I once took a test on which the grading worked as follows-

There are 120 multiple choice questions, each has five options to choose from :

A, B, C, D, E.

Each question can potentially get you 25 points (the maximal score is 3,000).

Potentially - because you can choose more than one answer to each question. You can circle A and B, or A D and E, so you basically can circle every subset of {A,B,C,D,E}.

Essentially, for each natural n≤5, n≠0, representing the number of options you chose, if one of them is right, you get 25/n points for that question.

If none of them are right, or you chose not to circle any option, the question will get you 0 points.

My questions are:

  1. assuming you haven't studied at all, is there a way for you to fill in the answer sheet, maximizing the score?

There is a huge amount of partitions for answering, from which we need to choose, and we have to take into account the probability of getting each question right or wrong. You can choose 3 options on 14 questions, choose 2 on 61 questions, and in the rest to choose to circle 4. Or, another partition, where you circle one option in the first question, and five on the rest.

  1. Given a similar test, with a maximal score of S, with q questions - each potentially worth S/q points, a options to choose as answers to each question, and an identical grading method. Is there a general formula describing how can one choose wisely?
  • 2
    $\begingroup$ It looks to me that no matter what, you get an average 5 points per question if you answer at all (the only bad decision you can make is not to answer a question). $\endgroup$
    – nicola
    May 7 '20 at 9:04
  • $\begingroup$ My hardship in understanding how to answer wisely is that you can choose to answer like so- choose 2 in 20 questions. Choose 3 in 40 questions. Choose one in 3 questions. Choose 4 in 7 questions, and at the rest choose five. Is that more efficient, score-wise? What partition should I choose taking into account the probability of getting it right? $\endgroup$ May 7 '20 at 9:31

I'll only answer your second question, since it is a generalization of the first.

Let's consider the case where the test only has one question, worth $S$ points, and with $a$ options. A strategy for choosing answers, whether it be deterministic or not, is equivalent to a probability distribution on $\mathcal P(\{1,\ldots,a\}$. Let us define two random variables: $X\subseteq\{1,\ldots,a\}$, your choice of answers, and $A\in\{1,\ldots,a\}$, the correct answer. Then the expected value (in terms of score) of your strategy is $$E[\text{score}]=\sum_{P\subseteq\{1,\ldots,a\}\\|P|\geq 1}\frac{S}{|P|}\Bbb P(X=P\;\wedge\;A\in P) \\ =\sum_{P\subseteq\{1,\ldots,a\}\\|P|\geq 1}\frac S{|P|}\frac{|P|}a\Bbb P(X=P) \\ =\frac Sa\sum_{P\subseteq\{1,\ldots,a\}\\|P|\geq 1}\Bbb P(X=P).$$ So in order to maximize your expected score, all you have to do is set $\Bbb P(X=\varnothing)=0$, and then your expected score is $\frac Sa$.

Some explanations for the calculation: $X$ and $A$ are independent, so $\Bbb P(X=P\;\wedge\;A\in P)=\Bbb P(X=P)\Bbb P(A\in P)$. Moreover, to calculate $\Bbb P(A\in P)$, we can calculate $$\Bbb P(A\notin P)=\frac{a-1}a\cdot\frac{a-2}{a-1}\cdot\cdots\cdot\frac{a-|A|}{a-|A|+1}=\frac{a-|A|}a,$$ so $\Bbb P(A\in P)=\frac{|A|}a$.

If there are multiple questions, the probabilities of getting each question correct are independent, so we can just add their expected values: your best strategy is to never leave a question blank, and then your expected score is $\frac{Sq}a$.

  • $\begingroup$ Excuse me for being ignorant, I never have studied probability. What exactly is setting X to be the empty set means? I see we sum all the probabilities of P=X (or, again, I am awfully wrong), and then we multiply the sum by S/a. So, what does the setting of P(X=Φ)=0, means? $\endgroup$ May 7 '20 at 11:28
  • $\begingroup$ If you've never studied probability, then the concept of "random variable" might be a bit complicated... Let's take a simpler example: let's say we have a dice, and $X$ is the random variable that takes the value of the dice when we roll it. If the dice is faire, then $\Bbb P(X=k)=\frac 16$ for each $k=1,\ldots,6$. But let's say we want to modify the dice so that it never lands on $1$. For example, we add magnets to the dice in such a way that the $1$ never lands on top. Mathematically speaking, what we have done is set $\Bbb P(X=1)$ to equal $0$. $\endgroup$
    – Isaac Ren
    May 7 '20 at 12:22
  • $\begingroup$ In the case of test-taking, one possible strategy is to pick a random subset of answers. How do we take a subset of answers? We can do it by rolling a dice with sufficiently many sides, each side labeled with the corresponding subset. Setting $\Bbb P(X=\varnothing)$ to $0$ is like modifying this big dice to never land on the $\varnothing$ side. $\endgroup$
    – Isaac Ren
    May 7 '20 at 12:24
  • $\begingroup$ Ok got it, so we never ever choose not to choose an option. Then, we choose to circle at least one answer each time. But then, what do we do? Is all I can know that I should never leave a blank answer? On what portion of the questions should I mark more than one answer? And, on what part should I mark less than five to maximize my score? $\endgroup$ May 7 '20 at 12:31
  • $\begingroup$ It turns out that it does not matter what you do, as long as you choose at least one answer. This is what I proved in my answer. Something important to understand is that the way you answer one question does not affect the way you answer another question: your strategy for each one is independent. So you don't need to worry about "pick 2 answers here, pick 3 answers there, etc." $\endgroup$
    – Isaac Ren
    May 7 '20 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.