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:
- 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.
- 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?
Best Answer
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 $\frac{Sq}a$.