# 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?

• Maybe try "punishment strategies" in the repeated games literature. A punishment strategy costs the punisher, but keeps the opponent from deviating to the bad outcome. There are surely a lot of models that would admit an interpretation you are looking for. – Pburg Nov 17 '14 at 19:09
• This is such a great question. I used to keep my dishes in my room so my room mates wouldn't leave them dirty in the sink. – dustin Nov 17 '14 at 19:33

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.

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.

• This misses the point about discouraging the roommate from being messy in the future by not cleaning and encouraging my cleaning. Hence, you'd need a repeated setting. – Pburg Nov 17 '14 at 20:43
• The question I offered shows why the optimal level of dishes might not be done in the first place. To go further into those incentives and punishment, as you suggest, I've amended my answer to offer further information on how to model it as a repeated game. – Shane Nov 17 '14 at 20:47

This is called "chore division". It's in some sense dual to the better known topic of "fair division" (dividing a cake, an inheritance, the contents of a treasure chest etc.). You can find out more via web search.

If we assume that

1. There are $n=2$ roommates (players),
2. After $N$ days the dishes will be washed by someone else (mother comes for a visit each N days for example) or by you yourself (so that the game necessarily terminates after $N$ days). I will assume the worst version for you, i.e. that at the N-th day you are the one who "has to" wash the dishes.

then we get a simplified version of the above situation which is a $N$-stage $2$ player game that can be solved recursively. Every player has two pure strategies $$S^I=S^{II}=\{\text{(w)ash, (n)ot(w)ash}\}$$ with (possible) payoff matrices when $0$ days remain $$Α_0=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&0\\(nw)&2&\color{red}{0}\end{array}\qquad \text{ and } \qquad B_0=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&2\\(nw)&0&\color{red}{2}\end{array}$$ (the payoffs in red are due to the assumption that you wash the dishes on the N-th day - you can of course change that) and when $n$ days remain $$Α_n=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&0\\(nw)&2&Γ_{n-1}\end{array}\qquad \text{ and } \qquad B_n=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&2\\(nw)&0&Γ_{n-1}\end{array}$$ The payoffs can be explained as follows

1. If both choose to wash - (w) vs (w) - then each has a profit of 1, because the dishes are clean and the effort has been split,
2. If you choose to wash - (w) vs (nw) - then your pleasure is offset by your effort, thus your profit is $0$ but your roommate takes a profit of $2$ (for obvious reasons).
3. If both choose not to wash - (nw) vs (nw) - then the game proceeds to the next day (i.e. there remain n-1 days).

On the last day (0 days remaining) if you choose not to wash (actually you have no choice) you will still be obligated to wash due to the assumption. Your roommate has clearly a dominant strategy on that day not to wash, so you will wash the dishes on the last day. This solves the last game and for $n$ you can solve recursively by substituting $Γ_{n-1}$ with $val(A_{n-1})$ and $val(B_{n-1})$ where with $val$ I denote the payoff of each player in the game $Γ_{n-1}$. So with this notation $$val(A_0)=0, \qquad val(B_0)=2$$ and for $n=1$ $$Α_1=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&0\\(nw)&2&0\end{array}\qquad \text{ and } \qquad B_n=\begin{array}{r|rr|}&(w)&(nw)\\\hline(w)&1&2\\(nw)&0&2\end{array}$$ and so on.

You can change the payoffs to make it suit better to your preferences. Moreover the assumption that there are only two players is not as restrictive as it seems, because from your point of view, you play a game against everybody else, so you can model everybody else as your "enemy" (of course even in that case it is a simplification). The assumption that the game terminates necessarily after $N$ days is due to my experience realistic... so you do not need to model it as an infinite game.

• This is a nice way of doing it also. In defense of modeling it as an infinitely-repeated game, that doesn't mean you're going to live with your roommate infinitely long, just that there's no deterministic day on which one of you will move out -- if there was, you would use backward induction and find the unique equilibrium being that nobody would ever do the dishes (at least if you model it with Prisoner Dilemma payoffs, which I think is a good approximation of the dynamics)! The discount factor $\delta$ in the infinitely-repeated game captures the idea that the game could end at any point. – Shane Nov 17 '14 at 20:53