Hi Presh~

Could you write a blog post on game theory tips for fairly dividing household chores/tasks (for roommates, partners, and perhaps coworkers)?

Bonus points for providing easy division methods that people will actually use. In any case, keep up the good work. Thanks.

I thought this was a great question and I wanted to take stab at it. Here are a couple of ideas I came across the following two ideas.

Method 1: “I cut, you choose”

This is one of the most famous examples from fair division. Let’s imagine that two people are trying to split up a triangular slice of cake, but they might have different preferences for the filling or whipped creme topping. What’s the fair way to divide the cake?

The answer is astonishingly simple: one person gets to cut the cake, but the other person gets to choose his piece first. The idea here is that the person cutting will be encouraged to make the pieces as equal as possible so that he ends up with a good piece. If he makes one piece too large, then the other person will just take it, leaving him with a small piece.

This cake-cutting problem is generalizable to all sorts of areas, as I have written about before in a peanut butter commercial and in terms of splitting a can of Coca Cola.

In cake-cutting, the goal is to get as large a piece as possible. For chores, the problem is exactly the opposite: you want to pick the smallest “piece” that corresponds to the least amount of effort.

The problems are analogous, and the “I cut, you choose” method works very well here. Here is how you can divide up chores between 2 people:

Let one person divide the chorses into two separate lists. Let the other person choose the which list to do.

The idea will work out in a very similar way. The person who makes the list of chores has to make the two lists as equal as possible. If not, the other person will choose the easier set of chores.

The “I cut, you choose” method is great for 2 people. It can be extended to 3 and 4 people, but those scenarios can prove to be a lot harder. For details, see this paper about fairly dividing up chores for 4 people.

Method 2: Tug-of-war method

The “I cut, you choose” method is great for infrequent chores like taking out the trash or cleaning the gutters. But it is not so good about dividing up daily chores like cooking.

If you’re in a house where you wish to split the job of cooking, what is the best way to do that?

The idea is you want to split up the amount of work and have some way that both people can agree to the commitment.

I came across a very interesting idea from a webpage posted in 1993. The author turned out to be the economist Robin Hanson who writes the interesting blog Overcoming Bias.

Here is Hanson’s advice found on this page:

When both my wife and I were employed, we split the task of cooking (or picking up) the evening meal with a “Tug-O-War” board, like:

      ---------------------------------------
      |h           /#\   |                 h|
      |i  0    0   ###   0    0    0    0  e|
      |m           \#/   |                 r|
      ---------------------------------------

This has a row of seven holes, with a peg sitting in one of them. The rule is that when you cook, you get to move the peg one step in your direction. If the peg gets all the way to your end, the other person *has* to cook. Now you can cook a few nights in a row if you feel energetic, or wait a few nights if you’re sick. If neither of you wants to cook, the person farther down is expected to cook.

The tug-of-war method is interesting to me for a couple of reasons.

First, with a daily chore it is very hard to remember who has done the work. Each person has a bias towards thinking they have done more of the cooking than they actually have. The board is an unbiased way to keep track of the work, and each person can see the results day by day.

Second, the board has a commitment rule so encourage the work split to stay near equilibrium. If one person has done too much of the work for the past few days, the other person knows it too, and will have to make up for the cooking.

The same technique can be generalized for more people. Instead of a single board, you make several columns with each person’s names. Everyone starts out in the middle, and each time you do a chore, you move your marker up and everyone else’s marker down. Anyone who is at the bottom has to do the chore, and anyone at the top does not have to.

I’m sure there are other interesting ideas for splitting up chores. How do you divide up work in your house?

Classroom discussion questions

1. Let’s say two roommates agree to a division of chores. Suppose one roommate decides to hire help rather than doing the work. The other roommate feels this is unfair because he cannot afford to hire help. Is this a valid complaint?

2. How might you penalize someone for not doing their chore?

3. One roommate feels the kitchen floor should be mopped every two weeks, while the other thinks a month is fine. What is a fair way to decide how often to mop, and who does the mopping?

It’s not exactly a scientific observation, but time and again, I have seen friends whipped by crazy girlfriends. (I would equally say there are a lot of crazy boyfriends, so don’t get caught up in the gender).

It turns out there is a game theory explanation for this phenomenon which is what I want to explore in this article.

You’ll see why crazy people get their way, and how you can use a similar strategy to fight back.

Battle of the sexes

To begin, consider the following model of conflict. This game is one of the simplest examples in game theory, often the first example presented in a game theory course.

Alice and Bob are planning to go out for an evening. Bob wants to go to the football game, but Alice wants to go to the opera. Still, both would prefer to be with the other person than going out to an event by themselves.

Let’s say each person gets 3 points to go to their favored event, 1 points to go to the other event but be with their significant other, and 0 points if the two do not go together.

The outcomes can be represented in the following matrix:

The question is: how will this game play out?

Solution 1: opera or football

This game is very easy to solve. We need to consider what each person’s best response is, given what the other person might be doing.

Here is how Bob thinks about the problem:

–If Alice goes to the football game, then I should also go to get 3 points rather than going to the opera for 0 points

–If Alice goes to the opera, I might not like the opera, but if I go I’ll at least get 1 point. If I watch football I would instead end up with 0 points.

Bob’s strategy is to guess what Alice wants to do and follow.

Alice’s reasoning is exactly the same: she wants to be with Bob, so she should pick the choice she expects him to do.

The outcomes that both go to football, or both go to opera are the solutions of the game.

These are the Nash equilibria of the game.

The pleasant finding is that both players end up choosing mutually beneficial outcomes.

But the annoying part is the game has two different solutions. I mean which one do they end up doing?? Do they go to football or opera? The concept of the Nash equilibrium falls short here in providing a specific prediction.

However, there are other ways to think about the game.

Solution 2: correlated equilibrium

If the game is repeated, as it would be in a relationship, there is a good compromise that Alice and Bob could agree to.

What they could do is flip a coin to decide whether they go to football or the opera. Over time the events will balance out, and they both guarantee that they are together. They will get an average payout of 2.

This seems like a very reasonable solution. It also makes a lot more sense they flip a single coin and coordinate their choices, rather than each player flipping a coin individually and hoping to end up at the same place (the mixed strategy Nash equilibrium doesn’t make much sense in this game: a good chunk of time the two end up in different places)

In a healthy relationship, there is compromise and people may choose the correlated equilibrium.

But a crazy girlfriend would not tolerate this. She wants to get her way, and she uses another strategy.

Solution 3: changing the game

Let’s suppose Alice really, really wants to go to the opera, and she wants to convince Bob that it’s in his best interest too.

Here is a drastic strategy that Alice could use to change the game.

Alice takes out a $20 bill from her purse, and then announces the following:

Look Bob, I really want to go to the opera. And if I think there is even a chance we are not going, that would stress me out. I’ll vent by burning this $20 bill.

I’m going to the room to think about what I’m going to do. Come knock on my door with your choice in one minute. But think about what’s best for both of us.

Alice has introduced another strategy to the game called burning money. In this game, Alice first chooses whether to burn her own $20 bill, and then both players choose where they would like to go.

Let’s say that burning a $20 bill will destroy 1 point of utility for Alice. How will this game play out?

The way to analyze this game is to write out a large matrix with all the choices and solve as one normally would.

Alice has two moves in her strategy: she can either burn the money or not, and she can either go to the opera or not. Let’s abbreviate Alice’s choice of “burn the money, go to opera” as BO, and similarly the rest of her choices as BF, NO, NF.

Bob also has two moves in his strategy. He needs to decide what he will do if Alice burns the money, and what he will do if Alice does not burn the money. Let’s abbreviate Bob’s choice of “if she burns the money I go to the opera, if she does not burn I go to football” as OF, and similarly the rest of his choices as OO, FO, FF.

The matrix of payouts is as follows:

We can now eliminate bad strategies by a concept of iteratively deleting weakly dominated strategies (see more in this game and this game).

To begin, Alice will realize that NO is weakly better than BF, so she will never play the strategy of burning the money and choosing football. Both players realize this and “cross out” that option in the matrix:

Looking at the matrix, Bob realizes that OO is weakly better than OF, and FO is weakly better than FF. So both Alice and Bob eliminate those options in their mind.

The process continues, and you can check the only strategies that remains are NO for Alice and OO for Bob.

The result is this: the equilibrium outcome is that Alice does not burn the money and goes to the opera, and Bob chooses to go to the opera whether Alice burns money or not.

That is, they end up going to the opera just like Alice wanted!

Just think about what happened: Alice got her way because she threatened to burn money. But she never actually has to burn the money: she gets her way because she threatens to torch her own utility.

This seems to capture an element of how spoiled brats in real life operate. They do not always throw tantrums. They only have to threaten to throw a tantrum and act unhappy to force everyone into their choice.

(While I find the solution interesting, I should mention there is controversy about the idea. It is odd that Alice can change the game by threatening to use bizarre behavior. This is an issue raised in this paper )

How to fight back

Bob has a couple of options for fighting back in this game. He can threaten to burn money pre-emptively too, which might get Alice to see his side and drop the pettiness.

Or he can play it safe and change the game once and for all. While crazy people do change and grow, it is a question of how fast and whether it is worth the effort.

Of course there are other ways to deal with crazy people, and I could go on and on. But in my opinion it is often not worth the time and effort. Sometimes breaking it off is the best move.

Let’s you and I play this very simple game and analyze the best strategy.

Imagine we are playing this game in a college experiment. We each have a chance to win money depending on how we play.

Here are the rules:

–You and I each secretly play “A” or “B”

–If we both pick “A,” then we each get $4

–If one person picks “A,” and the other “B”, the person picking “A” gets $1 and the person playing “B” gets $3

–If we both pick “B,” then we each get no money and leave with $0

Here is the matrix of payouts:

The game is played once. What option would you pick?

Analyzing the game

This game is a no-brainer: it is a dominant strategy to pick “A” and both of us should get $4.

Verifying this is an easy task. Each person thinks about the best response to the other player’s move. If the other player picks “A,” then it’s best to also pick “A” to get $4 rather than “B” to get $3. If the other player picks “B,” it is also better to pick “A” and get $1 rather than “B” to get $0.

The best strategy is to pick “A,” regardless of what the other person is doing. Both players should easily cooperate and get $4.

There is no sensible reason to pick “B.” And yet, that’s exactly what researchers found people doing over half a century ago, in a similar game played with pennies rather than dollars.

The results were astounding: more than 50 percent ended up playing the strategy “B”!

(The experiment is referenced in The Survival Game regarding this 1960 article)

We could be tempted to chalk up the result to the small payouts, or maybe people did not understand the rules. It is possible that people did not take the game seriously.

But the researchers also raised another possible, biological explanation that’s worth investigating.

The green-eyed monster of jealousy

As explained above, both players maximize their payout when they pick “A.” It should be obvious that picking “A” is the best thing to do. Except, perhaps we are thinking about the problem with the wrong motivation.

In game theory, economics, or business, we often choose the option that brings us the highest profit in absolute terms. All things equal, we would rather have $1,000 than $100.

But people do not always think in absolute success. They can sometimes think in terms of relative success: the goal is not to maximize payout, but rather, in the researchers’ words, “to maximize the difference between one’s self and the other player.”

There is further experimental evidence of this idea, as explained on Michael Shermer’s blog:

Would you rather earn $50,000 a year while other people make $25,000, or would you rather earn $100,000 a year while other people get $250,000? Assume for the moment that prices of goods and services will stay the same.

Surprisingly — stunningly, in fact — research shows that the majority of people select the first option; they would rather make twice as much as others even if that meant earning half as much as they could otherwise have. How irrational is that?

…In this case, relative social ranking trumps absolute financial status.[emphasis mine]

The point is that relative thinking, and jealousy of other people’s success, factors into how people think about money.

So let’s use this idea to re-analyze the game.

The destructive nature of relative thinking

The original payout matrix accurately reflected the total payouts for each player. It was apparent that “A” was the obvious choice.

But imagine that people were thinking in terms of relative rather than absolute payout. That is, they only cared about how much more they earned than the other player.

The game is now transformed into the following payouts:

–If we both pick “A,” then we each get $0 more than the other person

–If one person picks “A,” and the other “B”, the person picking “B” gets $2 more than the other player

–If we both pick “B,” then we each get no money and also leave with $0 more than the other person

Here is the matrix of relative payouts:

The payouts of this game are completely changed from the original one. Notice how the original game–which was non-zero sum and mutually profitable–is now suddenly a zero sum, and competitive, game.

The strategy becomes competitive too: in this game, it is a dominant strategy to pick “B,” meaning both players are expected to leave with nothing.

Rather than cooperating for mutual gain, both players “happily” end up with nothing to avoid letting the other person gain in stride.

Get over your money jealousy

This outcome is sadly not just theoretical: people gleefully act towards mutual destruction out of money jealousy.

I went over many such examples in a previous article. In that article, I made a plea for people to be more calm and not worry so much about other people’s success.

Here’s my closing advice from that article. I hope it will inspire people to be less jealous and look for mutual gain:

I have my own personal analogy to get over jealousy. I think about success as filling up water flowing from an ocean. Each of us has a different size glass that represents a personal level of achievement. There’s really no point worrying if your neighbor has a bigger glass than you since there is more than enough water to go around. If you want to get more, then focus on what you can do. Success will come from building your own glass and filling it, not from shattering what your neighbor has. It’s time to put the green eyed monster of jealousy to rest.

The other day I came across another game that dates all the way back to the ancient Greeks and Romans.

The game is called Morra. While there are many variations on the game of Morra, there are typically a few common rules.

You don’t need any special equipment to play Morra: you just use your hands like. In the two-person game, each person’s move is to extend a certain number of fingers on their hand and simultaneously guess something about what the other player will show (either the number of fingers or the sum of all fingers). For instance, you might show two fingers and guess the other person will show three. Points or money are awarded for correct guesses.

Because of its simplicity, Morra is a fun alternative to rock-paper-scissors. The interesting part is the strategy is slightly more nuanced. Today I want to analyze and solve a specific version of the game of Morra.

The rules of this version of Morra

As I said before, Morra is a game with many variations. I read about this version of the game from the book The Compleat Strategyst, a Rand publication that is a good read about zero-sum games.

The rules of the game are this:

–There are two players
–Each player can extend either 1, 2, or 3 fingers
–Each player has to simultaneously guess what the other will show
–If exactly one person guesses correctly, the person wins the SUM of the numbers showing
–If neither guesses right, or both do, then the payoff is zero

An example of gameplay can help illustrate the rules.

Suppose you and I play Morra. Let’s say that I extend 2 fingers and guess 2, and you extend 2 fingers but guess 3. In that case, I was the only person who guessed correctly so I would win. I get a payoff that is the SUM of both of our hands, so I win 4 points (or winnings could be money).

As stated, Morra is zero sum game that can be solved using game theory analysis. Let us model the game and figure out the optimal strategy.

Analyzing the game

Each person has to do two things in the game: each has to extend 1, 2, or 3 fingers, and each has to guess 1, 2, or 3 for what the opponent is holding.

We can model each person’s strategy as an ordered pair (fingers to show, number to guess). For compactness, we can drop the parenthesis and commas and write 12 to mean “I will show 1 finger and I will guess 2″.

Since there are 3 choices for what to show, and 3 choices for what to guess, each player has a total of 9 strategies: 11, 12, 13, 21, 22, 23, 31, 32, 33

Now we can take the next step and write out a payoff matrix. If you and I both play 11, then each of us guess correctly, so the payoff is zero. If you play 12, and I play 11, then only I guess correctly, so I get the payoff of 2.

As each player has 9 strategies, there will be 81 payoff cells to calculate. I will spare you the tedious details, but here is the payoff matrix:

Because this is a zero-sum game, we only need to write the payoffs for player 1. Positive numbers represent times player 1 wins, and negative numbers correspond to times player 2 wins. The payoff matrix for player 2 is the exact same matrix, with each entry multiplied by -1.

Finding the solution

It would be very difficult to figure out the solution to this game analytically. Luckily we have numerical methods to solve this game that use something called linear programming.

I am going to omit the details fo the process and just get to the solution. (I was unable to find a suitable online solver for this 9×9 game–please let me know if you are aware of one).

The best strategy turns out to be a mix of just three choices. You only want to play the strategies 13, 22, and 31, and you play them with the ratio of 5:4:3.

In other words, the best strategy is:

–play 13 (show 1 finger and guess 3) with probability 5/12
–play 22 (show 2 fingers and guess 2) with probability 4/12
–play 31 (show 3 fingers and guess 1) with probability 3/12

The game turns out to be a bit of a rock-paper-scissors dynamic after all: you end up having three choices that are worth playing, and you want to mix amongst them.

Understanding the solution

If both players choose these strategies, then the expected value to each player is 0. This is not a surprise at all: Morra is a fair game just like rock-paper-scissors, and you can’t expect to win if your opponent is playing correctly.

Of course, if your opponent does not play correctly, then you are in a great position to profit.

If you play the equilibrium strategy, and the other person plays 11 by mistake, then you can expect a profit of 0.17 (if the game is played in dollars, then you can expect a profit of 17 cents). The same is true if your opponent mistakenly plays 33.

Similarly, if your opponent mistakenly plays 21 or 23, then you can expect to profit by 0.08.

Notice it is not obvious that playing 11, 33, 21, or 23 is a mistake unless you compute the optimal strategy and see it.

So if you want to be a bit devious, then you should teach your friends Morra and not tell them the best strategy. As described in the Compleat Strategyst,

“It is probably a good game to teach your friends, since the solution is easy to memorize, and yet difficult to intuit.”

Competitive Morra

Morra is a social game and other versions are played with three or even four people.

I came across a couple videos where people are playing Morra. They are speaking in Italian so I don’t know exactly how they are scoring or which version they are playing. But the game has an incredible rhythm to it, and I love how quickly they are playing.

If you’re going to be playing this fast, you better know the best strategy and play it appropriately or you will lose very quickly!

Youtube Video: Morra

The article is about three friends who agree to share a cab, and the possible ways they can split the costs.

I highly recommend you read the article.

The thing I liked most is the article describes various fair division methods. As I have described before in my article about splitting restarant bills, fair division is not just a mathematical concept. Fair division depends on social norms and how people perceive fairness.

Therefore, it is useful to understand many methods of fair division and have them in your toolkit. Below I will describe some of the fair division methods mentioned in the article about splitting cab fares.

The details of the cab ride

The situation is a common one: three friends agree to share a cab to different destinations, and they need to split the costs fairly. How can you do that?

More specifically, let us consider the following situation:

Let’s say that passenger A’s usual fare would be $1, passenger B’s is $5 and passenger C’s is $9. If all three share a cab (and assuming A and B are allowed to hop out on the way to C’s destination, without incurring any special fees), the total bill would be $9 — rather than the $15 they’d have to pay, total, to ride alone. How should they divide up the cost of the shared $9 ride? Or, put another way, how do they share the $6 of total savings?

Think about how you might solve this problem if you were splitting a cab with your friends.

Division methods

The author of the article, Carl Bialik, tackled the question by asking several economists how they would solve the problem.

He additionally came up with his own solution of how to divide the trip.

Here is a table that summarizes the different methods. Below I will explain the logic in more detail.

Method 1: Proportional by time in cab

The author of the article took a stab at the problem before asking the economists. He suggested that each person should pay proportionally to the time they spent in the cab. That is, the first leg to passenger A’s house should be shared by all three, then the next leg to passenger B’s house should be divided by B and C, and finally the rest of the way should be paid by passenger C.

The logic here is that you are paying for the time you are in the cab. Under this method, passenger A pays 33 cents, passenger B pays $2.33, and passenger C pays $6.34.

The advantage of this method is it is simple to implement. After a passenger leaves the cab, the fare can be read, and each person immediately knows how much to pay.

The downside is that passenger C ends up saving less percentage-wise. In this example, passenger C only saves 30 percent compared to its normal fare, versus passenger B who saves 53 percent and passenger A who saves 67 percent. It seems a bit unfair that passenger C does not exactly share in the savings. One could argue passenger C should save the most as his trip home is most inconvenienced by having to drop off A and B.

Method 2: split the savings proportionally

Another way to think about the problem is to consider the savings surplus. Had each person gone home separately, the trip would have cost $15 = $1 + $5 + $9. By sharing the cab, the passengers only pay $9, which is a savings of $6.

This method frames the problem as how to split up the $6. The proposal is to split the $6 proportionally, using each person’s individual ride cost out of $15 for the ratio.

Since passenger A would have normally paid $1, passenger A gets a 1/15 share of the $6 savings, or 40 cents. Similarly, passenger B gets a 5/15 share of the $6 savings, or a $2 savings. Finally, passenger C gets the remaining 9/15 share of the $6 savings, or a $3.6 savings.

Ultimately, the ride is split with passenger A paying 60 cents, passenger B paying $3, and passenger C paying $5.40.

Sharing the savings proportionally makes a lot of sense, and this is a preferred method in many legal settings like in bankruptcy situations.

The downside is the answer is a bit harder to compute. The passengers need to know how much the final ride costs before any split can be made.

Method 3: Bargaining solutions

Another way to split the cab ride is by using a procedure known as the Nash bargaining solution. This is a precise mathematical formulation in which each player seeks to maximize his profit from agreeing to a deal, knowing that he can walk away and break the deal.

The amount a person saves depends on how important he is to the deal. Suppose that it is a busy night, and cabs are preferring to pick up groups of 3 passengers rather than individual fares. In that case, each person is vital to catching a cab and equally important.

The Nash bargaining solution, in this case, is that each person gets an equal share of the $6–each gets $2.

This is quite unrealistic as it means passenger C pays $7, passenger B pays $3, and passenger A profits by $1 for his role. The odd solution is because we assumed that 3 passengers were needed in order to hail a cab, and that means each person has equal bargaining power.

A more realistic assumption is that passengers can form subgroups, or coalitions, and continue to bargain. If passenger A is being annoying, then passenger B and C could just split off and take their own cab. In that case, the two would still end up paying $9, which means a saving of $5 plus the satisfaction of not having to travel with passenger A.

This means passengers B and C could argue $5 of the savings belong to them, split as $2.50 per person. They could then come to A and offer the remaining $1 of surplus from all three riding be split equally as 33 cents per person. This leads to passengers A paying 67 cents, passenger B paying $2.17, and passenger C paying $6.18.

Bargaining solutions are extremely important in game theory as they depend on how powerful each player is.

Method 4: Equal division of the contested sum

There is another method that is one known as an equal division of the contested sum.

The idea is you look at which portions are contested, and then you split that evenly. This method has historical importance and I have written about it before: how game theory solved a religious mystery.

I will not repeat the details of the algorithm here. The interesting part in this case is this method produces a exactly the same result of the bargaining solution.

Extension: my method to split cab rides

The problem leaves out an important detail in cab rides. Normally fares are quoted as some base cost, plus an amount that depends on mileage. Additional passengers usually cost extra too.

In the city of Chicago, the cab fares as of this writing are:

• The flag pull or initial charge is $2.25 for the first 1/9 mile.
• The additional fraction of a mile charge is $.20 for each additional 1/9 mile.
• Every 36 seconds of time elapsed is $.20.
• The flat fee for the first additional passenger over 12 and under 65 is $1.00.
• Each additional passenger after first passenger, over 12 and under 65 is .$50.

For three passengers, the initial cost is $2.25 and the two extra passengers add $1.50 to the fare.

From my perspective, the base amount of $3.75 should be borne by all passengers: each person should pay $1.25 minimum to ride. (You can also add in other surcharges if there are any, like tolls or fuel surcharges)

From there, I would split the cab fare proportionally based on time in the cab.

So my method is two steps:

1. Split the initial and extra passenger charges equally
2. Split the remaining charges proportionally

For instance, let’s say three people end up with a $12 fare with tip, and the passengers normally would have had $5, $10 and $15 cab rides.

The first step is to split up the initial and passenger charges. As explained above, that means each person pays $1.25 for a total of $3.75.

The second step is to split up the remaining $8.25 proportionally. Riding separately, the passengers would have racked up $30 in cab fares. Thus, the proportions are 5/30 for the first passenger ($1.38) , 10/30 for the second passenger ($2.75), and 15/30 for the third passenger ($4.12).

Therefore, the passengers pay $2.63, $4, and $5.38. To make it easier, people would probably round that to $3, $4, and $5.

How do you split cab rides?

Personally my cab splits are pretty easy. My friends and I usually take a cab to the same destination, and so we just split everything evenly.

The issue is when you are going to different destinations, as in the problem above. How would you split a cab fare when going to different destinations?