Dependent likelihood estimation I'm trying to solve an exercise from a past exam. The question is:

Suppose there is an urn with 10 balls, where $\theta$ are white and $10 - \theta$ are green. Two balls are then extracted, without replacement. Let $X_i = 1$ denote if the color of the $i$th extracted ball is white, and  $X_i = 0$ if green, for $i=1,2$. I) Obtain the maximum likelihood estimator for $\theta$. II) Verify if the estimator obtained is unbiased.

Here is my work so far.
Since there the first and second draw are not independent, the likelihood function is:
$$
\text{L}(\theta) = \text{P}(X_1 = x_1)\cdot\text{P}(X_2=x_2 |X_1 = x_1)
$$
For the individual cases of $X_1$ and $X_2$ I can see that:
$$
\text{P}(X_1 = 1) = \frac{\theta}{10} \\
\text{P}(X_1 = 0) = \frac{10 - \theta}{10} \\
\text{P}(X_2= 1|X_1 = 1) = \frac{\theta - 1 }{9} \\
\text{P}(X_2= 0|X_1 = 1) = \frac{ 8 - \theta }{9} \\
\text{P}(X_2= 1|X_1 = 0) = \frac{\theta}{9} \\
\text{P}(X_2= 0|X_1 = 0) = \frac{9 - \theta}{9} \\
$$
I can then develop each case of the likelihood function for the sample, but I can't really go from there. Deriving and equating to zero doesnt give me cases that are present in the parametrer space $\theta \in \{1, \dots, 10\}$.
Even with those separate cases, I don't know how evaluate the bias of the estimator when it is not a closed expression.

The discreteness does make this potentially interesting. Fortunately, there are only four possible outcomes: 2 white, white first, white second, 0 white.  It's pretty obvious that if you get 2 white the MLE is $\hat\theta=10$ (because increasing $\theta$ increases the likelihood) and if you get 0 white the MLE is $\hat\theta=0$.
If  you get one of each, you could argue by symmetry that if there's a unique MLE it has to be 5/10, but let's actually do the calculation
$$L(1,0;\theta)=\frac{\theta\times(10-\theta)}{10\times 9}$$
$$L(0,1; \theta) =\frac{(10-\theta)\times\theta}{10\times 9}$$
So it is the same either way (as it should be). Differentiating, the likelihood is maximised at $\hat\theta=5$, and since that's the maximum over continuous $\theta$ it must also be the maximum over discrete $\theta$.
If you code white as 1 and green as 0, $\hat\theta/10$ is just the sample mean, and we know that's unbiased for the population mean $\theta/10$ under simple random sampling, regardless of the distribution.  [Or, it's a straightforward calculation, if you don't want to just know that]

