A game with no pure or mixed strategy equilibrium?

Question

I'm trying to find any and all pure or mixed strategy Nash equilibria for the game

$$\begin{array}{|c|c|c|c|}\hline & L & C & R \\ \hline T & (6,2) & (0,6) & (4,4) \\ \hline M & (2,12) & (4,3) & (2,5) \\ \hline B & (0,6) & (10,0) & (2,2) \\ \hline \end{array}$$

By inspection I see no pure strategy Nash equilibrium. Not having a pure Nash equilibrium is supposed to ensure that a mixed strategy Nash equilibrium must exist.

However, when I go to solve for the mixed strategies I get one set of solutions that has a negative probability and in the set of equations for the other player I get an inconsistent system. My work is below.

If player I plays T, M, B with probabilities $p,q,1-p-q$ then for player II

$$E(L) = 2p+12q+6(1-p-q) = $$ $$E(C) = 6p+3q+0 = $$ $$E(R) = 4p+5q+2(1-p-q)$$

providing the system of equations

$$-4p+6q+6 = 6p+3q$$ $$-4p+6q+6 = 2p+3q+2$$

which simplify to

$$10p-3q=6$$ $$6p-3q=4$$

And subtracting equations you get $4p=2$ which implies $p=1/2$ but then plugging this into the first equation you find $10(1/2)-3q=6$ implies $q=-1/3$.

Did I do something wrong?

Answer 1

Your mistake is not considering mixed strategies that involve some, but not all, of the possible moves.

Suppose the row player plays $T$ and $B$ with probabilities $p$ and $1-p$, but never plays $M$ (note that $M$ is dominated by a $1/2-1/2$ mixture of $T$ and $B$, so this is reasonable). The expected payoffs for the column player are then $2p + 6(1-p) = 6 - 4p$ for $L$, $6p$ for $ C$, $2 p + 2$ for $R$. With $p = 6/10$ these are $3.6, 3.6, 3.2$, so the column player will never play $R$. Now determine probabilities for the column player playing $L$ and $C$ that make the row player's expected payoffs for $T$ and $B$ equal.

Answer 2

For the row player strategy $M$ is never BR and for the column player strategy $R$ is never BR. So, exclude these strategies and compute the mixed equilibrium which is $(T,M,B)=(3/5, 0, 2/5)$ and $(L,C,R)=(5/8,3/8,0)$.