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Consider 3-player game.

Players $x,y,z$, each player has two strategies. $x$: $x_1$ and $x_2$, $y$: $y_1$ and $y_2$, $z:z_1$ and $z_2$.

The outcome of the game are represented by the labels of the player who has playoff $1$, otherwise outcome is $0$.

The best way to represent the game is to depict two tables, the first one for the case when $c$ plays $c_1$ and the second one when $c$ plays $c_1$

$c$ plays $c_1$

╔════╦════╦═════╗
║ c1 ║ b1 ║ b2  ║
╠════╬════╬═════╣
║ a1 ║    ║ B   ║
║ a2 ║ A  ║ A;C ║
╚════╩════╩═════╝

$c$ plays $c_2$

╔════╦═════╦═════╗
║ c2 ║ b1  ║ b2  ║
╠════╬═════╬═════╣
║ a1 ║ C   ║ A;B ║
║ a2 ║ B;C ║     ║
╚════╩═════╩═════╝

The problem is to find all Nash equilibria (pure and mixed) and to show that there is no other Nash equilibria.

pure NE: There are three pure Nash equilibria $(x_2,y_2,z_1), (x_1,y_2,z_2), (x_2,y_1,z_2)$ (this is all I found, may be there is more).

mixed NE:Let's place equalities for mixed strategies. Assume that $p_x$ - the probability that $x$ will play $x_1$, therefore $(1-p_x)$- the probability that $x$ will play $x_2$ in the same manner define $p_y$ and $p_z$.

For $x$ to be indifferent between $x_1$ and $x_2$ - $(1-p_y)(1-p_z) = p_y \cdot p_z + (1-p_y)p_z$,

For $y$ to be indifferent between $y_1$ and $y_2$ - $(1-p_x)(1-p_z) = p_x \cdot p_z + p_x(1-p_z)$,

For $z$ to be indifferent between $z_1$ and $z_2$ - $(1-p_x)(1-p_y) = p_x \cdot p_y + (1-p_x)p_y$.

Apparently there is a symmetry, however it's not a symmetric game.

The main problem is how to proceed from this point. There are might be few cases either given one of the equalities we should consider only pure strategies of the rest two players or consider more complicated way when the rest two players play mixed strategies. May be because of the pattern we have symmetric solution? One more problem how to show that there are no other Nash equilibria.

Another approach: Let represent outcome in terms of vectors.

╔════════╦═════════╦═════════╗
║ number ║ outcome ║ winners ║
╠════════╬═════════╬═════════╣
║      0 ║     000 ║ -       ║
║      1 ║     001 ║ C       ║
║      2 ║     010 ║ B       ║
║      3 ║     011 ║ A,B     ║
║      4 ║     100 ║ A       ║
║      5 ║     101 ║ B,C     ║
║      6 ║     110 ║ A,C     ║
║      7 ║     111 ║ -       ║
╚════════╩═════════╩═════════╝

It looks like every player have exactly the same chances to win, every player has three permutations that leads to payoff 1. $A=\{3,4,6\}, B=\{2,3,5\}, C=\{1,5,6\}$, among them the following outcomes appear twice $(3) 011, (5), 101 (6) 110$.

The problem is I don't know how to approach the solution, I would appreciate for ideas.

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