Teacher: You got a 0.
Miles Morales: Is that...failed?
Teacher: Listen. If you answer all questions on a multiple choice test randomly, you would get a 25%. This means the only way for you to answer all questions wrongly is to...know the answer to all of them. Good job.
- From Spider-man: Into the Spiderverse.

Today, you had a probability exam. The teacher, wanting to spice up things, offered you another choice:

If you answered $k \%$ of the questions correctly, and $k$ is less than or equal to a predetermined value $p$ ($p < \frac{1}{number\ of\ answers}$), then the professor would assume that you answer the questions wrong on purpose, and your exam score would instead be $100 - k$ (For example, if $p$ is $20\%$, and you answered $10\%$ of the exam correctly, then your exam score will be instead $100 - 10 = 90$ out of $100$. If you answered $21\%$ of the exam correctly however, then tough luck, because your score would just be $21$ out of $100$).

You, knowing this, looked at all the questions first. You noticed the following thing:

  • There are $m$ multiple choice questions, each has $n$ different answers $(n \ge 2)$. Only one answer is the correct answer for each questions.
  • There are $X_1$ questions that you know exactly what the correct answer is.
  • There are $X_2$ questions that you can eliminate down to $2$ answers.
  • There are $X_3$ questions that you can eliminate down to $3$ answers.
  • $...$
  • There are $X_{n}$ questions that you can eliminate down to $n$ answers i.e. to you, any answer could be the correct answer. $X_1 + X_2 + ... + X_n = m$

The question is, what combination of $p$, $m$, $n$ and $X_i$ is needed for you to pursue the strategy of answering questions wrong on purpose and get the best expected score?

  • $\begingroup$ This looks and feels like a math assignment question, although admittedly you have a significant amount of rep.... did you make this yourself or did you find this somewhere? $\endgroup$
    – El-Guest
    Jun 15, 2019 at 2:36
  • 1
    $\begingroup$ @El-Guest I came up with this question myself. $\endgroup$
    – Chess960
    Jun 15, 2019 at 3:07
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, a response to the answerer to help steer them in the right direction would be helpful. $\endgroup$
    – Rubio
    Jul 2, 2019 at 0:26

1 Answer 1



If I try to get as many questions right as I can, my expected score is $\sum_{i=1}^n \frac{X_i}{i} = I$. If I attempt to intentionally fail, then my expected score is $m-\frac{X_n}{n}$ if I succeed and $\frac{X_n}{n}$ if I fail (to get my accuracy below $p$), since I know at least one incorrect answer for every question not among the $X_n$. If getting fewer than $mp$ of the $X_n$ questions correct occurs with probability $q$, then the expected value of failure equals $mq$, and I should try to fail if and only if this value exceeds $I$.

This of course assumes I happen to have a good calculator on me.

  • $\begingroup$ This doesn't seem to account for the possibility that you try to get as many questions right as you can and yet, by chance, end up guessing wrong on every one of them such that the teacher gives you a perfect score anyway. Right? (I think I mean, your expected value I is too low by some small but potentially interesting factor.) $\endgroup$ Nov 2, 2020 at 23:08

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