What is the MSNE for the following game?

enter image description here

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?

  • $\begingroup$ @denesp. I kind of suspected that. I have been unable to solve it without eliminating strategies... $\endgroup$ Apr 18, 2018 at 20:16
  • $\begingroup$ @HerrK. You are right. Sorry Пафну́тий, I shouldn't have been so hasty reading your question. $\endgroup$
    – Giskard
    Apr 19, 2018 at 4:50

1 Answer 1


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&=-1&&\Rightarrow\quad p=1/3\\ 3q+1-q&=1-q&&\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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.