Mixed Strategy Nash Equilibrium for this particular 3x3 matrix

Question

Suppose I am given the following matrix:

I would like to find all MSNE

I started by doing the double underline method to find any PSNE. I discovered that none exist. I then looked at which strategies are strictly dominated. I noticed that Player 1 will never play M, because player 1 can always get a better payoff by playing L or R. So I eliminated the middle row "m".

After doing this, I get the following reduced matrix:

I assigned Player 2 a probability of $p_1$ for strategy A, $p_2$ for strategy B, and $1-p_1-p_2$ for strategy C. Similarly I assigned Player 2 a probability of $q$ for strategy L and $1-q$ for strategy R.

So now, we can see that Player 1's expected payoff of choosing L, that is $E(L)$, is $(1/2)(p_1)+(2/3)(p_2)+(1)(1-p_1-p_2)$

= $(p_1/2)+(2p_2/3)+(1-p_1-p_2)$

= $(3p_1/6)+(4p_2/6)+((6)(1-p_1-p_2)/6)$

= $3p_1/6+4p_2/6+(6-6p_1-6p_2)/6$

= $(3p_1+4p_2+6-6p_1-6p_2)/6$

= $(6-3p_1-2p_2)/6$

= Thus, $p_1=(6-2p_2)/3)$ and $p_2=(6-3p_1)/2$

Player 1's expected payoff of choosing R, that is $E(R)$, is $(1)(p_1)+(2/3)(p_2)+(1/3)(1-p_1-p_2)$

=$3p_1/3+2p_2/3+(1-p_1-p_2)/3$

=$3p_1+2p_2+1-p_1-p_2$

=$2p_1+p_2+1$

=$p_2=-1-2p_1$

=$p_1=(-p_2-1)/2$

Did I do this correctly?

Similarly, Player 2's expected payoff for choosing strategy A, $E(A)$, is $(1/2)(q)+(1)(1-q)$

= $q/2-1-q$

= $q/2-2/2-2q/2$

=$q-2-2q$

=$-q-2$

=> $-2=q$

Player 2's expected payoff for strategy B, $E(B)$, is $(2/3)(q)+(2/3)(1-q)$

=$2q/3+(2-2q)/3$ =$2q+2-2q$ =$2$

Player 2's expected payoff for strategy C is $(1)(q)+(1/3)(1-q)$

=$q+(1/3)-(q/3)$

=$(3q+1-q)/3$

=$2q+1$

=$q=-1/2$

I am not sure if I'm doing this correctly.

Answer 1

You generally have the right approach with expected payoffs correctly, but there are some errors in your logic.

First, take a look at your reduced matrix. You're missing a way to reduce it further and simplify the math a bit! You've left an extra strategy in place for Player 2. Player 2 knows that Player 1 will never play $M$, so Player 2 only needs to consider best responses to $L$ and $R$.

Second, note that you're using Player 1's payoffs in your expected payoff calculations for Player 2. For example, you have

$$E[A] = \frac{1}{2}q + 1(1-q)$$

The payoffs of $\frac{1}{2}$ and $1$ are to Player 1, not to Player 2. You want to use $\frac{1}{2}$ and $0$ in your calculation.

Third, the way you find a mixed strategy Nash equilibrium is by setting players' expected payoffs to be equal. Player 1 and Player 2 each need to be indifferent between their strategies, and that occurs when they play their strategies with probabilities so that the other player's payoffs are equal.

So, for Player 2's equilibrium strategy, you'll take Player 1's expected payoffs and set them equal: $E[L] = E[R]$. If they aren't equal, then one's greater than the other. Say $E[L] > E[R]$. Then, Player 1 will play $L$, and Player 2 will want to play $C$, then Player 1 will want to play $R$, and you end up in a cycle (and not an equilibrium). When the payoffs are equal, you have a stable solution, and neither player has an incentive to deviate from that strategy.

The same rationale applies to Player 1's equilibrium strategy - it's the set of probabilities that result in all of Player 2's expected payoffs being equal, and your Nash equilibrium is the set of all the strategies and their corresponding probabilities.