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I hope this is the right place to ask this question as it is related to probability and stats.

Basically, we were supposed to have final exams in the form of a written one, yet due to the pandemic, it was changed to a remote multiple-choice test. The test had 30 questions, each question got 4 choices (only 1 correct answer). Choosing the wrong answer will cause a student 1/3 mark deduction. The thing is, once a student tick to any answer, they aren't allowed to untick to make that question become blank to avoid a possible 1/3 mark losing.

To me, I think it was an issue. But when I complained about this issue with the guy who's been in charge, he responded to me that the setup of not letting student uncheck their answers, on average, will not disadvantage students. He also said that statistically, choosing not to answer or choosing to give random answers would make no difference as the average total score would still be zero.

I found it not correct. I think the situation he explained to me only correct when the number of questions students randomly choose is a big number. In that case, choosing randomly answers or leaving the answers blank will be as he claimed. But in the real exam, students often had a small number of questions they were not sure about, like 5 or 6 questions, and randomly choosing with 1/3 mark penalty for the wrong answer would put students in the disadvantage than otherwise.

Please help me to explain to me if I get it right or not? Thank you.

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  • $\begingroup$ If you want to 'untick' only when--presumably after some informed reflection--you fear you marked the wrong answer, then it clearly makes a difference. (But then could you change to a 'better' answer?) The score averages 0 when a robot picks an answer or no answer at random $\endgroup$ – BruceET Jun 4 '20 at 21:57
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Your intuition is somewhat correct, and somewhat incorrect. The score from randomly guessing does tend to "even out" better the more questions there are. When guessing randomly on a test of 1 million questions, it's very unlikely that you'll get a score that's far from 0. When guess randomly on a test with only 1 question, it's not even possible to get a score of 0, you will either score a 1 or a -1/3. But the expected value of randomly guessing on a single question is in fact zero, so random guessing confers no benefit or disadvantage overall. Averaged over the whole class, students who choose to randomly guess for every question will perform no better or worse than students who leave every question blank. When answering only 1 question, 25% of students will score better than a non-answering student, and 75% will do worse, but the average score over all the students who guess will be 0, the same as students who don't answer the question.

On a test with one question where everybody randomly guesses, 1/4 of people will pick the right answer and get a score of 1, and 3/4 will pick the wrong answer and get a score of -1/3. In a way, whether this equivalent to leaving it blank and getting a score of 0 comes down to your level of risk aversion. If you need a score of 0 or higher to pass the class, random guessing is a bad strategy, since it's likely you will fail, while you could guarantee success by not answering. The fewer questions there are, the more likely it becomes that there will be random, nonzero, individual benefit/loss from guessing, but the average net benefit to the class from guessing or will be zero.

To think about it another way - consider a game where we roll a 4-sided die, and I pay you \$10 on a 4, but you pay me \$3.33 on a 1-3. The scoring system on the test assumes that you should be indifferent to whether you play this game or not, but depending on your risk-taking aversion, you may just prefer to not play and just stick with your \$0. The variance of the outcome is higher if you choose to play one round (you'll either wind up with \$10 or -\$3.33), but the expected value of the outcome is identical (\$0).

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  • $\begingroup$ This depends on there being ABCD answers to the questions, right, and not ABCDE? $\endgroup$ – Dave Jun 4 '20 at 20:03
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    $\begingroup$ @Dave Right, the penalty for a wrong answer should be -1/(N-1) if there are N choices, in order to make the expected value of random guessing equal to 0. As the likelihood of guessing wrong goes up with additional choices, the penalty for a wrong guess goes down. $\endgroup$ – Nuclear Hoagie Jun 4 '20 at 20:08
  • $\begingroup$ Great answer. That was what I tried to argue with him (the one in charge). But you made it very clear and comprehensive. Thank you. $\endgroup$ – Jin Jun 4 '20 at 23:47

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