1
    A_ub_col = A_shifted
    b_ub_col = np.ones(row_counts)
    res_col = linprog(c_col, A_ub=A_ub_col, b_ub=b_ub_col, method='highs')
    col_strategy = res_col.x * (1 / abs(res_col.fun))
# Final Game Value
    game_value = v_shifted - shift
    print(f"Game Value: {np.round(game_value,3)}")
    print(f"Rose Strategy: {np.round(row_strategy,3).tolist()}")
    print(f"Colin Strategy: {np.round(col_strategy,3).tolist()}")
    return None

example_game = np.array([[7,2],[-1,6]])

solver(example_game)

Game Value: 3.667
Rose Strategy: [0.583, 0.417]
Colin Strategy: [0.333, 0.667]

Hence, we have verified that our oddments solution above was indeed correct.

Additional Practice

Let’s now provide some game examples so that we can practice how to compute matrix game solutions using oddments.

Practice Problem 1#

\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B&\\ \hline A&1&3\\B&11&-1 \end{array}\end{array}\end{split}\]
# Do your work in this cell:

p1 = np.array([[1,3],[11,-1]])

solver(p1)

Game Value: 2.429
Rose Strategy: [0.857, 0.143]
Colin Strategy: [0.286, 0.714]

Practice Problem 2#

\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B&\\ \hline A&0&6\\B&2&0 \end{array}\end{array}\end{split}\]
# Do your work in this cell:

p2 = np.array([[0,6],[2,0]])

solver(p2)

Game Value: 1.5
Rose Strategy: [0.25, 0.75]
Colin Strategy: [0.75, 0.25]

Practice Problem 3#

\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B&\\ \hline A&-3&9\\B&0&-4 \end{array}\end{array}\end{split}\]
# Do your work in this cell:

p3 = np.array([[-3,9],[0,-4]])

solver(p3)

Game Value: -0.75
Rose Strategy: [0.25, 0.75]
Colin Strategy: [0.812, 0.188]
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