Htykuut

This is quite a simple question I heard and wondered what the mathematical absolute solution to it.

There are 2 students, the teacher saw their group project and graded it 100 points. Now he needs to grade each separately one for his work, Student A claims they both did 50% of the work while student B claims he did 100% while the other did nothing.

How should the teacher grade them separately and fairly?

Some answer I saw were (A, B): (75, 100), (75, 75), (50, 50), (100, 100)

We can setup the problem as follows:

1. The teacher decides on a functions $f(x, y)$ and $g(x, y)$ which describe how much of the credit each student should get based on their claims.


2. The students each report how much of the project they did. A claims A did $x$ of the project, and B did $1-x$ of it, while B claims B did $y$ of the project and $A$ did $1-y$. $0 leq x, y leq 1$.


3. The teacher awards the first student $f(x,y)$ of the total credit, and the second student - $g(x, y)$.

How, we can put restrictions on $f$ and $g$ to describe some properties we would like to impose. There isn't one correct set of restrictions, but for example to be this seems reasonable:

1. Be fair: if the students switched work, they should also switch grades, i.e. $f(x, y) = g(y, x)$.


2. Encourage doing the work: a student shouldn't be able to increase their grade by doing less, i.e. if $x < x^prime$, then $f(x, y) < f(x^prime, y)$, and similarly for $g$
   


3. Discourage lying: if A did $x$ of the task, he shouldn't get more credit for reporting $x^prime > x$, i.e. $forall x^prime > x, f(x^prime, y) leq f(x, y)$.


4. Encourage consistent reporting: if $x + y = 1$ and $x^prime + y^prime neq 1$, $f(x, y) + g(x, y) geq f(x^prime, y^prime) + g(x^prime, y^prime)$.

Clearly, to satisfy the above conditions, we need $f(x, y) = g(x, y) = c$ for some constant $c$, regardless of what they report. If makes sense to have $c = 0.5$. Note that this only nominally satisfies 4, because it's written with $geq$, but nobody gets anything from consistent reporting. We can change that by relaxing 2 and 3 to only hold almost surely. This means, for 2, $x < x^prime implies f(x, y) < f(x^prime, y)$ for all $y$ except a countable set of values. Then we have solutions like

$$f(x, y) = cases{x, text{ if } x + y = 1 \ 0.25, text{ otherwise}}$$

Not sure what you mean by "fairly". Here's one naive take:

The maximum individual grade is 100, and since this was a 2-person project, they get 100 if they did half the work. Assuming Student A is to be trusted, they are graded (100,100); And trusting Student B, (0,100).

Without any information on the credibility of their claims, there's no reason to trust one over the other. So we should take the average of the grades: $frac{1}{2}(100,100)+frac{1}{2}(0,100)=(50,100)$.

But there are flaws: Any student, regardless of their actual work done, can claim to have done half (or more) of the work and receive as high of a grade as possible!

Also, under the assumption that the students know they will be graded individually, the only incentive to doing more than half the work is for a higher group grade. Realistically, very few students would do more work than necessary for a maximum individual grade of 100 to pick up a noncontributing partner, just for a higher group grade.