5
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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?

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  • $\begingroup$ 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 at 2:36
  • $\begingroup$ @El-Guest I came up with this question myself. $\endgroup$ – Minh Tran Jun 15 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 at 0:26
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Answer:

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.

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