Find all mixed-strategy Nash Equilibria of 2x3 game. The setup of the game is as follows:
$$\begin{array}{ccc}
&\text{L}&\text{M}&\text{R}\\
\text{U}&5,10&10,15&5,0\\
\text{D}&0,20&5,5&10,25\\
\end{array}$$
I started with by doing:
$E[U] = 5q_1 + 10q_2 + 5(1-q_1-q_2)$ and
$E[D] = 0q_1 + 5q_2 + 10(1-q_1-q_2)$
then I reduce $E[U] = E[D]$ down to $2q_1 + 2q_2 = 1$.
Normally I would just solve for q but I'm not sure how what to do with two variables.
 A: Your $2q_1+2q_{2}=1$ tells you that if the column player mixes with probabilities $q_{1}$ and $q_{2}$ such that $q_{1}+q_{2}=\frac{1}{2}$, then the row player is willing to mix as well. When $q_{1}+q_{2}\neq\frac{1}{2}$, the row player strictly prefers $U$ or $D$.
Let's try to find all NE of the game. Standard argument shows that $(U,M)$ and $(D,R)$ constitute pure strategy NE profiles.
Let's try to find partially mixed strategy profiles that constitute NE (that is, one player mixing with strictly positive probability and the other player using pure strategy). Notice that the following is true for all $i\in\{r,c\}$: if player $i$ uses pure strategy, then player $-i$ has a best response that is a pure strategy. As a result, there are no partially mixed NE.
What remains are totally mixed NE. One condition that has to hold in this NE is your $q_{1}+q_{2}=\frac{1}{2}$. This implies that the column player plays $R$ with $\frac{1}{2}$ and that the sum of probabilities on $L$ and $M$ is $\frac{1}{2}$. In other words, the column player has to be indifferent between $R$ and either $M$ or $L$ (or all three).
You have, where $p$ is row's probability of $U$, $\mathbb{E}[u(R)]=25(1-p)$, $\mathbb{E}[u(M)]=15p+5(1-p)$, $\mathbb{E}[u(L)]=10p+20(1-p)$. The figure below plots these (blue, magenta, yellow for $L$, $M$, $R$).

What the figure shows is that when the column player is indifferent between $R$ and $L$ ($p=\frac{1}{3}$ by simple algebra) then $M$ is a worst option and thus $q_{2}=0$ so that you have one totally mixed NE with $(\frac{1}{3},\frac{2}{3})$ and $(q_{1}, q_{2}, 1-q_{1}-q_{2}) = (\frac{1}{2},0,\frac{1}{2})$.
When the column player is indifferent between $R$ and $M$ ($p=\frac{4}{7}$ by simple algebra) then $L$ is a best option and thus there is no NE in which the column player mixes between $R$ and $M$.
