Hey everyone I have a very unconventional multiple choice test format with negative marking and I'd appreciate your help in cracking it.
The rules are as follows:
- There are 10 different questions with various number of choices - minimum 5 (A,B,C,D,E) and maximum 8 (A,B,C,D,E,F,G).
- Each question may have multiple right answers, but bear in mind that the last choice is always "None of the above".
- All questions combined yield a maximum of 18 points. This means that for all 10 questions, there are a maximum of 18 correct answers.
- You receive 1 point for each correct answer, 0 points for non-selected answers, and -0.5 points for a wrong answer. Example - Question 1 has 6 choices and 3 right answers, you select 1 right answer and 1 wrong answer. In total you get 0.5 points on this question.
Now, I am pretty sure that this is statistically "unhackable" and guessing does not work in your favour. But what would happen if we include several other factors, such as the ability to eliminate one or more choices from each question and the possibility of "guessing" the remaining number of correct answers. For example I might be absolutely certain that the first five questions have only 1 correct answer and this leaves me with 13 correct answers for the remaining 5 questions.
My question is how do I best proceed in such a situation. Do I always leave a question unanswered assuming that I am not certain of the correct choice or do I approach it differently if I can eliminate half of the choices? For example if I am certain that the last question must have 5 correct answers and the choices are 6, including "None of the above", I would select A,B,C,D,E as correct.
Back of the envelope type calculation (expectation based). More of a very long comment:
There are $18$ correct answers and $10$ questions. So, one might expect about $1.8$ correct answers per question. Yes, totally wrong, but true in expectation.
For a specific question, suppose that there are $5\leq k\leq 8$ choices. Then, we're assuming that $1.8$ of them are correct and $k-1.8$ of them are wrong.
If you just select one answer, then you would expect about $$ 1\cdot\frac{1.8}{k}-.5\cdot\frac{k-1.8}{k}=\frac{1.8}{k}-\frac{1}{2} $$ points per problem. Since $k$ is more than $3.6$, you would expect to lose points on each guess. Now this doesn't take into consideration that if the last answer is true, then none of the others are true.
If you can eliminate enough answers to make this positive, then you can expect to gain points on a guess. If you want to do multiple guesses for a problem, a similar calculation should work (but things get more complicated).