Game theory problem about 2 students claiming different work amounts

Question

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)

Answer 1

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

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