Counter-intuitive mixed strategy Nash eq question I'm currently in intro game theory, and am having trouble with understanding (conceptually) these two games:
$\begin{array}
{c|c|c} \ & L & R \\
\hline
T & 2, 6 & 2, 2 \\ 
\hline
B & 2, 2 & 2, 4
\end{array}$
$\begin{array}
{c|c|c} \ & L & R \\
\hline
T & 2, 6 & 2, 2 \\ 
\hline
B & 1, 2 & 2, 4
\end{array}$
We have two Nash equilibria in pure strategies, TL and BR. Mathematically, I understand that row player in both games plays top 1/3 of the time, bottom 2/3 of the time, and that in the second game the column player will always pick R; conceptually, however, I am having trouble understanding why on earth this would happen: it seems to me that in game 1, row player can play any probability, and column player should accordingly just play L always (6 > 4), and in game 2, row player should always play T (weakly dominating) and column player should always play L (knowing what player 1 will do). Why doesn't this happen?
Any assistance would be greatly appreciated!
 A: Let $p \in [0, 1]$ denote the probability that row plays $T$ and $q \in [0, 1]$ that column plays $L$.
As every pure strategy equilibrium is a mixed strategy equilibrium we already figuerd that there exists two Nash equilibria for both games, namely $(p^*, q^*) \in \{(1, 1), (0,0)\}$.
Considering interior solutions, it is easy to show that the first game exihibts infinite equilibria $\{1/3\} \times [0, 1]$ and in the second game a single point $(1/3, 0)$.
The reason is, in order to motivate mixing each party needs to make the other party indifferent between playing one strategy or the other. While in the first game it does not matter what column does, as row always gets a payoff of 2 (and therefore $q^* \in [0, 1]$), row needs to play $T$ with probability $1/3$ as otherwise column would either play $L$ or $T$ with probability 100%. In the second game, column only can motivate mixing by playing $R$.
As far as I can tell you would like to add $[0,1] \times {1}$ for the first game and $(1,1)$ for the second game. As it was already mentioned above $(1, 1)$ is actually an equilibrium point. However, the set $[0,1) \times \{1\}$ does not contain any equilbrium, but $(1/3, 1)$ as otherwise column was not indifferent between playing $L$ and $R$ and therefore it would not be considered a (non-degenerate) mixed strategy equilibrium.
