How to correctly calculate the percentage of correct answers?

Question

Algorithms

There was a following problem:
There is a question to which 5 variants of the answer are attached. There are 2 correct answers in this question. An algorithm is needed to calculate the percentage of correctness of the answer, i.e.:
A person put 2 correct ticks and 0 incorrect ones - the percentage of fidelity is 100%.
A person put 1 correct tick and 0 incorrect ones - the percentage of fidelity is 50%.
A person put 1 correct checkmark and 1 incorrect one - what is the percentage of fidelity?
A person put 1 correct tick and 2 incorrect ones - what is the percentage of fidelity?
There should be some penalty for each incorrect tick, but how to calculate it? If the fine is too small, it will be easier for the user to tick all the boxes and get their own percentage. If the penalty is too high, the user will only check the box they are 100% sure of and will not look for possible second answers.
Perhaps there are any articles or known solutions?

Answer 1

You can, of course, not give any points at all for incorrect answers, but you can also do this:
Let prav be the number of correct answers, neprav - the number of incorrect ones.
Let's introduce the notation: X = (prav/2 - neprav/3), Q = (X + abs(X)) * 50%
Then the resulting Q value is the percentage to be set. In the case when the number of incorrect answers exceeds the number of correct ones, a zero percentage is set.

Answer 2

I propose such a solution. The weight of each item is the same (20% in case of 5 answers). For each item, you know whether it should be included in the answer (whether it should be checked). If the checkbox should be checked and the user has set it, points are awarded for this item. If the checkbox should be checked, the user did not check it - points are not awarded. Likewise for an item that should not be included in the answer.
Example: 5 points, 1 and 2 are correct, 3,4,5 are incorrect.
1 and 2 marked, 3,4,5 not marked - 100%
1 and 2 not marked, 3,4,5 not marked - 60%
1 marked, 2,3,4,5 not marked - 80%
1,3, 4 marked, 2.5 not marked - 40%
1.2 not marked, 3,4,5 marked - 0%
and so on (32 options in total).

A person put 1 correct checkmark and 1 incorrect one - what is the percentage of fidelity?

20% for one correct check mark and 40% for two correct unchecked check marks, total 60%

A person put 1 correct tick and 2 incorrect ones - what is the percentage of fidelity?

20% for a correct check mark, 20% for one correct unchecked check mark, 40% in total.

Perhaps there are any articles or known solutions?

Approximately so, in my opinion, testing is implemented in coursera.
UPD. As I understand it, you are confused by the fact that with some properties of the answers (number of correct / incorrect), the user can get an average score without ticking at all. If you think this is a problem, then I offer two solutions. The first is to compose questions accordingly, so that the number of correct answers is higher than the number of incorrect ones. The second is to set the checkboxes in random order at the start. In this case (again, with the right selection of parameters), it will be more difficult to get a high score without doing anything.

Answer 3

here is my version, see how many marked answers are in total, then divide 100 by the number of all marked answers, for example 100 / 4 = 25% per answer, then calculate the number of correct answers, if there are 2, then the percentage of correct answers will be 50%, with each incorrect the answer decreases the percentage for correctness, and no penalties are needed, just if the user clicks 10 out of 10 answers, and only one of them is correct, then he will receive 10%, and if he had checked only 2 checkmarks, he would have received 50%, if you need penalties so much, then make a coefficient that will be subtracted from the number of correct answers, let's say one percent is subtracted with each wrong answer, as a result, out of 10 if the user marks 10, then he will receive 1%, because I received 10% for correctness, - 9% for fines. something like this)