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I have often heard it said that if you are taking a multiple choice test and you have a block of problems you cant get to you should choose a random letter and fill it in for all choices.

I understand this, and it makes a lot of logical sense. After all if you choose random ones to fill in for the rest you could find yourself getting each one wrong, whereas there is likely to be at least one "c" in the section.

Still, something doesn't seem right about this logic sense tests have random answers. Personally I think that is possible that if the actual answers are chosen arbitrarily then it doesn't matter how you guess. While on some test you might get lucky and there might be a lot of c's on others there may not be any in the area you are guessing.

I think that the designation of an answer as "A" or "B" or "C" (etc) no longer matters if the key is created randomly, and you have no context with witch to guess which one is write. Essentially on a scantron the answer choices could be "ABCD" on one line and "ADCB" on another line. Without knowing the answers you would have the same chance of getting some more right if you bubbled every "c" as if you chose every answer that was one from the right on the scantron.

Let's make the following assumptions.

1. The test maker randomly decided the answer to every question, although some answers may come up more than others this was not purposeful.
2. There is only one correct answer to each question.
3. No context is given to the answers. You have no way of eliminating incorrect ones, or figuring out which one is most likely.
4. You are not guessing on all of the answers, only a few. However in your guessing you are not taking into consideration how many of each answer choice have come up before.

And we have the two strategies:

1. Choose a letter, and stick with it for every answer you don't know.
2. Choose answers randomly if you don't know.

Which strategy is the better one, or are they both equal? Does one get better then the other if you are guessing for a larger percentage of the test?

If you can prove this with math or a mathematical/probability theorem that would be preferable.

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2 Answers

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This depends on the test-maker's method of assigning letters to correct answers:

Consider first a test-maker that assigns letters to correct answers randomly with uniform probability (e.g. abcd each has 25% chance of being correct answer for each question, and these odds are independent of the correct choices for other questions).

In this case, randomly picking answers with uniform probability is no different than picking a 'letter of the day'. With either strategy your expectation and variance will be the same (25% chance, could get anywhere from 0% to 100%). This is because the letter choices are independent and have uniform probabilities.

Now consider instead a test-maker who (in say a 20 question test), chooses 5 copies of each letter and assigns them randomly to the 20 questions (so each letter appears 5 times in the answers, but which letter is correct for a given question is still 25% any of a, b, c, or d. In this case, the 'letter of the day' strategy is better, moreso, the more problems you our guessing on. At the extreme case, if you were guessing on every question, then you would guarantee yourself a score of 25% (!) (no variance at all)

The key thing that makes it so you can't do better (or worse guessing) than random guesses is that the probabilities are uniform and independent. Meaning that each letter has the same chance of being the correct answer to the question and the correct answers to other question do not influence the conditional probabilities of what the correct answer to the question being considered are (this is what independence means in this context).

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Both methods are equal. One of your assumptions is that the answer to each question is slected randomly and independantly from the answers to the others. Hence, what you choose on question $33$ only affects the outcome on question $33$. Since every choice is equally likely in question $33$ it doesn't matter what you choose.

The argument that at least one answer must be $c$ can be applied to any sequence of selections, if there are say, $20$ questions the odds you get all of them wrong when answering randomly is $(\frac{3}{4})^{20}$ which is less than one in a hundred.

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