If you give your future self a carrot

I spend a lot of time trying to figure out how to be productive. How can I fit more work, sleep, exercise, and leisure into my day? How can I overcome procrastination? How can I focus on a task for longer periods of time? For a long time, my strategy was something like the one illustrated in the following (brilliant) comic:

Image

Commitment devices like StickK received a lot of press when it was mentioned in Freakonomics – the premise was basically that people would commit to achieving a goal, like exercising, and pay a penalty when they didn’t stick to the task (the money could go to a person of their choice, a random charity, or an anti-charity – one whose cause the user opposes.) Users lost more weight when there was money on the line.

StickK gets its name from the carrot-or-stick analogy. The idea is that people might respond better to sticks (punishments) than to carrots (rewards), because they are loss averse: assuming that the income effect is negligible, losing $5 hurts a lot more than winning $5 is pleasurable.  So, when I create a StickK account and set a goal, I’m playing a game with my future self.  I commit to, for instance, working out every day, and if I don’t succeed that day, I have to give my roommate a $5. The commitment is self-executing – say my roommate wants those $5, so she’s definitely going to come and get the money from me if I deserve to lose it. Then, when my future self is debating whether to go to dance class, I’ll have to think, “Would I rather go to the class, or would I rather lose $5?” Of course, I’d rather not have to make the commitment at all, but Future Me won’t stick to the task if I don’t.

One problem is that I might value the time I would get back from being lazy far above $5. I might have to set the penalty at a price > the most I would be willing to give up to get that chunk of time back at the time that Future Me is making the decision (and that might be a pretty high number). Another issue is that for many people, succeeding in something like “not procrastinating” might feel like an even bigger loss than the procrastination itself – what if you don’t do as well as you would like at the task? What if you fail at it? Maybe you’d rather not find out – procrastinate instead. (In that particular case, the solution isn’t to penalize yourself with five pushups, punching yourself in the nose, and giving $1 to the NRA. You should probably figure out how to change how you evaluate your payoffs so that failing doesn’t hurt so much.)


What do kittens have to do with rising tuition?

If you read the Financial Times, you might suspect from an article on Monday that kittens have something to do with rising tuition and Prisoners’ Dilemmas. Let me assure you that they don’t.

A friend of mine sent me the article, which cites a model designed by a team of Bank of America consultants who use the Prisoners’ Dilemma to explain rising college tuition. Here is the graphic they used:

kitten

Fig. 1: Things that are pairwise irrelevant to each other:

a kitten, the Prisoner’s Dilemma, and rising tuition.

They explain that the college ranking system (assuming two colleges) is a zero-sum game. If one college moves up, the other one moves down. “A college can move up in the rankings if it can raise tuition and therefore invest in the school by improving the facilities, hiring better professors and offering more extracurricular activities.” And therefore, they conclude, this is why college tuitions have been rising and why student debt will continue to rise.

First glaring problem: (raise, raise) is a Pareto-optimal outcome as they’ve set up this game, but what they probably meant to say was that it is a Nash equilibrium. Or maybe they meant to say that “raise” is the best response for each college. Anyway, in this game, (don’t raise, don’t raise) is also Pareto-optimal (but not a Nash equilibrium)!

Secondly, they’re trying to illustrate a kind of ratcheting problem: both colleges raise tuition to raise the quality of the resources at the school, in order to maintain their rankings. But, this means it’s a repeated game. In repeated games that have a finite horizon, defection happens at every step, but at infinite horizon games, cooperation can occur. Now, let’s just assume that this is an infinite horizon game, which is what the folks at B of A are assuming when they predict that college tuition will keep rising indefinitely, beyond mere inflation. What incentive is there to cooperate and keep tuition low? According to this game, none.  And according to what you might expect in reality, none – is it plausible that, in the absence of antitrust laws, that colleges would want to collude to keep tuition low, and that because they can’t collude, they are doomed to raise tuition every year against their wills? Nope.

Then, we come to the matter that in fact this game can’t be infinite horizon as it is presented here.  The simple reason is that, even if education is becoming a larger and larger share of a household’s spending, and even if the student is taking out loans and borrowing against his future expected earnings, he still has a budget set that he can’t exceed. Furthermore, the demand for attending college at a particular university should drop as soon as the tuition exceeds the expected lifetime earning/utility advantage for whatever the student sees himself doing in 4 (or more) years over the alternative. So, there will be some stage at which the utilities change and it becomes a best strategy for neither school to increase its tuition. So, it’s a finite stage game and the increase will stop somewhere, namely, where price theory says it should. [1]

Finally, it’s not clear that increasing tuition actually has such a strong effect on school rankings or that colleges are in such a huge rankings race. And, even if students at colleges outside the very top schools tend to choose a college based on things like food quality and dorm rooms, students don’t demand infinitely luxurious college experiences at infinite prices. Evidence: Columbia students feel they’re overpaying for food, and feel entitled to steal Nutella.

The lessons here are these: It’s not a Prisoner’s Dilemma in a strong sense if the cooperative result isn’t strictly preferred to the Nash equilibrium. Don’t model a tenuous game where the game isn’t relevant to the ultimate result (tuitions will stop rising at some point). Don’t assume that trends are linear, when they are definitively not linear. And, don’t put a kitten on your figure just because you have some white space — it really doesn’t help.

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[1] Actually, the game doesn’t have to be finite horizon. Suppose the upper limit that the colleges know they can charge is A, and the current tuition is B_t. Then, at each stage, they could increase tuition by (0.5)*(A - B_{t-1}). But, as the tuition approaches A, the increases become smaller and smaller until they pretty much just vanish, and it would be the same as stopping, because there is a time at which the tuition would stop affecting rank (a college isn’t going to improve it’s rank by charging each student an extra cent.)


The Monty Hall Deception

When I was in middle school, I consumed a lot of typical nerd literature like Richard Feynman’s “Surely You’re Joking, Mr. Feynman” and anthologies of mathematics puzzles from the Scientific American by Martin Gardner. In the latter, I first encountered the Monty Hall Problem, and it goes something like this:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

It turns out that, yes, it is always to your advantage to switch your choice. This is a solution that has been notoriously difficult for people to wrap their heads around. After all, when you picked a door, the probability of having picked the door with a car was still 1/3, and after a door was opened, there would still be a car and a goat behind the remaining two doors – it seems as through the probability of choosing the door with the car ought to be ½ regardless of the door chosen.

The Monty Hall Paradox is in fact not a paradox at all, but rather just some clever sleight of hand. The trick is that people are drawn to the fact that there are only two doors rather than three doors remaining, and assume that the host’s having opened a door is favorable to the player. People tend not to realize that the game has imperfect information – the player does not know where on the game tree he is, whereas the host does. Additionally, people assume that the host has no stake in the game (and this is not unreasonable, because the problem does not explicitly describe a parsimonious host! On the other hand, intuitively, we know that the host isn’t going to ruin the game by opening the door with the car.) So, if we assume that the host is profit maximizing and we model the problem as an extensive form game with imperfect information, then the conditional probabilities would be easy to see.

Now, just for fun, we’ll assign some utilities to the outcomes. What is a goat worth? According to a popular Passover song in Aramaic, a (small) goat is worth about 2 Zuz, and according to the traditional Jewish prenuptial document, a wife is worth about 200 Zuz. So, a goat is worth about 1/100th of a wife. I asked my roommate, Anna, how many cars she thought a wife was worth, and she determined that a wife was worth three cars. By transitivity, then, a car is worth about 33 goats. (I think goats have become quite a bit less valuable since that song was written, or maybe goats back then were a lot better than our goats.) So, if the player wins the game, he will walk away with a utility of 33, and the host will walk away with the 2 goats.

monty hall

In this game, the light gray branches are dominated because the host has no incentive to open the door that the player has already chosen, and the dark gray branches are dominated because, of the remaining two doors, the host would not open the door that has the car. We can tell that in the top branch, the host as 2 possible choices for doors to open, whereas in the lower two branches, the host is constrained to only one door (since, if the player has chosen a goat door, there is only one goat door left to open.)

So, since the player has no idea after the first stage what door has the car, we assume he picks door No.1 (as in the game). If he observes that the host opens door 3, he would know that there are two cases where the host opens door 2: in the world where the car is behind door 2, the host chooses door 3 100% of the time, and in the world where the car is behind door 1, the host chooses door 3 50% of the time. It’s actually twice as likely that we are on the 100% branch as that we are on the 50% branch – and that’s the branch where the car is hidden behind the other door.

What if we know that the host has opened a door, but we don’t know which one? Then, we can’t condition on a prior, because we don’t know what the prior is – we don’t get any new information by observing which door was opened, and switching doors would not help.


Of skirts, judgement, and changing conventions

Jeff’s post last Sunday on Jewesses in Skirts got me thinking, how is it that we have Orthodox Jewish communities that are tolerant of pants-wearing by women, and communities that are not tolerant of pants-wearing, but rarely a community with large factions of each type? The question, of course, applies to more than just skirts worn by Jewish women – we can talk about many aspects of our culture, such as our changing views on LGBT people, or miscegeny, or a gold standard vs. silver standard using the same language.

We’ve established that, assuming that the types of women in the previous post are static (that is, Nature or Circumstance assigned you to one group and you can’t change allegiances), that it is optimal for the more conservative group to adhere to skirt-wearing, and the other group not to bother. Those static proportions of types in the population affect how the “others” in Jeff’s game form their prior beliefs about which type a woman is based on her choice of clothing. But, what if the women could choose or change their ideology, and what if we consider the effects of judgement and peer-pressure?

In the following model, we can look at a scenario where the proportion of each type of player in the population is endogeneous. Suppose that a new community forms, consisting of some random number of “conservative,” skirts-only types (Jeff puts them in class 1) and some number of “progressive” types who sometimes wear pants (Jeff puts them in class 2). This represents what we would expect to happen if everyone formed their own opinions and ideologies totally independently of everyone else. Each person will randomly encounter other members of the community on a one-on-one basis, and receive social payoffs from the encounter. If she encounters a likeminded person, they both feel validated in their choices, and if not, they feel judged. As before, we assume some disutility for a restriction on wardrobe.

 
Conservative Progressive
Conservative 1,1 0,0
Progressive 0,0 2,2

Now, say that the population starts out with a percentage p of progressive types and 1-p of conservative types. Then, assuming that the population is large, if you are one of the members, then of the people you meet, p will be progressive and 1-p will be conservative. Therefore, if you are a woman who chooses to be conservative and wear only skirts, your expected utility is 1(1-p) + 0(p) = 1-p and if you choose to wear pants, then your expected utility in any one encounter is (0)(1-p) + (2)(p) = 2p. You would be indifferent if 1-p = 2p — that is, if p (the fraction of progressive types) is 1/3.

Maintaining p = 1/3 is incredibly difficult, because it is so sensitive to shocks. If for any reason p becomes a little more or a little less — say, a contingent of pants-wearers suddenly move in — the balance would tilt and one of the ideologies would start providing the better payoff, the whole population would start snowballing in that direction, and it would become the predominant convention (or evolutionarily stable state) . That may be why these larger faction groups tend not to exist in real life: they are a lot like a flipped coin that lands on its edge.

So what kinds of things affect what p is? The payoff matrix, obviously, and how I’ve assigned the payoffs. No one said that I must assign those particular numbers (and indeed, I don’t. The numbers that go into those matrices don’t really matter. What matters is their order of size and relative distance to each other. Try it: multiply all of the numbers by a constant, or add a constant to all of them. The solution should be the same.) What if the inconvenience of wearing only skirts is very large? (Imagine replacing the 2s with, say, 5s.) Then, p (the tipping point) could be much smaller, and it would take a much smaller group of rebels to send the equilibrium going the other way. Issues like women’s suffrage are like this — they are so significant that a grassroots movement picks up momentum very quickly. If the inconvenience is less, then p would be greater, and if the existing equilibrium is skirts-only/conservative, it would be harder to change. Equivalently, we can think about the effects of mutually judgemental behavior (making the 0s in the matrix more negative). If people are less tolerant when they meet the other type, conventions are harder to change. If they are more tolerant, change is easier.

 

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If you enjoyed the ideas in this post, you may also enjoy When in New York, do as the New Yorkers do, which describes a special, symmetrical case of the kind of game we’ve discussed here.


Touch the Truck

There was a British game show in the early 2000s called “Touch the Truck,” where, according to the rules, the truck would go to whomever could keep his hands on it the longest. Similar games have been featured in a number of other shows, including Survivor, where the last person to keep his hand on the totem would win immunity and That 70’s Show, in the following episode (start at 3:37 — you don’t need to watch the whole episode, unless you want to):

The three games that I’ve just mentioned differ in a couple of ways. In the British show, the contestants are able to take breaks, and are disqualified if they fall asleep, so it is primarily a contest of sleep-deprivation. In Survivor and That 70’s Show, bathroom breaks are not allowed, so the contest is much shorter and contestants might gain an advantage by wearing (ew) diapers or catheters. Obviously, pulling a stunt like that incurs costs — does the contestant really want to wear a diaper?

We can also think of Touch the Truck as a war of attrition or all-pay auction — any amount of time that the contestants spend holding their hands to the truck is a sunk cost. They’re not getting that time back. In most all-pay auctions or wars of attrition, the outcome can someone be that the players bid more than the object is worth — suppose you auction off a $20 bill, and everyone pays their bid regardless of whether they win. Then, you can imagine a scenario in which someone has bid up to $20, and another player who already has $18 in the game might bid $21 so as to only lose $1 rather than $18. If he wins by bidding more than the value of the object, we call it a Pyrrhic victory.

However, there is a slight (and largely inconsequential) difference between Touch the Truck and most wars of attrition or all-pay auctions. In all-pay auctions, players can bid as high as they like, whereas in Touch the Truck, the human body can endure only so much. The world record for staying awake is something like 11 days, and in most of the latter 8 days or so the person is basically not really functioning. (The longest I’ve ever continuously stayed up is probably around 50 hours, and by then I pretty much can’t do anything.) The truck, however, we can say is worth at least $15,000 — a conservative estimate. So, even if someone were able to stay up for 11 days, and took another 4 days off to recuperate, he has still earned $15K in 15 days [1]. Not too shabby. And, since none of the contestants can actually stay up long enough for the time lost to be equal to $15K, there will always be additional rent that each contestant can earn (the winner will never have a Pyrrhic victory).

So what’s the equilibrium outcome? Well, the winner’s outcome is similar to the winner’s outcome in second-price auctions or English auctions. Since the contestants have differentiated levels of ability to stay awake, if you’re not the person who can stay awake the longest, then you might as well lose immediately. If you are the contestant who can endure sleep-deprivation the longest, then you should stay up as long (or \epsilon longer) than the second most fortitudinous contestant. You should choose to do it even if no one else is choosing to enter the contest, because if there’s a chance you’ll quit sooner than that, then someone else might be able to win and might want to enter the contest.

Of course, in this example, there is perfect information — everyone knows everyone else’s level of fortitude, etc. The entertainment value comes from everyone’s thinking that they might actually be able to endure longer than anyone else, and of course, this comes at the expense of the contestants. Everyone would be better off if he could reliably and accurately communicate how long he could stay touching the truck, but most efforts at communicating this information are cheap talk (like Daniel does in the show.)

[1] Generally we would have to account for different people’s different valuations of the truck, but, since this is a game show and the organization running the game wants to pick the most enthusiastic bunch of people it can find, we will safely assume that every contestant values the truck at market price.


How did the chicken cross the road?

While reading an open newspaper, according to my host for lunch a couple of weeks ago. Apparently this is the most efficient way for a pedestrian to cross a street in Boston, where signs are few and confusing and drivers aren’t fond of traffic laws. (Sound like any other city we know?) An open paper is a big gray signal to the driver that you have no idea she’s there, and if she doesn’t want to run you over, she has to stop. It’s not unlike ripping out your steering wheel and waving it out the window in a game of Chicken.

The city of Philadelphia is aware of this, of course, and the Philly police are cracking down on pedestrians who text and walk. Clearly a cell phone is just not a visible enough signal.

Disclaimer: Play at your own risk, and look both ways. We do not endorse breaking traffic laws.


The veto donation paradox

We think of the veto as a very powerful (perhaps even unfairly powerful) bargaining chip, but this is not always the case. Sometimes having a veto is not as good as giving it away.

In this example, you want to select a juror. Candidates arrive randomly — most are acceptable but mediocre for both sides, a few are great for one side and terrible for the other, and a few are pretty good for both.

This is a variation of the Secretary Game. The central question for secretary games is, “Since, once a candidate is rejected, he does not apply again, when should we stop interviewing?”

For simplicity, we assume that there are only three types:

Type by utility to (x,y) Probability of arrival
b, b where {1/2 < b < 1} 1 - 2\epsilon
1, 0 \epsilon
1- \epsilon, 1 - \epsilon \epsilon

[1]

1. If both players reject the candidate, he is rejected.
2. If both players accept the candidate, he is accepted.
3. If one player accepts the candidate and one player rejects, then the candidate is accepted unless someone uses a veto.

This is a sequential game, so it is a game of perfect information. It is also Markovian, which means that if the candidate is rejected, we return to the same state in which we started.

Suppose neither side has any vetoes. Then, X always accepts (1,0) and Y always rejects this, so it is accepted (since there are no vetoes). Y always accepts (1- \epsilon, 1 - \epsilon) since it is Y’s best outcome, so it is accepted. X rejects (b,b), but Y accepts (b,b) because b > 1/2 and so it improves Y’s average. (If Y also rejected it, then Y’s payoff would be the average of 0 and 1- \epsilon.)

Therefore, the expected utility is:

U(X) = (1- \epsilon)(\epsilon) + (b)(1- 2\epsilon) + (1)(\epsilon)
U(Y) = (1- \epsilon)(\epsilon) + (b)(1- 2\epsilon)

If X has exactly one veto and Y doesn’t, then X would use up the veto when Y accepts b in one round, and we return to the starting point. This makes X slightly better off and Y slightly worse off.

U^*(X) = (1- \epsilon)(\epsilon) + (U(X))(1- 2\epsilon) + (1)(\epsilon)
U^*(Y) = (1- \epsilon)(\epsilon) + (U(Y))(1- 2\epsilon)

However, if Y has one veto and X doesn’t, then X would need to reject (1,0) (since Y would veto this and X would end up with U close to b). If (b,b) arrives, X and Y both reject, and if (1- \epsilon) arrives, both accept. So, Y’s having the veto is actually better for both candidates than X’s having the veto (and, if X could, he should give the veto to Y). Additionally, since this arrangement guarantees a higher expected payoff for both sides, giving one side a veto can even be Pareto-improving.

Extension: What happens if both sides have a positive, finite number of vetoes?

It’s easy for X to guarantee himself an outcome of (1- \epsilon). He can simply accept (1,0) and (1- \epsilon,1- \epsilon) every time they appear and reject (b,b) up until Y only has one veto left. Then, play as in the previous case. He can’t do any better, since there are no cases where Y has vetoes and (1,0) is accepted, and there are no cases where Y has no vetoes and X’s expectation is greater than (1- \epsilon).

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[1] Mathematicians and other quantitative people talk about \epsilon (epsilon, pronounced either “EP-si-lon” in the US or “ep-SIGH-len” in the UK) a lot. We’ve certainly used it often. You can think of it as an arbitrarily small positive number, or “a number as small as you need it to be, but not 0.”

Example is based on Shmuel Gal, Steve Alpern, and Eilon Solan’s A Sequential Selection Game with Vetoes (2008)