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What is the optimal strategy for random guesses for multiple choice questions? (retrofed.com)

submitted 1 week ago* (last edited 1 week ago) by scallisto@retrofed.com to c/nostupidquestions@lemmy.ca

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If you know that each option (ABCD) is roughly 25% likely to occur, is it a better bet to guess all Cs (guaranteeing around 25%, probably a fail in most tests), guess 25% of each option randomly (so if there were 20 questions, 5 for each option), or do something in between?

Or what if you know that some option are more likely than others, let's say C has a higher chance of occurring than ABD. Should you go for all Cs or should you mix in some ABD?

And if you have already completed 50% (or any other amount) of the test already with presumably correct answers, is it better to guess all Cs for the ones you missed or to distribute them so that there are 25% ABCD each (or a different ratio if you know some are more likely on average)

edit: I'm not planning on random guessing every option on any future tests, and certainly not counting every option to ensure a spread out distribution (that would take up lots of time!). It's just a silly probability question

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[–] Godort@lemmy.ca 21 points 1 week ago* (last edited 1 week ago)

It depends on the answer key.

Guessing 'C' all the way down and expecting 25% assumes that the distribution of answers is roughly equivalent across the whole test. This is an extra variable you can't account for without knowing the answers.

Whereas, if you pick a random answer for each question, you will answer 25% correctly, assuming an infinite number of questions. However this strategy has error bars that grow as the number of questions gets smaller.(ie: you are far more likely to end up with more or fewer than 25% correct in practice)

The best strategy for any given multiple choice question that you don't know the answer, is to eliminate options that are definitely incorrect, leading to a much higher chance of success with a random guess.

Or just study the content more, as breadsmasher suggested.