Solving a 2*3 game with graphical method in game theory

Solve the following game. $$\begin{pmatrix} 1 & 2& 3 \\ 4 & 2 & 1 \\ \end{pmatrix}$$

Since this is a $2\times3$ matrix I used the graphical method to solve it.When I draw the graph I can see that the value of the game is $2$.
But how to find the probabilities of the strategies of Player $1$ ( Player who plays along the rows). Player $1$' s strategy is to maximize his minimum gain. The problem here is once the graph is drawn there is no single point representing the highest point in the lower envelope.

If player $1$ plays with probabilities $p_1$, $p_2$ and $p_1+p_2=1$,

Then the player $1$'s expected payouts are,
\begin{align*} E_1&= p_1 + 4 p_2\\ E_2&=2p_1+2p_2\\ E_3&=3p_1+p_2 \end{align*}

After drawing the graph what two equations should be equated to find the value of $p_i$'s.

Also when I come to calculate player $2$'s strategies a B has $3$ strategies played with probabilities $q_1$, $q_2$, $q_3$ this can't be done graphically? Can it?

Is there a method to find the player $2$'s strategies from the values of player $1$'s?
If not how can I find player $2$'s strategies?

Please help to solve this problem.

• This type of game is where the row player tries to choose the largest value in the matrix and the column player tries to choose smallest, right? – john.abraham Dec 24 '13 at 9:37

2 Answers

$$\begin{array}{c|ccc} & C_1&C_2&C_3\\\hline R_A &1&2&3\\R_B&4&2&1\\\end{array}$$

• Player $1$ choose row $A$ with probability $p_A$ and row $B$ with $p_B$ ($p_B=1-p_A$)
• Player $2$ choose column $1$ with probability $p_1$, column $2$ with $p_2$ and column $3$ with $p_3$ ($p_3=1-p_1-p_2$)

The expected outcome is then :

$$R=p_A\left( p_1+2p_2+3p_3\right) +p_B\left( 4p_1+2p_2+p_3\right)$$

This can be rewritten from player 1 perspective (eliminating $p_B$ and $p_3$) : $$R=\left(1+3p_1+p_2\right)+p_A\left(2-5p_1-2p_2\right)$$

This can also be rewritten from player 2 perspective : $$R=\left(1+2p_A\right)+p_1\left(3-5p_A\right)+p_2\left(1-2p_A\right)$$

Player 2 wants to minimize $R$. By studying the last expression you can see that :

• if $p_A\le\frac{1}{2}$, $R$ is minimized with $p_3=1$, but then player 1 have to choose $p_A=1$ to maximize $R$. This is not an equilibrium.
• if $p_A\ge\frac{2}{3}$, $R$ is minimized with $p_1=1$, but then player 1 have to choose $p_A=0$ to maximize $R$. This is not an equilibrium.
• if $\frac{1}{2}<p_A<\frac{2}{3}$, $R$ is minimized with $p_2=1$, but then player 1 can choose any value of $p_A$ to maximize $R$ ($R$ doest not depend of $p_A$). This is an equilibrium !

Hence the Nash equilibrium is $\frac{1}{2}<p_A<\frac{2}{3}$ and $p_2=1$.

• The given answer in the book is (3/5,2/5) for player 1 and for player 2 it is (0,1,0) . I understand how the 1/2<probability<2/3 is obtained and that's what I got when I did it graphically but that's not the given answer.I am trying to do this problem graphically as found in the last example of this:link . With this method how can I calculate player 2's probabilities.Sorry I don't know what an equilibrium is .This is a very basic preliminary course I am following as part of another course. – clarkson Dec 24 '13 at 15:52
• This is a good answer as $\frac{1}{2}<\frac{3}{5}<\frac{2}{3}$. They choose the value that negates the influence of $p_1$ for the second player. – Xoff Dec 24 '13 at 16:33
• @clarkson I read your link. This example is like the one on page 7. When you draw your graphics, you find a maximum segment (and not a point). Hence any value between those two points is good for $A$. For the second player, just remark that the (maximum) segment is on the line $p_2=1$. Hence, this is the best for player 2. This is in fact more simple than when you find a single point, and you must find how much of the two strategies that intersect on this point you have to take for the optimal. – Xoff Dec 24 '13 at 17:07
• Thanks for the explanation. – clarkson Dec 25 '13 at 2:57
• I thought I understood what you meant by " They choose the value that negates the influence of p1 for the second player ".But I haven't.Can you please explain it to me – clarkson Jan 6 '14 at 18:12

I'm assuming that the column player tries to minimize and the row player tries to maximize. Here's how I would think about the problem.

First the row player can choose a row, then the column player gets to choose a column.

You can think about it like this: When the row player chooses a row, he freezes that row, and the column player then chooses the best column for herself from that frozen row.

When the column player chooses a column, she freezes that column, and then the row player chooses the best row for himself within that frozen column.

If the row player chooses the first row, then the column player is going to choose column one, leaving a payoff of 1 to the row player.

If the row player chooses the second row, then the column player is going to choose column three, leaving a payoff of 1 to the row player.

So the row player's expected payoff is 1. Indeed, no matter which row he chooses he's going to end up with 1, so it makes sense that there is "no single point representing the highest point in the lower envelope" because he is indifferent. Any strategy will have an expected payoff of 1 for the row player.

Secondly we look from the column player's perspective: The column player chooses a column, and then the row player gets to choose a row.

If the column player chooses column one, then the row player is going to choose row two, yielding a payoff of 4 for the column player.

If the column player chooses column two, then the row player is going to either choose row two or row one, yielding a payoff of 2 for the column player.

If the column player chooses column three, then the row player is going to choose row one, yielding a payoff of 3 for the column player.

The column player is seeking to minimize the payoff, and she knows that the row player is seeking to maximize, so he's always going to choose column two.

The row player is seeking to maximize the payoff, but since he knows that the column player is going to choose the column that yields 1 no matter which row he chooses, he is indifferent between playing row one or row two.