# Game Theory; Finding the Nash Equilibria

How to proceed with such problems where there is no pure Nash Equilibrium? Even if we attempt to eliminate strictly dominated strategies, we find none.

• I don't think Nash will exist in this case, there will be no Nash equilibria. – Shubham Singh Sep 11 '18 at 12:59
• @ShubhamSingh We know there must be a Nash equilibrium because Nash's contribution was to prove that any finite game has at least one. – Ubiquitous Sep 11 '18 at 14:00

You can do some iterated deletion of weakly dominated strategies. Strategy $C$ for player $2$ is weakly dominated by $L$ played with $2/3$ and $R$ with $1/3$. Then, having eliminated $C$, we can see that $B$ is strictly dominated by (e.g.) $T$ with $3/5$ and $M$ with $2/5$ for player $1$. That leaves us with a $2 \times 2$ game, for which it is easy to solve for the M.S. Nash equilibrium.
It is $\big((1/2,1/2,0), (2/3, 0, 1/3)\big)$. It is unique.