Debating Superfreakonomics

Last month marked the release of Superfreakonomics, a sequel by economist Steven Levitt and journalist Stephen Dubner to the 2005 bestseller Freakonomics. The fanfare surrounding this prefix-enhanced release has been marred, however, by controversy surrounding a chapter on global warming. Starting with this entry on ClimateProgress.org, the debate has drawn a few responses on the Freakonomics blog, but nothing has seemed to blunt the allegations that Dubner and Levitt wrote the chapter from a contrarian perspective without understanding even the fundamental principles of climate science, and as a result, what they've written is garbage.

Much of the writing back and forth has been quite heated, and being a student of mathematics I am averse to conflict. However, one response resonated with me a great deal, and as a case study of the arguments that can be made using only simple calculations, it's quite effective. The response in question comes from RealClimate.org, and is titled "An Open Letter to Steve Levitt."

Written by fellow University of Chicago Professor Raymond T. Pierrehumbert, the letter takes Steve Levitt to task by harnessing the power of mathematics. After some opening remarks, Pierrehumbert sets the stage in the following way:

By now there have been many detailed dissections of everything that is wrong with the treatment of climate in Superfreakonomics , but what has been lost amidst all that extensive discussion is how really simple it would have been to get this stuff right. The problem wasn’t necessarily that you talked to the wrong experts or talked to too few of them. The problem was that you failed to do the most elementary thinking needed to see if what they were saying (or what you thought they were saying) in fact made any sense. If you were stupid, it wouldn’t be so bad to have messed up such elementary reasoning, but I don’t by any means think you are stupid. That makes the failure to do the thinking all the more disappointing. I will take Nathan Myhrvold’s claim about solar cells, which you quoted prominently in your book, as an example.

Myhrvold's claim in this context is essentially that using solar cells to fight global warming is not a good idea, because solar cells must be dark in order to absorb solar energy. However, only a fraction of that energy is converted into electricity, while the rest simply becomes waste heat that in turn will heat up the atmosphere.

Using nothing more than simple arithmetic, Pierrehumbert then tries to reason his way through such an argument to see if it makes any sense. As expected, it does not. My favorite part of the argument is the graphic that shows how many solar panels would be required to supply the world's electricity:


That black square in Saudi Arabia certainly doesn't look like it should make any significant contribution to the planet's heating, and indeed, Pierrehumbert uses mathematics to argue quite effectively that it wouldn't.

The letter is worth a read, not just for the strength of Pierrehumber's argument, but for the simple mathematics that gives his argument such strong support. Levitt offered a meek response in the comments (#47, I believe), which was then quickly rebutted. Since then, all's been quiet.

Of course, Levitt and Dubner may think that the math is on their side - since I haven't read the chapter in question, I can't comment. But the arguments put forth by Pierrehumbert are quite compelling, due in no small part to the simple calculations he performs. There's no doubt that mathematics can be used both for good and for evil, but Pierrehumbert, like Spider Man, seems to understand that with great power comes great responsibility.

Martin Gardner and the Three Way Duel

As you may have heard, last week Martin Gardner celebrated his 95th birthday. Gardner, who authored the "Mathematical Games" column in Scientific American for a quarter of a century, is often credited for introducing generations of young students to the beauty and charm inherent in mathematics. My favorite quote in this vein comes from professor Ron Graham, who is quoted in a recent New York Times article on Gardner as saying that "Martin has turned thousands of children into mathematicians, and thousands of mathematicians into children."

A warm brain is the key to mathematical dexterity.

Both Scientific American and Wired ran articles on Gardner last week, and each one used a different expression to represent his age. Scientific American congratulated him on reaching an age of 2⁵ x 3 - 1, while Wired proclaimed that Gardner had turned 5! - 2⁵. Upon reflection I think I prefer the latter expression over the former, since the exclamation point in the factorial makes his birthday feel just a little more exciting. No matter your preference, however, I hope this encourages others to write their ages using mathematical expressions - more complicated expressions would be especially useful for ladies who don't wish to reveal their age.

In any event, I'd like to take the opportunity to celebrate Martin Gardner's birthday in my own way, by discussing a problem credited to him. There are a few variations on this problem, but all fall under the general name of the Truel. I first encountered this problem during a job interview, and the variant I heard then is the variant I'll discuss now.

Suppose there are three men: Mr. White, Mr. Gray, and Mr. Black. These three men have a score to settle, and so they decide to engage in a three way duel (or a "Truel," if you want to be cute). Mr. White is a rather poor marksman who hits his target only 1/3 of the time, In contrast, Mr. Gray is successful 2/3 of the time, and Mr. Black is an expert marksman who never misses. Because of this disparity, they agree that Mr. White shall shoot first, followed by Mr. Gray, followed by Mr. Black, and this sequential rotation shall continue until only one man remains standing.

Setup of the truel.

Although Mr. White isn't so good with a gun, he is an expert strategist, and surmises that Mr. Black will dispose of Mr. Gray first, since Mr. Gray poses more of a threat than Mr. White. By the same reasoning, Mr. White assumes that Mr. Gray will shoot at Mr. Black before shooting at Mr. White.

Assuming that this analysis holds true, what should Mr. White do on his first turn to maximize the chance that he'll win?

I'll discuss the answer below, but you should try this out for yourself if you've never seen this problem before. So that you don't peek at the answer accidentally, let's pause for a moment and look at some drawings courtesy of M.C. Escher.

Relativity

Drawing Hands

Circle Limit III

So, what should Mr. White do? Should he shoot first at Mr. Gray or at Mr. Black?

Phrased this way, it's a bit of a trick question, since the answer is that Mr. White shouldn't shoot at either one. Instead, he should intentionally miss!

Intuitively, this seems reasonable if you reflect on it. The reason is that it's too risky for Mr. White to shoot at either of his opponents in the first round, because if he actually hits one of them, the remaining opponent will immediately turn and shoot at Mr. White. For example, if Mr. White shoots and hits Mr. Grey, he's done for, since Mr. Black will shoot at Mr. White and is guaranteed to hit. Similarly, if Mr. White hits Mr. Black, Mr. Grey will turn to shoot Mr. White. If, however, Mr. White shoots neither one, then either Mr. Black or Mr. Grey will be eliminated by the other one, leaving Mr. White with only one opponent to worry about.

For those looking for a more quantitative argument, we can use the odds given in the problem to calculate the probability that Mr. White will win if he a) shoots first at Mr. Black, b) shoots first at Mr. Gray, or c) shoots first at neither one.

As part of the analysis, we'll need to determine the odds that Mr. White will win when he's paired one-on-one against either one of his adversaries. For Mr. Black this is easy: if Mr. White faces off against Mr. Black, he has a 1/3 chance of winning, since if he misses Mr. Black will win with certainty.

In a face-off between Mr. White and Mr. Gray, however, things are more complicated. This is because there are many ways for Mr. White to win. He could win by hitting Mr. Gray on his first shot, he could win if both men miss and he hits Mr. Gray on his second shot, he could win if both men miss twice and he hits Mr. Gray on his third shot, and so on. In general, since the probability that both men will miss in a given round is 2/3 x 1/3 = 2/9, the probability that Mr. White wins in the first round is 1/3, the probability that he wins in the second round is (2/9) x 1/3, the probability that he wins in the third round is (2/9)² x 1/3, and so on.

In general, we find that the probability Mr. White wins when facing Mr. Gray is 1/3 x (1 + 2/9 + (2/9)² + (2/9)³ + ... ). The value of the geometric series is 9/7, so the probability is just 1/3 x 9/7 = 3/7. That is, when facing off against Mr. Gray directly, Mr. White has a 3/7 chance of winning (assuming Mr. White shoots first).

How are these calculations relevant to the matter at hand? Well, we can analyze the outcome of the truel depending on Mr. White's initial action by using some tree diagrams. Let's first see what happens in the simplest case when Mr. White intentionally misses. In this case, Mr. Gray will shoot, and either he will hit Mr. Black or he will miss:

Here, GHB and GMB stand for "Gray hits Black" and "Gray Misses Black," respectively.

We know that if Gray hits Black, we will now be in a face-off between Gray and White. Similarly, if Gray misses Black, Black will in turn kill Gray, putting us in a face-off between Black and White. Since we calculated the probability of White winning in either one of these face-offs, and since Gray has a 2/3 chance of hitting Black, this diagram shows that the probability White will win must be 2/3 x 3/7 + 1/3 x 1/3 = 25/63, which is roughly 39.7%.

What if White shoots at Gray instead of shooting at Black? In that case, the picture gets a bit more complicated:
In this case, if White hits Gray the truel will end, since Black will hit White with certainty. Therefore, the only way White can win is if White misses Gray - but in this case, the situation is just as it was before, when white shot to miss. Since White has a 2/3 chance of missing, and from above we know that White has a 25/63 chance of winning if he misses, this shows that the probability of white winning is 2/3 x 25/63 = 50/189 in this case, or roughly 26.5%.

Finally, if White shoots at Black, we get an even more complicated tree diagram:

Notice that the probabilities in the bottom half of this diagram are the same as in the earlier diagram, and therefore it suffices to consider what happens in the top half. Here we see that White can only win if he hits Black, if Gray misses White, and then White wins the face-off against Gray. Following the probabilities on the tree, we see that the odds of this happening are 1/3 x 1/3 x 3/7 = 1/21. Therefore the total probability that White will win if he shoots at black is 50/189 + 1/21 = 59/189, which is roughly 31.2%.

From this analysis, we see that White's odds are significantly increased by shooting to miss: 39.7%, versus 31.2% if White shoots at Black and a paltry 26.5% if White shoots at Gray.

Another interesting feature of this problem is that by shooting to miss, not only does White maximize his chances of winning, but he also becomes the most likely person to win. Using similar arguments, one can show when White shoots to miss in the first round, the probability that Gray will win drops to 38.1%, and the probability that Black will win goes down to just 22.2%. In other words, using this strategy not only makes White the most likely to win, it makes Black the least likely to win, even though Black is the strongest marksman!

As discussed in the Wikipedia entry on truels, there are other variants one can consider as well. One could change the order of the shooters, for example, or toy with the probabilities involved. One could also consider more realistic situations in which the players have only a finite number of bullets, or think about what happens if the players fire simultaneously rather than sequentially. These variants will lead to generalizations of the phenomena observed here.

As with much of mathematics, this recreational problem shows that once you start digging, there's practically no limit to the questions you can ask and answer with a bit of mathematics. Here's to Martin Gardner, for providing us with such a delicious taste of the bounty mathematics has to offer.

Math Goes Trick Or Treating Again

Around this time last year, I wrote up some suggestions for math-themed Halloween costumes. Based on the traffic I received from that article, I can tell that many people are desperate to integrate their holiday festivities with mathematics. For this reason, and in the interest of not breaking tradition, I thought it would be fitting to suggest a few more ideas for this year.

1) Mathemagician.

In the strictest sense, a mathemagician is simply a mathematician who does magic. Or, perhaps it is a magician who does mathematics. You may (rightfully) be tempted to say that every mathematician does magic, but the tricks of the mathemagician are geared more towards a general audience, although they do often feature mathematics in a starring role. Sadly, the same cannot always be said for the typical magician.

There are examples of mathemagicians in real life, including Arthur Benjamin, who has been the subject of an earlier article (unfortunately not for his skills in mathemagic). However, the mathemagician I have in mind is closer to the kind portrayed in the episode "Grade School Confidential" from season 8 of The Simpsons. Sadly, a clip of the scene in question isn't available online - the best I can do is show this picture, which should be enough to jog the memory of any fan of the series.

Thanks for the screen cap, University of Connecticut Mathematics Department.

2) Mathematician Who Died in a Spectacular Fashion

Some people may think it sufficient to simply dress as their favorite mathematician for Halloween, and depending on the party, that may be true. But to give your costume that extra bit of Halloween pizazz, I recommend restricting yourself to deceased mathematicians. Not only are dead people big hits at Halloween parties, but in some cases you can incorporate the cause of death to create an even more compelling costume.

Regarding mathematicians who met an untimely demise, my top two choices would be Évariste Galois, who died in a duel under mysterious circumstances at the tender age of 20, and Jørgen Gram, who was killed by a bicycle.

Jaunty joy-rider, or cold-hearted killer?

3) Quant for Hire

This is a good costume if you are going to a party with friends who think themselves political. It's also good if you really are an unemployed quant. The key here is to wear an outfit that at one point could have been quite valuable, even if it now looks like something you pulled out of the garbage. Failing that, you could always see if they make the costume below in adult sizes.

That's right, Jimmy - never give up on your dreams.

4) The Monster Group

The monster group is a very large finite simple group. How large? As that Wikipedia link will tell you, it contains 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,
000,000,000 elements. In case you are curious, this reads as eight hundred eight sexdecillion, seventeen quindecillion, four hundred twenty-four quattuordecillion, seven hundred ninety-four tredecillion, five hundred twelve duodecillion, eight hundred seventy-five undecillion, eight hundred eighty-six decillion, four hundred fifty-nine nonillion, nine hundred four octillion, nine hundred sixty-one septillion, seven hundred ten sextillion, seven hundred fifty-seven quintillion, five quadrillion, seven hundred fifty-four trillion, three hundred sixty-eight billion.

If you wanted your costume to be completely impenetrable, you could just make a t-shirt with the two 196,882 x 196,882 matrices that generate this group. Alternatively, you could simply dress up like a monster. Although technically, I suppose you'd need to dress up as a group of monsters.

I think this would be suitable.

5) Your favorite knot.

While many people no doubt reserve a special place in their hearts for the trefoil knot, there is certainly room for creativity here. For example, check out this dude!


This is but a small sample of what you can do to combine Halloween with mathematics. I would encourage you to think of your own ideas as well. As I'm sure you know, every holiday can be made better with just a splash of math.

More on Football Pools

This post is a follow-up to an earlier post that looked at betting squares for football scores. In particular, we analyzed the distribution of the second digit of final football scores, and compared that to the digital root of final football scores (recall that the digital root of a number is found by iteratively calculating the sum of the digits in that number until you come up with a single digit number from 1 through 9). We found that on average, the final digits of football scores do not distribute themselves evenly - a score ending in 2 or 5 is much rarer than a score ending in 7 or 0, for example. However, the analysis of the digital root suggested that digital roots may become evenly distributed on average.

We now turn to a related question of independence, which was mentioned at the end of the previous post. More specifically, we address the following question: is the second digit of the home team's football score independent of the second digit of the away team's score? What if we replace "second digit" by "digital root" in the above question?

A more basic question that is related to both of these is whether or not the two final scores in a football game are independent. The answer to this question is not entirely obvious. On the one hand, knowing that one team scored 7 points doesn't tell you anything a priori about the second team's score. On the other hand, one team's score during a game will certainly influence the strategy of the opposing team: if one team is ahead by a wide margin, for example, the losing team will likely play riskier in the hopes of making some big plays. Conversely, if the difference in scores is very close, teams are less likely to make plays that are as risky unless it's late in the fourth quarter.

Alternatively, what if you are told that one team scored 7 points, but that the game went into overtime? Now you know a great deal about the other team's score: since overtime in the NFL is sudden death, and since the only way to score 7 points is by a touchdown or the (very unlikely) combination of a field goal and two safeties, you know that the other team must have scored 4, 5, 7, 9, 10, or 13 points.

From these examples, it seems that football scores aren't entirely independent - there is often some relationship between the two. However, it's natural to think that perhaps on average, we do have independence. Similarly, we may hope that by looking at several games, we have independence of the second digit or the digital root between home and away teams.

One reason to hope that we would have independence is that it allows us to calculate probabilities of winning squares with much less input. If the probability of the home team having a score that ends in a 7 is 17%, for example, and the probability of the away team having a score that ends in an 8 is 7%, we'd like to be able to say that the probability of the (7, 8) square winning in a football pool is .17 x .07. Otherwise, to estimate the probability of any particular square winning we'd have to rely on historical data, and create a running tally of how each square has performed. Needless to say, this is more work.

Unfortunately, it seems that this is necessary work. By looking at data from 4,642 preseason and regular season games from 1994-2008 (the 1994 season was the first in which the two-point conversion was instituted), I tallied the counts for each winning square in the case of both the traditional football pool and the digital root pool. Here is the data:



Recall that for the digital root pool, the digit 0 only occurs if one team doesn't score - to even things out, I have assigned a score of 0 with a digital root of 9 (I discussed reasons for this in my earlier post).

Using a standard statistical test for independence, we conclude that there is essentially no way that the digital root of one team is independent of another team, or that the last digit of one team's score is independent of another team's score.¹ To illustrate one example, 10.8% of the home team's scores had a digital root of 2, while 9.26% of the away team's scores had a digital root of 2. Therefore, assuming the digital roots of the away team and home team's scores were independent, the probability of winding up in the (2,2) square would be 0.108 x 0.0926, which is approximately 1%.

However, as the data shows, only 12 games ended in the (2,2) square this gives a much smaller probability of 12/4,642, which is approximately 0.26%. In other words, if you know that the score of the home team's digital root is 2, it is suddenly much less likely that the score of the away team's digital root is also 2.

Some other observations:

1. Using this larger data set, we also find that it's actually unlikely that the digital roots become equally distributed among the digits from 1 to 9. While the distribution is certainly more uniform than in the case of the second digit, we also see that certain digital roots occur significantly more frequently than others (5 occurs much less frequently than 1, for example).

2. Even though away team and home team 2nd digits and digital roots are not independent, one may expect that the probability of having a score with a given 2nd digit or a given digital root should be independent of whether the team is the home team or the away team. Interestingly, in the case of the 2nd digit, this is not true. In particular, it is more likely for an away team's score to end in a 3 than a home team's score, and it is more likely for a home team's score to end in an 8 than an away team's score.² In the case of the digital root, however, the data is insufficient for us to reject the hypothesis that the probability that the away team's score has a certain digital root should be different from the probability that the home team's score has a certain digital root.³

In considering the relative merits of the traditional football pool versus the digital root method, one question from my earlier post remains unanswered: which method will hit the most squares during a typical football game? It seems reasonable to expect that the digital root method will hit a larger number of squares than the traditional method. For example, if the score is 13-7, and the home team scores a field goal followed by a touchdown, two boxes will be hit by the traditional pool: (3,7), (3,0), followed again by (3,7). However, the digital root method will hit three boxes: (3,7), (3,1), and then (3,8).

This heuristic is certainly reasonable, but does it hold when looking over a large number of games? I'll return to this question in a later post.

1. In the case of the last digit, we obtain a χ² value of 425.50 with 81 degrees of freedom, which gives a probability that is essentially 0. Similarly, in the case of the digital root, we obtain a χ² value of 326.52 with 64 degrees of freedom, again yielding a probability that is essentially 0.

2. Testing the Null Hypothesis that the probability that a home team's score ends in a 3 is equal to the probability that the probability that an away team's score ends in a 3 yields a χ² value of 33.62 with 1 degree of freedom. Replacing 3 with 8 gives a χ² value of 11.86. At a 0.01 significance level and 1 degree of freedom, we reject the null hypothesis if χ² is larger than 6.6348.

3. In other words, the χ² values in the digital root case are never larger than 6.6348. The case of the digital root equal to 5, however, comes close - in this case, the χ² value is 6.6130.

Math on TV: Halls Refresh

If you've watched any television recently, you may have noticed the following ad for Halls Refresh. I strongly encourage you to watch it, even if you've seen it before, because it's basically fantastic.


A tremendous ad, to be sure. However, if you didn't watch closely, you may be wondering what such a sensual commercial has to do with mathematics. Watch again if you missed it - it may help to watch it full screen, although the quality gets muddy.

Did you catch it the second time? When the camera cuts to the Asian kid sitting at his desk, right before he starts to charm Mrs. Hunter, you'll notice that he has a poster on his wall filled with mathematics. There are 5 equations on the poster, but most are probably too difficult to make out from the Youtube copy. I was fortunate enough to see this ad on television, and after a few replays I made out 4 of the 5 equations. In order, they are as follows:

1. x/x = x^x.
2. (10+x+x^x)^1/x/x^x/41/x = x.
3. (x^x-1-1)^1/x + tan(π/(x+1)) = x.
4. This one I didn't get, but I can tell you it was long and involved a logarithm. Bonus points to anyone who can fill in the blank here.
   
5. x^x - x³ = 4(x^x-1 + x^x-2).

At first this poster made me a bit upset. Like other math jokes I've discussed before, throwing together mathematically complicated equations just for the sake of it seems lazy, when one could instead try to make some kind of joke. It's not as if the equations above are famous, so initially it may seem like there's nothing going on under the hood.

Upon further investigation, however, I discovered that somebody involved in the production had a sense of humor. Let's see what happens when we look for solutions to each of the equations.

The first one is easy: since x/x = 1, we're looking for a solution to x^x = 1. This happens, of course, when x = 1.

The equations that follow are too difficult to solve by hand, but this is where we let computers do the work for us. If you graph these equations, you'll see that the solution to the 2nd is x = 2, the solution to the 3rd is x = 3, and the solution to the 5th is x = 5. Even though I wasn't able to read that fourth equation, I'm fairly confident that the solution to it is x = 4.

It's not a great joke, I know, but I appreciate the fact that there is a payoff for those who are willing to dig a little bit. It's not perfect (for example, the third equation has a second solution near x = 0.33), but it's certainly better than many attempts. Kudos to you, Halls Refresh. This almost makes up for your use of the stereotypical nerdy Asian dude who is good at math and likes dragons.

For more surreal advertising, there's always this classic. No math involved, unfortunately, but I'm willing to let it slide.

Reforming Education through Geek Chic

Earlier this month, Wired published an article written by Daniel Roth, enticingly titled "Making Geeks Cool Could Reform Education." It serves as an interesting counterpoint to the commonly used argument that the best way to reform education is to better integrate it with the most current technology, so that going to school feels less like going to school and more like playing video games (family friendly ones, of course).

Sorry, Typing of the Dead, but you're a little too creepy.

The essay in Wired takes a slightly different approach - it profiles schools that have successfully channeled the inner geeks of their students, the argument being that the geek subculture rewards intelligence with popularity. To do this, schools must make learning seem cool. This is a feat which is easier said than done, because, as we all know, there's no better way to convince a teenager that something is uncool than to repeatedly say how cool it is.

One way in which the schools were able to motivate students to embrace their inner geek was to surround them with older people - teachers, parents, and working professionals. One school in particular forces students to present their work to groups of outsiders. The effect here is to downplay the importance of youth culture: if students can see what their education can do for them down the road, they're more willing to value it in the present.

Other schools have taken different measures, but the goal of curbing a focus on youth is the same. For example, at Roxbury Prep, Roth tells us that "Kids eat lunch in the classroom, they're not allowed to talk in the halls, and they're disciplined for using the word nerd." Certainly social time with peers is important, but this added emphasis on academic performance appears to be paying off, because students in these schools value learning for its own sake, and are rewarded for their efforts not just by their teachers, but by their peer group as well.

Applied to mathematics, this philosophy could have a significant impact. After all, many students will tell you they hate math because they don't see the value in it. But if students were able to interact with people who used mathematics in their everyday lives (aside from their classmates and their math teacher), one hopes they would be motivated to learn the material. Or, even better, even for students who don't plan to make a career out of mathematics, in a culture where learning is perceived as cool, one would hope that students would take advanced mathematics just to get a taste for what it's like.

If only...at least a man can dream. Perhaps one day we really will see the triumph of Geek Chic at all levels of education. Certainly, this is a good sign (thanks Michelle). Once we see some modern pocket protectors, I think we'll have reached the tipping point.

Math in the Movies: Hodgepodge Edition

Most of the time I write about films where math takes a central role, but it is just as often the case that mathematics is at work in more of a supporting capacity. There are many examples of this phenomenon, even if we restrict our attention to movies that are fairly recent. To catalog each such instance would no doubt be fairly time consuming, but thankfully someone has already begun the task. It comforts me to know that I am not the only one who takes pleasure in seeing mathematics on the big screen.

Last week the Boston Globe ran an article that discusses the appearance of mathematics in a variety of recent films. In addition to mentioning the recent work on zombie dynamics, the article also discusses the link to mathematics as found in films like Casino Royale, Harry Potter and the Half Blood Prince, and The Dark Knight.

It's a short article and worth a read, but here are some highlights:

1. The ferry scene in The Dark Knight gives a modern twist on the classic Prisoner's Dilemma. The Dilemma is best understood by means of the following table (courtesy of Wikipedia):


Suppose two men have been arrested for a crime. Each is separated, and is asked by the authorities what happened. If one man betrays the other, the betrayer will go free while the betrayed will serve a long sentence. If both men betray, however, both will serve a moderate sentence, while if both men stay silent each will serve a short sentence.

One easily sees from the table that independent of Prisoner B's actions, Prisoner A will receive a better payoff by betraying (it's the difference between going free versus serving six months if Prisoner B stays silent, and serving 5 years versus serving 10 years if Prisoner B betrays). However, if both prisoners act in this way, they will both betray and end up serving 5 years, far longer than they would serve if both of them had remained silent in the first place. Therein lies the dilemma.

Of course, what happens in The Dark Knight isn't quite so interesting. In this case, the Joker hijacks a pair of ferries and tells the passengers on each that they have the controls to blow up the other boat. If neither boat is destroyed by midnight, the Joker will destroy both.

In other words, we have the following diagram:

In this case, the strategy is clear: you want to blow up the other boat before your own boat is blown up. This strategy, of course, ignores the difficulty inherent in deciding to blow up a boat full of people - the situation is made more interesting by the fact that one of the boats is filled with convicts.

No doubt it would've made for a better social experiment had the Joker gone with a more traditional Prisoner's dilemma, since here there is no advantage if both parties remain silent. One could argue that this lack of appreciation for the underlying mathematics was an early indication that the Joker's plans would ultimately be foiled.

Criminal mastermind, or mediocre math student?

2. Even spies enjoy a bit of math here and there. The introduction to 2006's Casino Royale uses a bit of fractal geometry, in the form of self-replicating spades:



3. 6 Degrees of Kevin Bacon is, in fact, an easier game with a celebrity other than Kevin Bacon. Surprisingly, the game is easiest when played with Dennis Hopper - in a ranking of most connected actors, Kevin Bacon came in 507th.

Sorry Kev, but even a fancy tie and a delicious last name can't change the fact that there are actors with more connections than you.

For more links between math and the movies, I recommend taking a look at the full article. The moral here is that you can run, but you cannot hide from mathematics.

(Hat tip to Caroline for the article link.)