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I am looking at a symmetric zero-sum game with 4 strategies and the following payoff matrix that has no pure Nash Equilibrium since it forms a circulant graph.$$\begin{pmatrix} 0 & 1 & 0 & -1\\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1\\ 1 & 0 & -1 & 0 \\ \end{pmatrix} $$I entered the game in a state of the art calculator (bimatrix solver, by Avis, Rosenberg, Savani, and von Stengel) and obtained the result as follows: $$p_1=0.5,\quad p_2=0,\quad p_3=0.5,\quad p_4=0,\quad q_1=0.5,\quad q_2=0,\quad q_3=0.5,\quad q_4=0 $$ $$p_1=0.5,\quad p_2=0,\quad p_3=0.5,\quad p_4=0,\quad q_1=0,\quad q_2=0.5,\quad q_3=0,\quad q_4=0.5 $$ $$p_1=0,\quad p_2=0.5,\quad p_3=0,\quad p_4=0.5,\quad q_1=0.5,\quad q_2=0,\quad q_3=0.5,\quad q_4=0 $$ $$p_1=0,\quad p_2=0.5,\quad p_3=0,\quad p_4=0.5,\quad q_1=0,\quad q_2=0.5,\quad q_3=0,\quad q_4=0.5 $$ What I do not understand is why the mixed strategy with all strategies played with probabilities $\frac{1}{4}$ not a MNE for this game? Is the calculator missing some solutions or am I missing something?

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    $\begingroup$ Isn't it the case that if you have two equilibria in a two-player zero-sum game, then any convex combination of those equilibria is also an equilibrium? $\endgroup$ May 17, 2023 at 17:06
  • $\begingroup$ How do you say that the strategy in question is not an equilibrium? $\endgroup$ May 17, 2023 at 17:26

1 Answer 1

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When you enter your bimatrix in the solver, you get the following result:

4 x 4 Payoff matrix A:

   0   1   0  -1
  -1   0   1   0
   0  -1   0   1
   1   0  -1   0



4 x 4 Payoff matrix B:

   0  -1   0   1
   1   0  -1   0
   0   1   0  -1
  -1   0   1   0

EE = Extreme Equilibrium, EP = Expected Payoff

Decimal Output

  EE  1  P1:  (1)  0.500000  0.000000  0.500000  0.000000  EP=  0.0  P2:  (1)  0.500000  0.000000  0.500000  0.000000  EP=  0.0
  EE  2  P1:  (1)  0.500000  0.000000  0.500000  0.000000  EP=  0.0  P2:  (2)  0.000000  0.500000  0.000000  0.500000  EP=  0.0
  EE  3  P1:  (2)  0.000000  0.500000  0.000000  0.500000  EP=  0.0  P2:  (1)  0.500000  0.000000  0.500000  0.000000  EP=  0.0
  EE  4  P1:  (2)  0.000000  0.500000  0.000000  0.500000  EP=  0.0  P2:  (2)  0.000000  0.500000  0.000000  0.500000  EP=  0.0

Rational Output

  EE  1  P1:  (1)  1/2    0  1/2    0  EP=  0  P2:  (1)  1/2    0  1/2    0  EP=  0
  EE  2  P1:  (1)  1/2    0  1/2    0  EP=  0  P2:  (2)    0  1/2    0  1/2  EP=  0
  EE  3  P1:  (2)    0  1/2    0  1/2  EP=  0  P2:  (1)  1/2    0  1/2    0  EP=  0
  EE  4  P1:  (2)    0  1/2    0  1/2  EP=  0  P2:  (2)    0  1/2    0  1/2  EP=  0

Connected component 1:
{1, 2}  x  {1, 2}

I believe the way to read this is the strategy $1$ is given by $s_1 =(0.5,0,0.5,0)$ and the strategy $2$ is given by $s_2 = (0,0.5,0,0.5)$.

The very last line says there is a "connected component", and in this case it means that both players playing any convex combination between $s_1$ and $s_2$ is also a Nash equilibrium. In particular, the strategy profile $((0.25,0.25,0.25,0.25),(0.25,0.25,0.25,0.25))$ is a Nash equilibrium.

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  • $\begingroup$ I am confused of the meaning of having an infinite number of equilibrium. Does stability matter in those scenarios? I guess I should read on Equilibrium Selection. $\endgroup$
    – Max M.
    May 22, 2023 at 15:55

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