Could the Monty-Hall Problem be applied to multiple choice tests?
Instinct tells me no - switching shouldn't make a difference, but instinct is what gets most people the incorrect answer for Monty Hall.
However, I'll try to explain it:
In the Monty hall problem, the host chooses which doors to show you (or to show you which answers are incorrect).
In your exam, the person who can "show you" incorrect answers is you. You do not, unlike the host, have knowledge of what the incorrect/ correct answers are which I think is the fundamental difference between the situations. Therefore, finding out whether an answer is incorrect is random. You have an equal chance of finding out that the answer you selected was incorrect as finding out one of the other answers was incorrect. In Monty Hall, the host will never tell you that the door you selected was a goat.
No, you can't apply the Monty-Hall problem to a multiple choice test.
The difference is, that in the Monty-Hall problem, there is a person which knows where the winning door is, and always opens a door which you didn't select and which contains a goat after you made your first choose.
In the situation you describe, you assume to know one incorrect answer, which allows you to "open a door with a goat". However, it is possible that this exactly the answer which you blindly selected first, a scenario which isn't possible in the Monty-Hall problem, because the other person chooses the door he opens depending on your choice, which is not the case in the multiple choice test.
Mathematical explanation
Let's assume we have a multiple choice question, with four alternatives. You chose an answer at random, without reading the question. After that you read the question with the answers.
We assume that you can exclue one possible answer with certainty, but you have no knowledge concerning the other three answers, they are all as likely to be correct. There are now two possibilities:
Alternative 1: You initially chose the answer that surely isn't correct (chance of this happening is $\frac14$)
You obviously want to switch. The chance of getting the right answer is $\frac13$, assuming that each answer is just as likely to be right.
Alternative 2: You initially chose another answer than the one that surely isn't correct (chance of this happening is $\frac34$)
Now we can analyze this like the Monty-Hall problem.
Possibility 1: you initially chose the correct answer (chance is $\frac13$). You switch answers and are now incorrect.
Possibility 2: you initially chose a wrong answer (chance is $\frac23$). You switch answers and are now correct with chance $\frac12$).
So your chance of being correct for alternative 2 is: $\frac13 * 0 + \frac23 * \frac12 = \frac13$.
Conclusion
Your chances of getting the right answer are always $\frac13$. The same as if you immediately had read the question, eliminated the answer you know is incorrect, and chose one of the remaining at random.