How to decide on the MLE when pmf is 0? Suppose you have $\theta=\{1,2\}$ and the sample of (0,1,2) with the task of finding MLE:
\begin{array} {|c|c|c|}
\hline
x & p(x|\theta=1) & p(x|\theta=2) \\
\hline
0 & 1/2 & 1/4 \\
\hline
1 & 0 & 1/4 \\
\hline
2 & 1/3 & 0 \\
\hline
\end{array}
I can't compare $\prod_i p(x_i|\theta)$, because of the zeroes. Even though I can compare probability for a single observation, I'm still having trouble quantifying the difference. $p(x=1|\theta=2)>p(x=1|\theta=1)$ for x=1, but can I do a similar comparison for the whole vector?
Edit
The original question is

One observation is taken on a discrete random variable X with pmf f(x|θ), where θ∈{1,2,3}. Find the MLE of θ.

This is followed by a 4x5 table which includes zeroes in every column. I think the idea behind this question is that we have a distribution that doesn't describe population well. Do we just say that the method of MLE is not applicable here?
 A: Now we have the original question, the mystery is no mystery. 
In that question, a single observation will be consistent (to greater or lesser degree) with some of the $\theta$ values and inconsistent with others (the ones that have no chance of giving that observed value).
So for example, if we look at your probability distributions, and possible samples of a single observation:
$x=0:\, $ the ML estimate is $\theta=1$  (since $\frac{1}{2}>\frac{1}{4}$)
$x=1:\, $ the ML estimate is $\theta=2$  
$x=2:\, $ the ML estimate is $\theta=1$
So whichever outcome is observed the ML estimate is obvious. This (possibly in table form) is what the original question is looking for.
If instead of one observation, you have a sample of several (as in your version of the question), then a value of $\theta$ will be "ruled out" (have likelihood 0) by any observation in the sample having 0 probability under that $\theta$. If, as above, at least one sample value has likelihood zero under every value of $\theta$, all $\theta$ values have likelihood 0, as was previously suggested in comments.
A: If $P(x=1|\theta = 1) = 0$, then (0,1,2) is impossible, since it has a 1 in it. If $P(x=2|\theta = 2) = 0$, then (0,1,2) is impossible, since it has a 2 in it. So, your likelihood calculation is telling it like it is. It is impossible to get that sample, given your probability distributions and thetas.
