Mixed strategy Nash equilibrium in 3x3 game What is the MSNE for the following game?





I think you can eliminate strategies $A$ for player 1 and $C$ for player $2$, as these will are weakly dominated by all other strategies. Then, the game becomes a 2x2 game with $B,C$ for $1$ and $A,B$ for $2$.

let $q$ and $1-q$ be the probability $2$ plays $A$ and $B$ resp, and $p$, $1-p$ the probability $1$ plays $B$ and $C$. Then in equilibrium $q=1/4$ and $p=1/3$. Then the equilibrium is 

$$(0,1/3,2/3); (1/4,3/4,0).$$

Is this correct?
The general procedure to solve for a MSNE in a 3-by-3 (or larger) game is always a bit tricky and involves some trial and error


Step 1: Conjecture (i.e. guess) a subset of strategies that will be used in equilibrium
Step 2: Calculate their probabilities using the indifference condition
Step 3: Verify that the equilibrium payoff cannot be unilaterally improved upon; that is, no player has a strict incentive to deviate to another strategy 


Suppose your conjectured strategies are $\{B,C\}\times\{A,B\}$ (it doesn't really matter what the basis for your conjecture is; you're going to find out one way or another whether that's correct). Next, calculate the probabilities using players' indifference conditions. Let $p=\sigma_1(B)$ and $q=\sigma_2(A)$, we have 
\begin{align}
-3p&amp;=-1&amp;&amp;\Rightarrow\quad p=1/3\\
3q+1-q&amp;=1-q&amp;&amp;\Rightarrow\quad q=0.
\end{align}
[This suggests that your calculation for $q$ was incorrect.] 

Lastly (this is the most easily forgotten step), check that no one has an incentive to deviate from this equilibrium. In this case, player 1's payoff is $1$, which is already highest given player 2's strategy of choosing $B$ with probability 1. He'd be indifferent between mixing in other proportions over $B$ and $C$, and his payoff is strictly lower if he plays $A$ with positive probability. 

Player 2's expected payoff in this equilibrium is $-1$, which is also the highest given player 1's mixed strategy. She's indifferent between mixing over $A$ and $B$ with any other proportions and is strictly worse off if $C$ is played with positive probability. 

So, one MSNE is $((0,1/3,2/3),(0,1,0))$. This is only borderline consistent with your initial conjecture because $\sigma_2(A)=0$. But it's nonetheless a MSNE. In fact, there are infinitely many MSNEs of this form: $((0,p,1-p),(0,1,0))$ where $p\ge1/3$. This is a full description of all equilibria (including the pure one) in this game.