# Found two Nash equilibria in $3 \times 3$ game but the writer says there is a third one

I actually find the two Nash equilibria, which are the pure $(a_3,a_6)$ and the mixed

$$\left(\frac 17 a_1 + \frac 17 a_2 + \frac 57 a_3, \frac 17 a_4 + \frac 17 a_5 + \frac 57 a_6\right)$$

but I don't have any idea how to get the mixed equilibrium

$$\left(\frac 12 a_1 + \frac 12 a_2, \frac 12 a_4 + \frac 12 a_5\right)$$

The book "Algorithms for Strong Nash Equilibrium with more than two agents" by Gatti and Rocco says there are three Nash equilibria. Let me know why they could find $3$ equilibria.

Here is my solution. For the player1, the best response to player 2's action $a_6$ is $a_3$ and similarly, the best response to player 1's action $a_6$ is $a_1$ for player2. the pure Nash equilibrium is $(a_3,a_6)$. In case of Mixed strategy, $x_{1} = (p_1,q_1,1-p_1-q_1)$ and $x_2=(p_2,q_2,1-p_2-q_2)$

$U_1(a_1) = p_1 * 5 + 0 + 0=5p$

$U_1(a_2) = 0 + q_1 * 5 + 0=5q$

$U_1(a_3) = 0 + 0 + (1-p_1-q_1)*1=1-p-q$

Since $p_1=q_1$, $1-p_1-q_1=5p$ => $p_1=q_1=1/7$ and

$1-p_1-q_1=5/7$.

the same thing adopts to $x_2$. Therefore, NE1= $(1/7a_1+1/7a_2+5/7a_3,1/7a_4+1/7a_5+5/7a_6)$

But I don't have any idea to get another mixed strategy $(1/2(a_1)+1/2(a_2),1/2(a_4)+1/2(a_5))$

rOes.jpg

• The print-screen could be larger... – Rodrigo de Azevedo Oct 15 '16 at 7:01
• I changed it! can you read the print-screen? – Mathpractictioner Oct 16 '16 at 8:45
• How did you find those two NE? Show your work. – Rodrigo de Azevedo Oct 16 '16 at 8:47
• I tried to solve it plz check my work – Mathpractictioner Oct 17 '16 at 8:58
• When I took a course in game theory, I computed Nash equilibria using quadratic programming. Did you solve any quadratic programs? – Rodrigo de Azevedo Oct 17 '16 at 9:25

To find the equilibrium in fully mixed strategies, set $U_1(a_1) = U_1 (a_2) = U_1 (a_3)$ and find $p_2 = q_2 = 1/7$, leading to the solution above. (Your reasoning starts by assuming $p=q$ but this is unjustified; and $U_1 (a_i)$ for $i=1,2,3$ is a function of 2's probabilities, not 1's.)
To find the other equilibrium in mixed strategies, note that the support for 1 is $\{a_1, a_2 \}$ and the support for 2 is $\{a_4, a_5 \}$. So, $\sigma_1^* = (p a_1 + (1-p)a_2 + 0 a_3)$ and $\sigma_2^* = (q a_4 + (1-q)a_5 + 0 a_6)$. Equate $U_1 (a_1) = 5q$ and $U_1 (a_2) = 5 (1-q)$ to get $q=1/2$. Repeat by inverting players' roles to find $p=1/2$. Finally, check that U_1 (a_3) = 0 < 2.5 = U_1 (a_1)\$.