Maximum Likelihood Estimate of a a discrete r.d - I spent more than 4 hours on this questions, help!! Suppose $X$ is a discrete r.d with the following p.d.f:
$$
\begin{array}{c|lc}
X & \text{0} & \text{1} & \text{2} & \text{3} \\
\hline
p(x) & 2\theta/2 & \theta/3 & 2(1-\theta)/3 & (1-\theta)/3
\end{array}
$$
where $0\leq\theta\leq1$ was taken from such a distribution $(3,0,2,1,3,2,1,0,2,1)$. What is the MLE of $\theta$?
Likelihood function:
$$(2\theta/2)^2.(\theta/3)^3.(2(1-\theta)/3)^3.((1-\theta)/3)^2$$
Log-Likelihood function:
$$2\log(2\theta/2)+3\log(\theta/3)+3\log(2(1-\theta)/3)+2\log((1-\theta)/3)$$
Set derivative to zero:
$$2/\theta+3/\theta-6/(2-2\theta)-2/(1-\theta)=0$$
$$\frac{5-5\theta}{\theta}-\frac{6+6\theta}{2-2\theta}=2$$
$$5-5\theta=2\theta+\frac{6\theta+6\theta^2}{2-2\theta}$$
$$2-2\theta(5-7\theta)=6\theta+6\theta^2$$
Finally I got something like this:
$$10+8\theta^2=30\theta$$
Don't know how to find $\theta$. But I guess I have made mistake because the answer is $\theta=0.5$.
I spent more than 4 hours on this, tried different ways, like simplify the likelihood function before $"\log"$ them and still cannot work out, anyone help please?

Your $p(x)$ is wrong: its sum is $4/3$, not $1$.  And who would write $2\theta/2$ instead of $\theta$?  I suspect that $2\theta/2$ should be $2\theta/3$ and $2(1-\theta)/2$ should be $2(1-\theta)/3$.
