ML estimation of a multinomial proportion with constraints The Question
A researcher surveys $n$ college students and counts how many support, how many oppose
and how many are undecided about a recently introduced federal policy. Letting $X_1, X_2, X_3$ denote these counts, she model $X = (X_1, X_2, X_3)$ as $X \sim \mathsf{Multinomial}(n,p)$. 
I want to find the restricted MLE of $\mathbf p = (p_1, p_2, p_3)$ for a null hypothesis that the actual proportions of supporters and opposers
in the entire college are equal. My mathematical formulation of this null was that this MLE would be the vector $\bf p$ that maximizes the likelihood function over $\Delta_3^0 \subsetneq \Delta_3$ where $\Delta_3$ is the 3-dimensional simplex of all probability vectors i.e. vectors that sum componentwise to 1. Then $\Delta_3^0$ would contain all $p$ of the form $\mathbf p = (a, a, 1 - 2a), a\in(0,1)$. 
Note
I realize that this a simple question for such mathematical abstractions, I have simply tried to follow my teacher's notation. 
My Work
When I went through and tried to write out the likelihood function, I got $$L_{x, H_0} = \text{const. }\times a^{x_1}\cdot a^{x_2}\cdot (1 - 2a)^{x_3}$$ which should maximize at $\hat a = {x_1 + x_2 \over x_1 + x_2 + x_3}$. However, when I actually calculated for data of counts $(X_1, X_2, X_3) = (140, 165, 195)$ I got $a = 0.61$, which gives me a negative probability in my probability vector which doesn't make sense. Thanks for any help. 
 A: An alternative way to find the ML estimator is to note that the constraint $p_1=p_2$ reduces the number of relevant outcomes to two: decided and undecided. The vector of probabilities is thus $(2p_1,p_3)$. We have $$(X_1+X_2,X_3)\sim\mbox{Multinomial}(n,(2p_1,p_3))$$ which is equivalent to $$X_1+X_2\sim\mbox{Binomial}(n,2p_1).$$
This transforms the problem into a simpler one. Deriving the MLE of $p$ in the binomial distribution is perhaps the standard example for ML estimation in discrete distributions: denoting $2p_1=r$ one finds that $$\hat{r}=\frac{X_1+X_2}{n}=\frac{X_1+X_2}{X_1+X_2+X_3}.$$
By the functional invariance of maximum likelihood estimators, $$\hat{a}=\hat{p}_1=\hat{p}_2=\frac{1}{2}\hat{r}=\frac{X_1+X_2}{2(X_1+X_2+X_3)}$$
and
$$\hat{p}_3=1-\hat{r}=\frac{X_3}{X_1+X_2+X_3}.$$
A: The estimator for $a$ should be
$$
\widehat{a} = \frac{X_1+X_2}{2(X_1+X_2+X_3)}
$$
You can derive this by writing the log-likelihood as (up to a constant)
$$
\ell_n(a \mid X_1, X_2, X_3) = (X_1+X_2)\log a + X_3 \log (1-2a)
$$
so that the first order conditions are
$$
\begin{align}
\left.\nabla_a\ell_n(a \mid X_1, X_2, X_3)\right|_{a=\widehat{a}} &=0 \\
\widehat{a} &= \frac{X_1+X_2}{2(X_1+X_2+X_3)}
\end{align}
$$
For your dataset, this then gives you $\widehat{a} = 0.305$.
