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In $3$ players game like one in image, how to check if there is an equilibrium when only one player plays mixed strategy and others play pure strategies 3 players game image
\begin{align}3\text{ plays s }&\begin{array}{r|c|c}1\backslash2 & s&n \\\hline s &1,1,1 &-1,-2,-1 \\\hline n&-2,-1,-1 &1,1,-1 \end{array} \\[0.2cm] 3\text{ plays n } &\begin{array}{r|c|c}1\backslash2 & s&n \\\hline s &-1,-1,-2 &1,-1,0 \\\hline n&1,-1,1 & 0,0,0\end{array}\end{align}
Yes, there are infinitely many, with player 2 mixing.
Assume that player 2 mixes: Here it may be possible because the payoff for player 2 in $(s,s,n)$ and in $(s,n,n)$ is equal. You only need to check if $(s,(y,1-y),n)$ is an equilibrium for $0<y<1$. Given that $3$ plays $n$, $s$ is a best response for player $1$ for any $y$ of player $2$ such that $$(-1)y+(1)(1-y)\ge (1)y+0(1-y)\implies y\le \frac13$$ Now, given that $1$ plays $s$, $n$ is a best response for player $3$ for any $y$ of player $2$ such that $$(-2)y+0(1-y)\ge(1)y+(-1)(1-y)\implies y\le \frac14$$ Hence, any strategy profile of the form $$(s,(y,1-y),n)\quad \text{with } 0<y\le \frac14$$ is a Nash equilibrium where only player 2 uses a mixed strategy.
To see that there are no other such equilibria:
Assume that player $3$ plays the mixed strategy $(z,1-z)$ where $0<z<1$ is the probability of playing $s$. If players $1$ and $2$ play the pure strategy profile $(s,s)$ then player $3$ has an incentive to choose $z=1$, hence this is not an equilibrium. If players $1$ and $2$ play any other pure strategy profile, then player $3$ has an incentive to choose $z=0$, hence there is no equilibrium where $3$ uses a mixed strategy and $1$ and $2$ use pure strategies.
Repeat for player 1: Assume that player $1$ plays the mixed strategy $(x,1-x)$ where $0<x<1$ is the probability of playing $s$. If players $2$ and $3$ play the pure strategy profile $(s,s)$ then player $1$ has an incentive to choose $x=1$, hence this is not an equilibrium. Similarly, if players $2$ and $3$ play any other pure strategy profile, then player $1$ has an incentive to choose either $z=1$ or $x=0$, hence there is no equilibrium where $1$ uses a mixed strategy and $2$ and $3$ use pure strategies.
