Graphical method to solve $m\times 2$ game
A game where one player has more than $m$ course of action and the other player has only two course of action is called $m\times 2$ game.
Graphical method can be used to solve a $2\times n$ or $m\times 2$ game or a game reducible to either $2\times n$ or $m\times 2$ after applying a dominance property.
In this tutorial we will discuss about the graphical method to solve a $m\times 2$ game or a game reducible to $m\times 2$ after applying dominance property.
Step by Step Procedure to solve m by 2 game
Consider the $m\times 2$ game. In this game player $A$ has $m$ strategies and player $B$ has $2$ strategies. Assume that the game does not have a saddle point.
| Player A \ Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | $a_{11}$ | $a_{12}$ |
| $A_2$ | $a_{21}$ | $a_{22}$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
| $A_m$ | $a_{m1}$ | $a_{m2}$ |
Following are the steps to solve $m\times 2$ game using graphical method.
Step 1
Construct two vertical axis, axis 1 and axis 2, by taking suitable scale. (Graph is just a sample)
Step 2
Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.
Step 3
Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$. (Graph is just a sample)
Step 4
Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$ and it identifies the two critical moves (strategies) of player $A$.
Suppose the upper envelope is created by $A_1$ and $A_m$ strategy. The lowest point on the upper envelope gives the minimax point. This shows that the two critical moves for player $A$ are $A_1$ and $A_m$.
Using the above selection the reduced game becomes
| Player A \ Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | $a_{11}$ | $a_{12}$ |
| $A_m$ | $a_{m1}$ | $a_{m2}$ |
Step 5
Solve the reduced game (i.e. $2\times 2$) using maximin criterion. (if saddle point exists then find the value of the game otherwise use mixed strategy technique to find the value of the game.)
m by 2 game by graphical method Example 1
Solve the game with the following payoff for player A.
| Player A / Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | -2 | 5 |
| $A_2$ | -5 | 3 |
| $A_3$ | 0 | -2 |
| $A_4$ | -3 | 0 |
| $A_5$ | 1 | -4 |
Solution
Given game is $5\times 2$. That is for Player $A$ has five strategies and player $B$ has two strategies. Let us check using maximin-minimax principle whether the game has a saddle point or not.
| B1 | B2 | |
|---|---|---|
| A1 | -2 | 5 |
| A2 | -5 | 3 |
| A3 | 0 | -2 |
| A4 | -3 | 0 |
| A5 | -1 | -4 |
Step 1
Apply maximin-minimax principle to check whether saddle point exists or not.
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | -2 | 5 | -2 |
| A2 | -5 | 3 | -5 |
| A3 | 0 | -2 | -2 |
| A4 | -3 | 0 | -3 |
| A5 | -1 | -4 | -4 |
| ColMax | 0 | 5 |
Thus $Max(min) = Max(-2, -5, -2, -3, -4, )=-5$ and $Min(max)=Min(0, 5)=0$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.
Step 2
Draw two vertical axis, axis 1 and axis 2. Mark appropriate scale on each axis. The two axis represents strategies for player B. Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.
Let the player B play the strategy $B_1$ with probability $q$ and strategy $B_2$ with probability $1-q$.
Expected payoff for player B for any pure strategy of player A would be given by
| A's Strategy | Expected Payoff for B |
|---|---|
| $A_1$ | $E_1=-2q+5(1-q) = 7q+5$ |
| $A_2$ | $E_2=-5q+3(1-q)=-8q+3$ |
| $A_3$ | $E_3 = 0q+(-2)(1-q)=2q-2$ |
| $A_4$ | $E_4=-3q+0(1-q)= -3q$ |
| $A_5$ | $E_5 = -1q+(-4)(1-q)=3q-4$ |
Step 3
Draw the graph of expected payoff $E_1,E_2,E_3, E_4, E_5$ against $q$
Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$.
Step 4
Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$.
The upper envelope is created by $A_1$ and $A_5$ strategy.
The two critical moves (strategies) of player $A$ are $A_1$ and $A_5$.
The lowest point $P$ on the upper envelope gives the minimax point.
Step 5
The reduced game (i.e. $2\times 2$) is
| Player A / Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | -2 | 5 |
| $A_5$ | 1 | -4 |
Apply maximin-minimax principle on the reduced $2\times 2$ game.
| B1 | B2 | |
|---|---|---|
| A1 | -2 | 5 |
| A5 | 1 | -4 |
For each row of the payoff matrix, select the minimum element and call it RowMin.
$$ \begin{equation*} \text{i.e., } \min_{j} a_{ij}, i=1,2,\cdots, m. \end{equation*} $$
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | -2 | 5 | -2 |
| A5 | 1 | -4 | -4 |
For each column of the payoff matrix, select the maximum element and call it ColMax.
$$ \begin{equation*} \text{i.e., } \max_{i} a_{ij}, j=1,2,\cdots, n. \end{equation*} $$
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | -2 | 5 | -2 |
| A5 | 1 | -4 | -4 |
| ColMax | 1 | 5 |
From each RowMin, obtain the maximum value, i.e., $Max(RowMin)$.
$$ \begin{equation*} \text{i.e., } \max_{i}\min_{j} a_{ij}=\underline{v}. \end{equation*} $$
Thus $Max(min) = Max(-2, -4)=-2$
For each ColMax, obtain the minimum value, i.e. $Min(ColMax)$.
$$ \begin{equation*} \text{i.e., } \min_{j}\max_{i} a_{ij}=\overline{v}. \end{equation*} $$
Thus $Min(max)=Min(1, 5)= 1$.
Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.
Hence the optimal strategy for the reduced game can be obtained by algebraic method.
$$ \begin{aligned} p_1 &=\frac{d-c}{(a+d)-(b+c)}\\ &=\frac{(-4)-1}{(-2)+(-4)-(5+1)}\\ &=\frac{5}{12}\\ p_2 &= 1-p_1\\ &=\frac{7}{12} \end{aligned} $$
and the optimal strategies for player B can be determined by
$$ \begin{aligned} q_1 &= \frac{d-b}{(a+d)-(b+c)}\\ &=\frac{(-4)-5}{(-2) +(-4) -(5+1)}\\ &=\frac{9}{12}=\frac{3}{4}\\ q_2 &= 1-q_1\\ &=\frac{1}{4}. \end{aligned} $$
The optimal strategies for player A can be written as
$$ \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2 & A_3 & A_4 & A_5\\ p_1 & 0 & 0 & 0 & p_2 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2 & A_3 & A_4 & A_5\\ \frac{5}{12} & 0 & 0 & 0 & \frac{7}{12} \end{bmatrix} \end{aligned} $$
and the optimal strategies for player B can be written as
$$ \begin{aligned} S_{B} &= \begin{bmatrix} B_1 & B_2\\ q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2\\ \frac{3}{4} & \frac{1}{4} \end{bmatrix} \end{aligned} $$
And the value of the game for player A is given by
$$ \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{(-2)(-4)-(5)(1)}{(-2)+(-4)-(5+1)}\\ &=\frac{3}{12}=\frac{1}{4}. \end{aligned} $$
Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{5}{12}$ and strategy $A_5$ with probability $\dfrac{7}{12}$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{3}{4}$ and strategy $B_2$ with probability $\dfrac{1}{4}$. And value of the game for player A is $\dfrac{1}{4}$ and for player B is $-\dfrac{1}{4}$.
m by 2 game by graphical method Example 2
Solve the game with the following payoff for player A.
| Player A / Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | 2 | 4 |
| $A_2$ | 2 | 3 |
| $A_3$ | 3 | 2 |
| $A_4$ | -2 | 6 |
Solution
Given game is $4\times 2$. That is for Player $A$ has four strategies and player $B$ has two strategies. Let us check using maximin-minimax principle whether the game has a saddle point or not.
| B1 | B2 | |
|---|---|---|
| A1 | 2 | 4 |
| A2 | 2 | 3 |
| A3 | 3 | 2 |
| A4 | -2 | 6 |
Step 1
Apply maximin-minimax principle to check whether saddle point exists or not.
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | 2 | 4 | 2 |
| A2 | 2 | 3 | 2 |
| A3 | 3 | 2 | 2 |
| A4 | -2 | 6 | -2 |
| ColMax | 3 | 6 |
Thus $Max(min) = Max(2, 2, 2, -2, )=2$ and $Min(max)=Min(3, 6)=3$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.
Step 2
Draw two vertical axis, axis 1 and axis 2. Mark appropriate scale on each axis. The two axis represents strategies for player B. Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.
Let the player B play the strategy $B_1$ with probability $q$ and strategy $B_2$ with probability $1-q$.
Expected payoff for player B for any pure strategy of player A would be given by
| A's Strategy | Expected Payoff for B |
|---|---|
| $A_1$ | $E_1= 2q+4(1-q) = -2q+4$ |
| $A_2$ | $E_2= 2q+3(1-q)=-q+3$ |
| $A_3$ | $E_3= 3q+2(1-q)=q+2$ |
| $A_4$ | $E_4=-2q+6(1-q)= -8q+6$ |
Step 3
Draw the graph of expected payoff $E_1,E_2,E_3, E_4$ against $q$
Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$.
Step 4
Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$.
The upper envelope is created by $A_1$ and $A_3$ strategy.
The two critical moves (strategies) of player $A$ are $A_1$ and $A_3$.
The lowest point $P$ on the upper envelope gives the minimax point.
Step 5
The reduced game (i.e. $2\times 2$) is
| Player A / Player B | $B_1$ | $B_2$ |
|---|---|---|
| $A_1$ | 2 | 4 |
| $A_3$ | 3 | 2 |
Apply maximin-minimax principle on the reduced $2\times 2$ game.
| B1 | B2 | |
|---|---|---|
| A1 | 2 | 4 |
| A3 | 3 | 2 |
For each row of the payoff matrix, select the minimum element and call it RowMin.
$$ \begin{equation*} \text{i.e., } \min_{j} a_{ij}, i=1,2,\cdots, m. \end{equation*} $$
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | 2 | 4 | 2 |
| A3 | 3 | 2 | 2 |
For each column of the payoff matrix, select the maximum element and call it ColMax.
$$ \begin{equation*} \text{i.e., } \max_{i} a_{ij}, j=1,2,\cdots, n. \end{equation*} $$
| B1 | B2 | RowMin | |
|---|---|---|---|
| A1 | 2 | 4 | 2 |
| A3 | 3 | 2 | 2 |
| ColMax | 3 | 4 |
From each RowMin, obtain the maximum value, i.e., $Max(RowMin)$.
$$ \begin{equation*} \text{i.e., } \max_{i}\min_{j} a_{ij}=\underline{v}. \end{equation*} $$
Thus $Max(min) = Max(2, 2)=2$
For each ColMax, obtain the minimum value, i.e. $Min(ColMax)$.
$$ \begin{equation*} \text{i.e., } \min_{j}\max_{i} a_{ij}=\overline{v}. \end{equation*} $$
Thus $Min(max)=Min(3, 4)= 3$.
Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.
Hence the optimal strategy for the reduced game can be obtained by algebraic method.
$$ \begin{aligned} p_1 &=\frac{d-c}{(a+d)-(b+c)}\\ &=\frac{2-3}{(2+2)-(4+3)}\\ &=\frac{-1}{-3}\\ &=\frac{1}{3}\\ p_2 &= 1-p_1\\ &=\frac{2}{3} \end{aligned} $$
and the optimal strategies for player B can be determined by
$$ \begin{aligned} q_1 &= \frac{d-b}{(a+d)-(b+c)}\\ &=\frac{2-4}{(2+2)-(4+3)}\\ &=\frac{-2}{-3}\\ &=\frac{2}{3}\\ q_2 &= 1-q_1\\ &=\frac{1}{3}. \end{aligned} $$
The optimal strategies for player A can be written as
$$ \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2 & A_3 & A_4 \\ p_1 & 0 & p_2 & 0 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2 & A_3 & A_4 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 \end{bmatrix} \end{aligned} $$
and the optimal strategies for player B can be written as
$$ \begin{aligned} S_{B} &= \begin{bmatrix} B_1 & B_2\\ q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2\\ \frac{2}{3} & \frac{1}{3} \end{bmatrix} \end{aligned} $$
And the value of the game for player A is given by
$$ \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{2*2-4*3}{(2+2)-(4+3)}\\ &=\frac{-8}{-3}\\ &=\frac{8}{3}. \end{aligned} $$
Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{1}{3}$ and strategy $A_3$ with probability $\dfrac{2}{3}$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{2}{3}$ and strategy $B_2$ with probability $\dfrac{1}{3}$. And value of the game for player A is $\dfrac{8}{3}$ and for player B is $-\dfrac{8}{3}$.
Endnote
In this tutorial, you learned about graphical method to solve $m\times 2$ game and how to use graphical method to solve $m\times 2$ game with illustrated example.
To learn more about different methods to solve a game please refer to the following tutorials:
Let me know in the comments if you have any questions on Graphical method to solve $m\times 2$ game and your thought on this article.
