Is there a game theory for doing the dishes in a shared living situation? It occurred to me this morning (when I was intentionally not tidying up my flatmate's dishes) that doing the dishes in a shared living situation, such as at an office, or living with housemates, might be subject to a kind of game theory. 
The idea being, that there's a conflict between doing the dishes yourself and immediately producing value for yourself, and the others, but at the sametime enabling the lazy housemate, and decreasing the likelyhood that they'll do the dishes in the future. 
Are there any studies or research or theories around this?

You could take the approach @Pburg suggests and view it in the repeated games context, but I think it's simpler to view it as a positive externality question. Here's a related question (with solutions) from a game theory class (where improving one's garden, I argue, is comparable to doing the dishes):
Question:
Alice and Bob are neighbors, and each maintains his/her own garden. Each enjoys looking at the other's garden, as well as his/her own. This enjoyment is increasing in the quality of the gardens, but a higher quality garden requires more effort. Alice has Saturday off from work, while Bob has Sunday off, so Alice works on her garden before Bob. That is, Bob observes the quality of Alice's garden in deciding how much effort to put into his own, but Alice does not observe the quality of Bob's garden before making her decision. For all $i \in {A,B}$, $u_i(e_i)=e_i(c+e_{-i}-e_i)$, where $c>0$ is a constant and $e_i$ is $i$'s choice of effort. Find the subgame perfect equilibrium effort levels. Is there a first- or second- mover advantage?
Solution:
When Bob makes his decisions, he will best respond to Alice, whatever her effort choice was. That is Bob solves
$$
\max_{e_B} e_B(c+e_{A}-e_B)
$$
Taking the FOC yields $e_B^* = \frac{c+e_A}{2}$. Alice anticipates this, so she solves
\begin{align*}
&\max_{e_A} e_A(c+e_{B}-e_A) \\
=&\max_{e_A} e_A(c+\frac{c+e_A}{2}-e_A) \\
=&\max_{e_A} e_Ac + e_A\frac{c}{2} - e_A^2\frac{1}{2}
\end{align*}
Taking the FOC yields $e_A^*=\frac{3}{2}c$. That means that $e_B^*=\frac{5}{4}c$. Then $u_A=\frac{3}{2}c\left(c+\frac{5}{4}c-\frac{3}{2}c\right)=\frac{9}{8}c^2$. $u_B=\frac{5}{4}c\left(c+\frac{3}{2}c-\frac{5}{4}c\right)=\frac{25}{16}c^2>u_A$. Bob has a second-mover advantage as his payoffs are higher than they would be in the static game.
Comments:
The related concepts here are dynamic games (i.e. players make decisions sequentially) and subgame perfect equilibrium. The method to solve it was backward induction, in which we first find what the last player to move (Bob) would do, then work out that the first person will do given she knows how the second will respond.
If you really want to capture the notion of punishments and cooperation, however, it is probably best to model this as an infinitely-repeated prisoner's dilemma and look at grim-trigger and limited punishment subgame perfect strategy profiles. The gist of that line of thinking is that your housemate will do the dishes if he believes that failing to do so will yield a punishment (perhaps you not doing your own dishes for a week, or forever in the grim trigger case), and therefore there can be an equilibrium where all parties do their dishes understanding this potential punishment.
