Now, a show such as this might seem to have very little to do with mathematics.  But look, and ye shall find.  In the second episode of this past season, the chefs were paired up for one of the challenges.  There were 16 chefs at the time, combining to make 8 pairs.  The pairing was determined by drawing knives: 16 knives were presented in a knife block, and each had a number on it from 1 to 8.  The number was printed on the blade, so each chef would walk to the block, draw a knife, and read the number.  The knives were not replaced afterwards.  Pairs were formed by people who drew the same number.

This dude loves the number 3.

In this particular episode, the first six numbers drawn were 2, 1, 3, 6, 7, and 7.  In particular, the first pair was formed on the 6th draw.  This leads to a natural question: how long would you expect it to take before the first pair is formed?  Six draws seemed a bit long to me (I would have expected the first pair to have been formed sooner), so I immediately set about trying to understand the answer to this question.

To ease ourselves into it, let’s simplify things.  Instead of 8 pairs, suppose there were only 3.  And instead of knives, which are dangerous and pointy, let’s suppose people were choosing balls from a bag.  Rather than differentiating the balls by writing numbers on them, let’s differentiate them by color.  So suppose you have a bag with 3 pairs of balls: one pair red, one pair green, and one pair blue.

The game is this: you draw a ball from a bag and put it aside.  You keep doing this until you have drawn a pair.  The question is how long it will take before the first pair is drawn.

Right away we see that you will get a pair some time between your second and your fourth draw.  Obviously you can’t get a pair after only drawing one ball, so you need a minimum of two draws.  On the other hand, if you don’t have a pair after drawing three, you must have one ball of each color, which means your fourth draw MUST give you a pair of some color.

Given this observation, we can now start to calculate probabilities.  What is the probability that you will have a pair after two draws?  Well, this happens precisely when your first and second draw are the same color.  The probability of this happening is equal to 1/5, since there is no restriction on the first ball you draw, but then there is a 1 in 5 chance that the second one you draw will be the other ball with the same color.

You have a 1 in 5 chance of picking the second green ball after picking the first one, for example.

What about the probability that you’ll have a pair after exactly three draws?  In order for this to happen, your second draw must be a different color than the first, and your third draw must be the same color as either your first or second draw.  Of the 5 balls remaining after your first draw, 4 will have a different color from the first, meaning that the probability of drawing a second ball which is a different color than the first is 4/5.  Similarly, the probability of drawing a third ball which is the same color as either the first or the second ball is 1/2 (see the picture below).  Thus, by the laws of conditional probability, the odds that you will have a pair after your third draw is 4/5 x 1/2 = 2/5.

You have a 2 in 4 (i.e. 1 in 2) chance of pulling a blue or green ball given that the results of your first two draws were blue and green, for example.

The same argument works when calculating the odds that the pair will come on the fourth draw.  There is no restriction on the first draw, there is a 4/5 chance that your second draw will be a different color from the first, there is a 1/2 chance that the third draw will be a different color from the second, and there is then a 100% chance that your fourth draw will be the same color as one of your earlier draws.  This again gives a probability of 2/5.  We see that the probabilities add up to one, as they should.

Given these probabilities we can also calculate the expected value: on average, how many draws will you need before you get a pair?  Since the probability of two draws is 1/5, the probability of three draws is 2/5, and the probability of four draws is 4/5, we see that the average is

2 x 1/5 + 3 x 2/5 + 4 x 2/5 = 16/5 = 3.2.

In other words, on average you will need 3.2 draws before you come up with a pair.

It’s more interesting, of course, to deal with c different colors, rather than just 3.  We can still perform this analysis, and try to find probabilities and expectations. Suppose we have c pairs of balls, each pair of a different color. We draw the balls without replacement from a bag until we find a pair of the same color, then we stop. We can define a random variable to be the draw on which we complete our first pair. For example, in the case c = 3 above, we saw that with probability 1/5, with probability 2/5, and with probability 2/5.

As before, notice that must take a value between 2 and c + 1.  This is because we can’t draw a pair before our 2nd draw, and after c draws, the worst case scenario is for us to have each ball of a different color.  Since we have exhausted all color possibilities, the c + 1st draw must give us a pair (in essence we are applying the Pigeonhole Principle).  So, to describe the behavior of we need to calculate the probability that $Y_c = k$ for k between 2 and c + 1.

The same sort of argument as in the simple case c = 3 works here.  Suppose you want to calculate .  In order to find your first pair on the kth draw, you need to NOT draw a pair on your 2nd, 3rd, 4th, …, or k – 1st draw, and then have the color on the kth draw match one of the colors you have already drawn.  Since there are a total of 2c balls in the bag to begin with, we see that the odds of not drawing a pair on the 2nd draw is , the odds of not drawing a pair on the 3rd draw is (since there are 2c – 2 balls remaining, and you want to avoid 2 that are colors you’ve already drawn, leaving you with 2c – 2 – 2 = 2c – 4 options), and so on, so that the odds of not getting a pair on the k – 1st draw is .  Meanwhile, in order to draw a pair on your kth draw, you must pull one of the k – 1 colors that have already been pulled.  Since there are 2c – k + 1 balls remaining, the probability that this happens is .

Combining these, we see that

(Recall that the binomial coefficient is given by , and as usual n! = n x (n-1) x … 3 x 2 x 1 is the product of all integers from 1 to n.) With this formula, we can now see how likely it was for the first pairing on Top Chef to have occurred on or after the 6th draw.  In this case there are 8 pairs, so c = 8, and we see that

which comes out to 16/39, or roughly 41.03%.

Of course, there’s still the question of expectation: approximately how large do we expect to be (remember we saw that )?  I’ll spare you the details, but one can show that for general c, the expected value is given by

In particular, in the case c = 8 from Top Chef, we find that , which is approximately 5.09.  So on average, for c = 8 we expect to find a pair after a little more than 5 draws.

One final question: what happens to the expected value as c grows large?  As it turns out, we can write the expected value in a very nice form in terms of the Gamma function (which one can think of as a generalization of the factorial to the entire real line).  Using the doubling formula , the interested reader can show that

If one then uses Stirling’s formula to approximate the Gamma function, it follows that

in other words as   What a wonderful asymptotic!  This tells us that the number of draws we will need from a bag of c pairs before obtaining our first pair grows like the square root of c times a factor of .  We can compare the estimate given here for with the exact value computed above – in doing so, we find that .  So indeed, the approximation is fairly close to the true value (and the approximation will only get better as c grows).

There are many related questions one could ask.  For example, what if instead of pairs, we look at collections of triplets, or quadruplets?  What if we consider formation of the 2nd pair or 3rd pair instead of only considering the 1st pair?  What if we allow for different numbers of balls of each color (e.g. 2 red balls and 3 green balls)?  But I’ve already gone on too long, so I will leave these questions for another time.  I don’t know if these questions go by a certain name or not – I couldn’t find this particular problem anywhere.  If anyone knows of a paper or book where these problems are discussed, I would be much obliged.

In the mean time, I will close with a picture of Tom Colicchio looking like a badass.  Clearly the worlds of chefs and rock stars have collided – will mathematicians be next?

(Kudos to Dr. Moore for some helpful commentary.)

]]> http://www.mathgoespop.com/2010/07/top-chef-mathematics.html/feed 0 http://www.mathgoespop.com/2010/06/the-twilight-saga-a-mathematical-perspective.html http://www.mathgoespop.com/2010/06/the-twilight-saga-a-mathematical-perspective.html#comments Wed, 30 Jun 2010 15:00:54 +0000 Matt http://www.mathgoespop.com/?p=71 Living in Los Angeles, it’s hard not to be aware of the fact that the new Twilight movie, Eclipse, arrives in theaters today.  The series has developed an insatiable fan base of people willing to spend thousands of dollars to fly here in the hopes of scoring tickets to the premiere, which certainly indicates . . . → Read More: The Twilight Saga: A Mathematical Perspective]]> Living in Los Angeles, it’s hard not to be aware of the fact that the new Twilight movie, Eclipse, arrives in theaters today.  The series has developed an insatiable fan base of people willing to spend thousands of dollars to fly here in the hopes of scoring tickets to the premiere, which certainly indicates the film will be a success.  But of course, the film’s success was never in question: with the first two movies having grossed over $1 billion worldwide, the success of this latest entry in the franchise is a foregone conclusion.

Of course, the success of this franchise should not be viewed in isolation, but as just a part of the larger vampire pop culture renaissance.  HBO’s True Blood, also based on a book series involving a girl who knocks boots with the undead, is going strong into its third season this summer, and the CW’s Vampire Diaries will return for a second season this fall.  And just when I thought the market for vampire-themed programming had become saturated, ABC premiered its own summer show featuring blood suckers called The Gates.  Clearly there is a trend here, with the ever-growing popularity of the vampire at its center.  No doubt Eddie Murphy is rolling in his undead grave for not releasing Vampire in Brooklyn 10 years later.

While there are many words that could be used to describe these shows and movies that place supernatural love triangles at their center, “realistic” is not one of them.  Nevertheless, there are a handful of people who have taken a critical eye to the vampire phenomenon and have used mathematical models to gain insight into how the populations of such creatures might behave in real life.  Just like the fights between Team Edward and Team Jacob, however, the debate over whether vampires could actually exist rages on.

Not long ago, an article went around the web purporting that a team of physicists had proven that vampires could not exist.  The physicists, Costas Efthimiou and Sohang Gandhi, posted a paper to the arXiv in which they purport to use physics to dispel pop culture portrayals of ghosts and zombies, in addition to vampires.  Their argument for debunking vampires rests on the following assumptions:

1. When a vampire bites a human, that human becomes a vampire (we will return to this assumption later).
2. Vampires need to feed on a human once every month (a conservative estimate when compared to what popular culture would have us believe).
3. Assume the first vampire came into existence in 1600, when the human population was roughly 500 million.
4. Ignore human mortality rates due to other factors, and ignore human birth rates as well.

With these hypotheses, they show that vampires would wipe out humanity in just 2 1/2 years.  In fact, no matter the size of the initial human population, their model will lead to humanity’s extinction in a short amount of time.  This is true even if we assume a more conservative estimate on the length of time vampires can go between feedings.

The reason is simple.  Using their model, the first vampire will feed after one month, creating a new vampire (by assumption 1).  After 2 months, those 2 vampires will each feed, giving us a total of 4 vampires.  After 3 months, those 4 vampires will feed, giving us a total of 8 vampires.  The pattern continues – after n months, the vampire population will be 2ⁿ. In other words, the population of vampires will grow exponentially.  Moreover, because of the assumption on the birth and mortality rate of mankind, we see that as the population of vampires grows exponentially, so too must the population of humans shrink exponentially.  This means that at some point (sooner than you might think), humans would be wiped out.

The careful reader, however, will note a number of problems with this analysis.  For one, ignoring the birth rate of humans means that the model’s date of extinction is premature.  However, Efthimiou and Gandhi point out that even if we include the birth rate, that rate would not be high enough to counteract the explosion in the vampire population.  A more serious flaw, however, is in not considering the mortality rate of the vampires themselves.  After all, once people realize there are vampires in their midst, wouldn’t they fight back, or at least defend themselves so that not all of the vampires could feed?  Assuming that every vampire would be able to feed whenever necessary seems unrealistic.

What’s more, assuming that vampires can only satisfy themselves with human blood, it seems unreasonable to assume that vampires would feast so carelessly, without regard to the diminishing supply of their food.  If vampires killed all humans, they in turn would die (again), and so it seems reasonable to expect that vampires would apply a better strategy, one in which they kept the human species afloat so that they could themselves continue to exist.  Just ask Ethan Hawke.

In a brief article for Math Horizons, mathematician Dino Sejdinovic addresses these issues and highlights an article from 1982 that modeled the vampire outbreak more realistically, by including the human birth rate and vampire mortality rate.  In doing so, the mathematics becomes less fit for a general audience, but it also gives us a more interesting picture – regardless of the collective desire for human blood, vampires can act in a way that the ratio of vampires to humans reaches an eventual equilibrium.  In other words, it doesn’t seem right to throw out the idea of vampires based on purely mathematical arguments.

Interestingly, though, all of these analyses rest upon assumption (1), which states that humans always become vampires once bitten.  In the modern incarnation of these creatures, however, this assumption no longer appears to be valid.  For example, in both True Blood and The Vampire Diaries, the process of turning into a vampire requires consent (I guess it’s more romantic that way); not only must the vampire drink the human’s blood, but the human must also drink the vampire’s blood.  In this case, it is possible for vampires to satiate themselves without killing humans (provided the vampires can show enough restraint) or increasing their own population.

There also appear to be rules governing population control in vampire communities.  For example, in an episode of True Blood, one vampire is tasked with creating a new vampire as penance for murdering one of his own kind.  Are such rules keeping the population stable widespread?  How might such rules, in conjunction with a weakening of assumption (1), alter the vampires’ optimal strategy?  I will leave it to the curious reader to discover the answer.

Suppose it’s late and you’re looking for a good time. Temptation runs rampant in the midnight hour of a Saturday night, especially for those of us in the fast-paced world of graduate school. But before you open that bottle, or pick up the phone to talk to live singles in your area, let me take the opportunity to inform you of a new way to spend your time during the late night wind-down: starting this weekend, you will be able to relive your childhood through the nostalgia-inducing satire that is Look Around You.

That’s right: starting this weekend, Adult Swim will begin airing the former BBC series on Sunday mornings (or late Saturday nights, depending on who you ask) at 1 a.m. The show lasted only two seasons, although I am of the firm belief that the first season was the more inspired of the two – it took it’s motivation from science videos shown to children during school on those rare days when teachers “rewarded” us by letting us watch an educational video during class time. The show’s parody of these old videos is nearly dead on, and the dry delivery with which the program makes its absurd statements only makes the show that much better. Plus, those of you with a keen eye may notice some extras who are now clearly much more famous than they were when these programs were produced.

You may be wondering what this all has to do with mathematics. The answer is that one of the educational videos is all about maths. The video is pretty fantastic, and shows that it is possible to make math funny to a general audience and a mathematical audience at the same time – something that most t-shirt designs fail to do.

The clip itself is on YouTube, but because the show is coming to Adult Swim, I’m not sure how long the video will stay up. If you haven’t seen it, take 10 minutes and watch – I promise it will be worth your time. Just make sure you have your copy book nearby.

For those of you who may not know, the titular character is the daughter of Santa with Muscles star and All-American hero, Hulk Hogan. In the show, Brooke lives in an expensive looking condo in Miami, goes to the beach, and sings her own theme song. This is about as much as I know. I swear. For those of you who are curious, the following video gives a good sense of what this show is all about.

The particular episode to which I would like to draw your attention (Episode 4) involved Brooke traveling to Panama City, Florida, to host some Spring Break parties. And I bet you thought math and spring break were incompatible.

Because of his overprotective nature (or because the producers thought it would make better television), Hulk Hogan decides that it’s a good idea to tag along on these spring break adventures. He even brings his chubby, mullet-sporting friend in tow, and the two of them cause all kinds of PG-13 hilarity. It is as if they are the love children of the love children of Abbott & Costello and Schwarzenegger & Stallone.

Hulk Hogan sees dead people.

One day, Hulk suggests that the group go visit The World’s Largest Human Maze (or, as he calls it, “The World’s Largest Human Maze in the world”). The maze in question is the Gran Maze of Panama City. Calling it the World’s Largest Human Maze is a bit deceptive, as there are hedge mazes which are larger – however, to the maze’s credit, it does not seem to refer to itself as the World’s Largest Maze in any of its information.

In any event, they go to this maze: Hulk and his buddy as one team, along with Brooke and her two roommates as another. At some point during their travels through the maze, they decide to split up and make a competition out of it: whichever team can make it through the maze first will get to plan the agenda for the rest of the day. Hulk and his friend make it through first, but only by squirming underneath the wall panels. The other team, once they learn of this deception, declares Hulk’s victory null and void.

Hulk Hogan: setting a bad example for maze solvers worldwide.

There is certainly a moral here: don’t cheat. But there is another important moral here, one I think may be lost on Hulk Hogan, even to this day. That moral is this: Hulk, you should have brushed up on your maze-solving algorithms!

Indeed, there are a number of procedures one can follow in order to try and solve a maze. Perhaps the most well known is the Wall Follower algorithm. In this procedure, you walk through the maze keeping either your left or right hand in contact with the wall of the maze at all times. If you do this, without removing your hand from the wall, you will eventually find the exit. Here is an example:

Solving a Maze: The Wall Follower algorithm.
This maze was constructed using the Maze Maker.

Now, the astute reader may realize that unfortunately, this method will not always lead you to the end of the maze. Even in a simple example, such as the one below, it is easy to see that regardless of whether you follow your left hand or your right hand, you will merely circle the entrance and never reach the end.

Failure of the Wall Follower algorithm

So, how do we know if the Wall Follower algorithm will work? This algorithm will always lead us to the exit (provided there is one) as long as the maze itself is simply connected. Less rigorously, when the maze cannot be solved using this algorithm, it is because the maze is in separate pieces. In the example above, we see that the piece of the maze that surrounds the start is disjoint from the rest of the maze.

Why, then, does this algorithm work if the maze is simply connected? Because every simply connected maze can be continuously deformed to a circle (i.e. such mazes are homeomorphic to the circle). Once we deform the maze to a circle, it is obvious how the wall follower algorithm works: it is equivalent to simply tracing your way along a circle between two points. This video does a better job explaining what goes on:

So, had Hulk confirmed that this human maze was, in fact, simply connected, he could have used this algorithm to try and beat his daughter fair and square. However, what if the maze is not simply connected? Indeed, a zoomed in view of the maze, courtesy of Google Maps, is somewhat inconclusive. How then, could Hulk guarantee that he would eventually make his way out?

The algorithm can be described according to the following rules:

1. Walk down the maze, drawing a line behind you.
2. When you come to the first intersection, choose a path at random and follow it.
3. When you come to a dead end, turn around and return to the last intersection.
4. If you are walking down a corridor that you have not been down before, and you come to an intersection you have already visited, treat the intersection as a dead end and turn around.
5. If you are walking down a corridor that you have been down before, and you come to an intersection (necessarily one you have already visited), then go down a path which you have not visited yet, if possible. If this is not possible, go down a path you have only been down once.

Rule 5 may look restrictive, but in fact, with this algorithm you will never need to walk down the same corridor more than twice. If the maze has a solution, this algorithm will find it. Moreover, once you have used this algorithm to find a solution, the corridors marked only once will trace out a direct path to the finish.

Solving a Maze: Tremaux’s algorithm. Notice that no
corridor is traveled more than twice, and the ones
traveled exactly once trace a direct path to the finish.

So there you go, Hulk. Two methods you could’ve used to try and defeat your daughter fair and square. Granted, there is no guarantee that following these procedures would have gotten you through any faster than your daughter, but they certainly wouldn’t have disqualified you from victory. Instead of being able to savor your win, you were called a cheater, and your prize was revoked.

Let this episode serve as a lesson to aspiring professional wrestlers everywhere: math can be of service, even to you.

For more on maze algorithms (both algorithms for creating mazes, or for solving them), this website offers a treasure trove of useful information. Hopefully Hulk Hogan will do his research the next time he makes such a mathematical challenge.

Hulk Hogan is ready to listen to what
mathematics has to say. Are you?

Since premiering in January of 2005, Numb3rs has been a consistent performer for CBS, in spite of (or because of, depending on your assumptions about the makeup of the show’s audience) its Friday night time slot. For those of you who may have never seen the show, the following synopsis should help give you some perspective:

Body counts, multiple criminal masterminds, and perpetrators who are likely to act again … this is the world of NUMB3RS. FBI agent Don Eppes (Rob Morrow) couldn’t be more different from his younger brother, Charlie (David Krumholtz), a brilliant math professor at a California university. Don deals in hard facts and evidence, whereas Charlie thrives in a world of mathematical probability and equations. But despite their disparate lives and career paths, Don and Charlie often combine their areas of expertise to solve a wide range of challenging crimes in Los Angeles. (Courtesy of the Numb3rs Season 2 DVD Box)

Indeed, mathematicians would be nowhere without their probabilities, or their equations. Startlingly, one could replace the word “equations” by the word “witchcraft” without at all effecting the tone of the above synopsis.

Let’s take a look at the man who provides the center for the show, Professor Charlie Eppes.

When you’re this good at math, you get
to wear blazers made entirely of gold.

Watch a few episodes of Numb3rs (or read this Wikipedia entry), and you will likely learn the following about this darling mathematician:

1. Charlie graduated from Princeton when he was 16, and is a young math prodigy.
2. Charlie enjoys chess.
3. Charlie loves blackboards.
4. Charlie has a beautiful girlfriend (a former graduate student, no less) named Amita Ramanujan, who may or may not be related to this famous (and awesome) mathematician.

5. Charlie enjoys socializing, and appears to shower regularly.
6. Charlie loves to explain mathematical concepts using real world examples, such as vacuums or spiders (see below).

For the record, let it be known that you don’t have to graduate from college when you’re 16 to be a good mathematician (although it certainly doesn’t hurt). Moreover, not all mathematicians are good at, or even enjoy chess.

By and large, though, we do enjoy a good blackboard.

Points 4 and 5 signify a departure from the math nerd we all know and love. A sexy, brilliant mathematician with an equally sexy, brilliant mathematician? Neither of them even wear glasses! Not to mention the fact that Charlie has more charm than most of the other characters on the show. Is he a mathematician, or a rock star? Or, even better, is he merely a prophet for the future, in which mathematicians and rock stars will be one and the same?

Of course, when you are centering your show on a mathematician, you had better make that mathematician marketable. So in a sense, avoiding certain stereotypes becomes a necessity. Still, having a positive mathematics role model like Charlie Eppes certainly can’t be bad for the math community. In fact, if the below video is any indication, people LOVE mathematicians.

Mathematicians may be portrayed relatively favorably, but what about the math itself? Is it legit? Well, it’s hard to say, really, since when all is said and done, there’s not a whole lot of math on display. Certainly there are a lot of scenes with people waving their hands and discussing math, or scenes with chalkboards that have math on them, but these scenes are often placeholders in between scenes with guns or explosions or good looking government employees (here I use “or” in the inclusive, mathematical sense of the word). This, however, is expected, again because of the mass market nature of the program. Overall, I think you’d find it difficult to do math any better after having watched an episode of Numb3rs. On the plus side, they do emphasize that the story lines are based on actual cases, so viewers can take comfort in the fact that even out in the real world, math is helping to bring in the bad guys.

The show certainly doesn’t do any damage to math’s reputation. By making a protagonist who is smart and has a winning smile, the creators seem to be doing their part to show that math needn’t be as scary as it’s made out to be. While there are plenty of moments where it can be hard to suspend your disbelief (he may be smart, but I don’t think even Doogie Howser can solve the Riemann Hypothesis, contrary to what Prime Suspect would have us believe), as an overall ambassador to the universe of mathematics, the show gets a pass. If nothing else, it teaches America that not all mathematicians are completely socially inept, even if we do live at home and put the moves on a few of our advisees.