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.

Taken from Wikipedia

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.


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: $\frac13 * 0 + \frac23 * \frac12 = \frac13$.


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.


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.

  • $\begingroup$ This is a good counter-argument; could you provide more proof to support it? $\endgroup$
    – Klik
    Mar 5 '16 at 22:10
  • $\begingroup$ Are you looking for a mathematical one? $\endgroup$
    – Shuri2060
    Mar 5 '16 at 22:17
  • $\begingroup$ Yes, something more definitive that can remove all doubt. $\endgroup$
    – Klik
    Mar 5 '16 at 22:22
  • $\begingroup$ I'm afraid I can't think of anything for now. But perhaps a better answer will come along which can. Part of the problem for me is understanding the exact conditions of the situation you provide - is the student only able to find out if his non-selected answers are incorrect (and never his selected one)? Can he only find a maximum of one definitely incorrect for every question? Because unless if you put something like those conditions, this situation is slightly different from Monty Hall, but if you do put conditions like those, then the situation is non-realistic (compared to an actual exam). $\endgroup$
    – Shuri2060
    Mar 5 '16 at 22:26
  • $\begingroup$ The situation should model an actual exam. It should be as I've described it above where any answer could be determined as being incorrect by deduction (with 100% accuracy) and where an answer was originally randomly selected. $\endgroup$
    – Klik
    Mar 6 '16 at 0:29

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


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