Could the Monty-Hall Problem be applied to multiple choice tests? Given a multiple choice test where each question contains 4 possible answers, what would happen if before beginning the test (before reading the questions), someone were to make a random selection for each question?

At this point it seems logical that for a given question the student has a 1/4 chance of their choice being correct and a 3/4 chance of one of the other choices being correct.

Let's say that they now begin to read the questions and in some cases they can deduce that one of the provided answers which was not the one that they picked is not correct (let's assume that there is no error in this deduction). In the scenario with the Monty Hall Problem, the probabilities did not change once the door was opened, they just shifted. 



By applying the same logic, the original selected answer has a 1/4 chance of being correct and the other three have a 3/4 chance of being correct, except that since one was deduced to be incorrect, the two remaining options have a 3/4 chance of being correct and so switching answers would increase the odds of being correct to $\frac{1}{2} * \frac{3}{4}$.

Is this an accurate assumption or are there pitfalls in doing this?

If this is the case, then what happens if another deduction is made such that their original answer was determined to be incorrect? It seems that there would be no change in the odds, but that seems unlikely.
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.
You are assuming you read the question and answer before picking a answer. This makes the probability void. 
FIRST you pick an answer randomly, THEN you check if it may or may not be correct.
4 possible answers, P(Correct)= 1/4 P(FALSE)= 3/4
meaning there is 75% chance that the answer is in the other 3 answers (you have not looked at the content as of yet!)
Now you read. Assuming that you KNOW that 1 of the other 3 answers is wrong, then there is only 2 answers left to split 75% -> (3/4)*(1/2)=3/8 37.5%
Assuming you can tell the answer YOU randomly picked is wrong, you can just pick one of the other 3 anyway, 33% chance of correct answer

This only works when you picked an answer under BLIND circumstances, and that the answers left have NO info what so ever, that they are indistinguishable from each other, which in MCQs are almost never correct.

NOTE*** if you can not gleam ANY info from the question, stick to your original answer, with gives P(C)= 25%, since picking erratically is known to lower the your score