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


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 $(1/2(a_1)+1/2(a_2),1/2(a_4)+1/2(a_5))$

enter image description hererOes.jpg

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

The game is symmetric, so I focus my answer on Player 1. (Repeat the reasoning for Player 2.) A mixed strategy is a best reply when the player is indifferent between the strategies he plays with positive probability and no other pure strategy yields higher utility.

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)$.


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