[Math] Solving a 2*3 game with graphical method in game theory

game theorylinear programming

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.

Best Answer

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

  • 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 $p_2=1$.

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