Calculating the gamble chance for correction on a multiple choice question with multiple answers usng negative marking I'm trying to figure out the following, and I can get some results but I would like some input from more knowledgeable people.
So I have a multiple choice with 8 total answers, 3 correct answers and 5 incorrect answers (actually these numbers are all variable and I'm looking for a formula to fit all if possible).
I will be using negative markings so every wrong answer will lower their score.
Say they normally have a score of 1, and for every incorrect answer points are subtracted. We don't want to reward an answer that has, for instance, more than 3 incorrect answers in it, so that would be the falloff point that 0 points are rewarded. 
Now my question would be twofold. Should I even be using a correction value for guessing when using negative marking? And if so, what would be a way to deal with this? How would I calculate such a thing?
To clarify:
I'm trying to calculate how hard it is to guess the answers, and find some meaningful number I can use in my formula to calculate the actual score of the test. In order to calculate that I need to know first how to calculate the result of a single question.
So what are the odds to guess a multiple choice question with multiple answers correct. This number together with a proposed knowledge value will be used to calculate the score the user will actually get. 
Now because I am using negative marking, I am still doubting if I need to use a guessing correction. I was hoping somebody here had some experience working with calculating scores on MC questions with multiple answers using negative markings might know the best deal way to deal with guessing.
 A: I'm not exactly sure what you are asking but here's the math I can figure out so far.
So suppose who have $n$ questions and the falloff threshold is $k$.
Each correct answer scores $a$ marks and each incorrect answer deducts $b$ marks.
Let $X$ denotes the number of correct answers. Suppose you have $c$ choices per question. 
So we know that $X\sim\mathrm{B}\left(n,\frac{1}{c}\right)$ if the paper is completely randomly attempted.
Let $Y$ be the score for that test. So we have
$$Y=\begin{cases}aX-b(n-X) &\text{if } &X>k\\
0&\text{if } &X\leq k\end{cases}$$
This simplifies to
$$Y=\begin{cases}(a+b)X-bn &\text{if } &X>k\\
0&\text{if } &X\leq k\end{cases}$$
The maximum possible mark would obviously be $aX$.
The expected mark from a student that completely guessed the answers will be
\begin{align}
E(Y)&=\left(\sum_{i=k+1}^{n}{((a+b)X-bn)P(X=i)}\right)+0P(X\le k)\\
&=\sum_{i=k+1}^{n}{((a+b)X-bn)\binom{n}{i}\left(\frac{1}{c}\right)^i\left(1-\frac{1}{c}\right)^{n-i}}
\end{align}
which we can just use a software to compute since the exact value is not something we need here.
