Fair division and game theory in a Jif Peanut Butter commercial

I was recently pointed to a Jif Peanut Butter commercial that touches on game theory. It deals with a mom trying to resolve a dispute between her two kids and the last slice of bread.

Here is the commercial on Youtube (my transcription of it follows):


Link to Youtube video

Jake: Mom it’s the last slice [of bread].
Mom: Hmm…Well then let’s share. We’ll cut it in half.
Cody: His half can’t be bigger than mine!
Mom: All right. I’ll tell you what. Jake gets to cut.
Jake: Yes! [cuts a big piece for himself]
Mom: But…Cody gets to choose.
Jake: [sad about this new information, passes the plate]
Cody: Nice…[after a moment, takes bigger piece for himself]. I got a pretty big half!
Jake: [smiles despite ending up with a smaller piece]

While the commercial is cheesy, it does raise some interesting topics from game theory. Here are some of the things it got me thinking about:

“His half can’t be bigger than mine”-preferences are crucial

While the statement that one half is bigger than the other is absurd mathematically, it is not absurd practically. The reason is that some “halves” are more valuable than others.

Consider splitting a slice of triangular pizza. The mathematical way to split the pizza in half is by dividing along the line of symmetry so each side gets equal crust and equal middle portion. But this is not a true halfway point if the toppings are spread unevenly. If the people dividing the pizza care only about toppings or have other preferences, then other divisions may be considered equal halves depending on who is playing. I, for instance, would be fine with a division that included the crust but less than 50 percent of the cheese.

This same issue of “different halves” comes up during many other divisions, such as cake-cutting (frosting and fillings are not evenly spread), car pooling (some days have worse traffic), and assigning tasks in an office (do you split tasks by time or by “unpleasantness”?).

Conflicting preferences can make the task of division very difficult. But amazingly there is often a method that can create agreeable divisions quickly.

“I pour, you choose”-a shortcut to agreeable division

I wrote about this method in my article relating finances to mechanism design. Here’s the relevant excerpt on the splitting method:

I have to thank my fifth grade math teacher for unintentionally introducing me to game theory. The game theory is hidden in the following extra-credit problem that he asked us:

My mother would often give a can of Coke to me and my two brothers and tell us to split it. Naturally, we all wanted more Coke, but our Mom told us to be fair and split it-without arguing. After we failed, she came upon a solution that suited all of us. What method did we use to split the Coke?

Most of us in the class thought mathematically and submitted answers about pouring 1/3 of the volume into each glass. My teacher told us these answers were incomplete because they described an outcome but not how the outcome would be achieved. Who would pour the Coke? And what order do people pick? And how do you make every one in the group trust each other?

Here is the solution the mom devised: one person was chosen to do the pouring. After the Coke can was empty, the person who did the pouring would be the last to choose his glass. The method proved to be successful-the Coke was always split evenly.

Why does the method work? It is because the method gives the person pouring an incentive to make the glasses as even as possible. If he does not pour the Coke evenly, he will suffer because the other brothers pick the fuller glasses first. Another way of thinking about the solution is that the other brothers are made to trust the person pouring. And this is a remarkable trait because the brothers’ interests are diametrically opposed.

The mother in the Jif commercial attempts to implement this method, but you’ll notice the kids don’t end up with equal halves.

What did she do wrong?

The rules must be stated in advance!

The “I’ll pour, you choose” method depends crucially on knowing the rules in advance. In the Jif commercial Jake thinks he gets to cut and choose first, which is why he made the pieces unequal to begin with.

But perhaps something else is going on? While the mom may have failed in fair division, she may have taught a bigger point about sharing. By withholding the rules, she essentially was baiting Jake into being greedy so she could penalize him. Mom signals that she won’t allow unfair division among brothers. This lesson may serve well for future times when Jake and Cody share and Mom isn’t around.

Further reading

Fair division is a topic I’ve touched on many times before. Here are some of my favorite articles which include some amazing reader comments:

How do you divide restaurant bills fairly? (62 comments)

How game theory solved a religious mystery (32 comments)

Game theory in The Dark Knight (50 comments)

How can you stop free riders and games of chicken? Try changing the game (18 comments)

The tragedy of the commons: Working during the holidays and why Thanksgiving almost didn’t happen (21 comments)

Why patience pays off in negotiations (11 comments)



Share this post:

| More

Previous post:

Next post:



  • Scott

    I’ve always enjoyed this strategy, if only for the fact that, when dealing with similar situations with my brothers as a child, I independently derived this solution.

    The first thing I thought when I first saw this commercial was what you state here: She didn’t tell the older brother the rules!

    But, being a commericial and, thus, a fantasy, there was no subsequent argument.

    In the real world, the older brother would no doubt argue that it wasn’t fair that he didn’t know the rules. Of course the mother could explain that that is what he gets for being greedy in the first place. I’m sure that, at that age, there would have been a non-zero amount of sulking.

  • Scott

    How does having more than 2 players affect the situation?

    Initially I didn’t think it would since the only way for the divider to maximize his portion is to create equal portions.

    But once I started thinking about it, I started wondering if there could be ways for the divider to get a larger portion.

    At first I thought this could be accomplished through deals or pacts, but I couldn’t find one that wouldn’t create an incentive for betrayal.

    Will this strategy always result in equal divisions, or are there ways to game the system for parties of more than two choosers?

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Good question. I am doing some more research on this topic of more than two players and have come across an interesting idea called an “envy-free” division.

    There is an online applet that allows you to experiment:
    http://www.math.hmc.edu/~su/fairdivision/calc/

    Here is a technical paper on the topic:
    http://www.math.hmc.edu/~su/papers.dir/bfe.pdf

    I’ll probably post about this as I learn more!

  • Keith

    My mother and father used this method to divide things up between my sister and I and I was always the divider. It was always quite bothersome to me, because no matter how hard I tried, I could never divide things 100% perfectly. I would always end up with like 49.75% of the total amount, which if you have ever had a sibling rivalry, seems like you got shortchanged by quite a lot. Still though – it was by far the best method of division possible.

  • DrObviousSo

    The three person split require a little bit of rules clarification, but it works. Suppose you, Presh and I are splitting a delicious pie, and I am to cut and choose last.

    Random first pick: I will guarantee one pie piece is bigger than the other two if you both agree to take your bigger piece and cut off anything over 40%. I cut the pie into 50%, 25% and 25%. You are chosen to pick first, take the 50%, and turn over 10% to me. You end up with 40%, I get 35%, and Presh is stuck with 25%. This can be optimized so I win by demanding anything over say 34%.

    Agreed first pick: Before the cut, we know Presh will pick first. I choose to cut the pie into 99%, 1%, 1% if you agree to (cut your part in half OR hand over all but 34%). Thus, I get either 50.5% or 66% of the pie.

    This resolution works because I am bribing one or both of you to act as my proxy in the selection process while still retaining my cutting rights.

    I would be willing to be this problem and solution can be reduce to a well known game theory problem and solution, I just don’t know a lot of game theory (I know enough to have used this on my siblings when I was younger though!)

    Yum. Pie.

  • DrObviousSo

    Erm, that should be “willing to bet”

  • Scott

    “Random first pick: I will guarantee one pie piece is bigger than the other two if you both agree to take your bigger piece and cut off anything over 40%. I cut the pie into 50%, 25% and 25%. You are chosen to pick first, take the 50%, and turn over 10% to me. You end up with 40%, I get 35%, and Presh is stuck with 25%. This can be optimized so I win by demanding anything over say 34%.”

    In this example, what’s my incentive to follow through?

    From my point of view, here are my options:
    Don’t agree, and you perform an even cut and I get 33.3333%
    Agree and follow through, and I get 40%
    Agree, but betray, and I get 50%

    There is no incentive for me to keep my bargain.

    “Agreed first pick: Before the cut, we know Presh will pick first. I choose to cut the pie into 99%, 1%, 1% if you agree to (cut your part in half OR hand over all but 34%). Thus, I get either 50.5% or 66% of the pie.”

    I’m not sure I completely understand this (not to mention that 99% + 1% + 1% is 101%), but it still seems that there is every incentive to betray.

    So, it seems that, from a single trial standpoint, you can’t prevent betrayal.

    But what about multiple trials with memory and possible punishment?

    Well, that depends on how the orders are determined.

    Scenario 1, static order:
    Brother A cuts and chooses last.
    Brother B chooses first
    Brother C chooses second

    Brother’s A and B collaborate. Whatever they choose, they agree to cut their piece in half, then swap halfs.

    Under this scenario Brother A cuts the pie into 98%, 1% and 1%.
    Brother B chooses 98%
    Brother C chooses 1%
    Brother A chooses 1%

    Brothers A and B cut their slices in half (0.5%, 0.5%; 49%, 49%) and swap halfs giving each brother 0.5% + 49% = 49.5%

    Compared to an even cut strategy brothers A and B are +16.1667% while brother C is -32.333%.

    What if Brother B decides to betray? Well this depends on how Brother A reacts in subsequent trials. Unfortunately, due to the nature of the game the only person in a position to punish anyone is the cutter and they can punish everyone but the first person to choose and, in doing so, will also punish themselves. In this scenario, Brother A cannot punish Brother B. Brother A’s only recourse is to resort to an even cut after the betrayal, thus abandoning the deal:

    Scenario 1b:
    Brother’s A and B collaborate. Whatever they choose, they agree to cut their piece in half, then swap halfs. Brother B agrees, but betrays on the first trial, afterwards Brother A resorts to an even cut.

    Under this scenario Brother A cuts the pie into 98%, 1% and 1%.
    Brother B chooses 98%
    Brother C chooses 1%
    Brother A chooses 1%

    Brother B betrays and subsequent trials proceed as:

    Brother A cuts the pie into 33.333%, 33.3333%, 33.3333%
    Brother B chooses 33.3333%
    Brother C chooses 33.3333%
    Brother A chooses 33.3333%

    As we increase the number of trials, Brother B’s average will approach 33.3333% thus, at some point, will drop below the 49.5% he would have in the previous scenario. So, with enough trials, betraly becomes less profitable. When does that happen?

    When there are 5 or more trials. When Brother B gets 99% on the first trial, and 33.3333% for 3 more trials, his average is 49.75%

    If there is another trial where Brother B gets 33.3333% his average drops to 46.4667%

    Scenario 2, round robin:
    Brother A cuts and chooses last.
    Brother B chooses first
    Brother C chooses second
    After each trial, the brothers move up in the order. Thus the selecting orders will be:
    B, C, A
    C, A, B
    A, B, C
    B, C, A
    etc.

    This is not significantly different than before. If Brother B betrays, then Brother A is still never in a position to punish him.

    Scenario 3, random ordering.

    This is the only scenario when Brother A will possibly be in a position punish Brother B for betray and, with betrayal involved (for example, Brother A cuts 99%, 1%, 1% when Brother A cuts and Brother B chooses second) Brother B would be better off cooperating. The problem is, it is never in Brother A’s best interest to punish, because he will always punish himself at least as bad as anyone else. If it is Brother A’s purpose to maximize his pie intake, he CAN’T punish anyone.

    Basically it depends on how many trials you are performing. In order to get Brother B to cooperate there have to be enough trials that betrayal will result in less pie, on average, than cooperation and this will have to be explained to him, of course. For single trial runs, however, Brother B should always betray.

  • DrObviousSo

    I guess I always just assume that an agreement must be followed through, which is probably a bad assumption in most cases.

  • malachi

    If you like board games, check out San Marco (http://www.boardgamegeek.com/game/1041), it uses the “I divide, you choose” mechanism. It is best with 3 players.

    Each round the divider changes. The divider gets a set of cards she has to split up, with some good cards and some bad (“limit”) cards. When one player gets to a certain limit number, the round is over and they pay a penalty, so the preferences are fairly apparent and change over time making the divider’s job fairly tricky.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    All good comments…I’ll do some research and write up my thoughts in a future article

  • Huey

    Kids shouldn’t be eating so many carbs anyway. :)

  • Benjamin Vitale

    A question on the cake cutting problem:
    How to divide fairly a two-(or more)flavored cake?
    Suppose you and your friend have a chocolate-vanilla cake.
    Division must be envy-free.
    Suppose that your friend likes cakes with lots of frosting, but you don’t care for it.
    So, we need to quantify the problem in order to solve it.
    To quantify how much you and your friend want chocolate or vanilla you have assigned monetary values in dollars to each section of the cake.
    Could u show an example how to work it out?

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Good question Benjamin…I’ll try to think of an example for a future article.

  • Jud

    This is actually the essence of old Celtic inheritance laws. Youngest son divides up the father’s property, but the choosing is done oldest first. The idea, then, is a pretty old one indeed.

  • Pingback: The ultimatum game played by children - Mind Your Decisions

  • Pingback: How to divide up chores fairly using game theory - Mind Your Decisions

  • Pingback: How would you explain game theory to a ten year old? - Quora

Previous post:

Next post:

This site is for recreational and educational purposes only (privacy policy)

Powered by WordPress