Finding the MLE of a multinomial distribution (uneven probabilities) I am trying to simulate loaded die where the face probabilities are:
$$
p_1=p_2=p_3=p_4=1/6+\theta\text{ and }p_5=p_6=1/6-2\theta
$$
And so using the multinomial distribution I have:
$$
\binom{n}{x_i}\prod_{i=1}^6 p_i^{\displaystyle x_i}=\binom{n}{x_i}\left ( \frac{1}{6} + \theta  \right)^{\displaystyle\sum_{i=1}^4 x_i} \left(\frac{1}{6}-2\theta \right)^{ \displaystyle\sum_{i=5}^6 x_i}
$$
How do I find the MLE w.r.t $\theta$?
 A: Your notation is rather confused.  You have $\displaystyle\prod_{i=1}^6 p_i^{x_i}$, and that means
$$
p_1^{x_1} p_2^{x_2} p_3^{x_3} p_4^{x_4} p_5^{x_5} p_6^{x_6}.
$$
In the first factor, $p_1^{x_1}$ you have $i=1$, in the next you have $i=2$, and so on.  But what is $i$ equal to in the factor $\dbinom{n}{x_i}$ that you've written in front of the product??
What needs to be there is not $\dbinom{n}{x_i}$, but $\dbinom{n}{x_1,\ldots,x_n}=\dfrac{n!}{x_1!\cdots x_n!}$, i.e. a multinomial coefficient.
You need the value of $\theta\in[0,1/12]$ that maximizes the product. First check that $\theta$ must be between $0$ and $1/12$. The multinomial coefficient does not depend on $\theta$.  So you can just maximize
$$
L(\theta)=\left(\frac16+\theta\right)^{x_1+x_2+x_3+x_4}\left(\frac16-2\theta\right)^{x_5+x_6}.
$$
Since $\log$ is an increasing function this is the same as maximizing
$$
\ell(\theta)=\log L(\theta) = (x_1+x_2+x_3+x_4)\log\left(\frac16+\theta\right) + (x_5+x_6)\log\left(\frac16-2\theta\right).
$$
We have
\begin{align}
\ell\,'(\theta) & = \frac{x_1+x_2+x_3+x_4}{\frac16+\theta} - \frac{2(x_5+x_6)}{\frac16-2\theta} \\[8pt]
& = 6\left( \frac{x_1+x_2+x_3+x_4}{1+6\theta} - \frac{2(x_5+x_6)}{1-12\theta} \right) \\[8pt]
& = (\text{a positive number}) \cdot \left( (1-12\theta)(x_1+x_2+x_3+x_4)-(1+6\theta)(x_5+x_6) \right) \\[8pt]
& = (\text{pos.}) \cdot (x_1+x_2+x_3+x_4-(x_5+x_6) - 6\theta(2(x_1+x_2+x_3+x_4)+x_5+x_6)).
\end{align}
Find the value of $\theta\in[0,1/12]$ for which $\ell\,'(\theta)=0$, and show that for smaller values of $\theta\in[0,1/12]$ this derivative is positive and for larger ones it is negative.
