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?

• 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? – El-Guest Jun 15 '19 at 2:36
• @El-Guest I came up with this question myself. – Minh Tran Jun 15 '19 at 3:07
• 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. – Rubio Jul 2 '19 at 0:26

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 $$I$$.