A friend of mine received free tickets to a Kings game when I was living in LA several years ago. He invited my now-wife and me along, and the price was certainly right, so four of us went to the Staples center one Saturday afternoon.

I don’t remember much about the game (though I do recall that the Kings emerged victorious). What I remember most vividly was that during one of the breaks between periods, a new car was brought onto the ice and there was a contest to give that car . . . → Read More: Give it Away, Give it Away, Give it Away Later]]>

A friend of mine received free tickets to a Kings game when I was living in LA several years ago. He invited my now-wife and me along, and the price was certainly right, so four of us went to the Staples center one Saturday afternoon.

I don’t remember much about the game (though I do recall that the Kings emerged victorious). What I remember most vividly was that during one of the breaks between periods, a new car was brought onto the ice and there was a contest to give that car away. Sort of like what happens in this video, but the rules were a little different:

In the game I attended, six contestants were given a key to a new car, but they were told that only one key would start the vehicle. One at a time, each person had to enter the car, put the key in the ignition, and try to start the engine. If they succeeded, they won the car; otherwise, they had to exit the vehicle and give the next person a turn.

There’s a nice little probability model in here. Let’s generalize it a bit and suppose *n* people are vying for the car. The probability that the first person wins is then equal to 1/*n*. If that person doesn’t win, the next person steps into the car; since we know the first key wasn’t a winner, the second person’s probability of winning is 1/(*n* – 1). And so on: assuming the first *k* people don’t win the car, the probability that the (*k + *1*)*^{st} person is the winner equals 1/(*n* - *k*).

This leads us to an intuitively obvious conclusion: if none of the first *n* – 1 people have the winning key, the last person must be the winner: 1/(*n* – (*n* – 1) + 1) = 1. And this is precisely what happened on that fateful day at the Staples center: five people entered the car, and five people exited without winning. The audience cheered for the last contestant, so clear was it to everyone that she was the winner.

But then something unexpected happened. That last person went into the car, turned her key, and…nothing. Her key didn’t work either! In a supremely anticlimactic end, the car was moved off the ice, and the announcer told us that they would figure it out backstage. We never found out who claimed the prize. Life is full of mystery.

This led me to another question: would it have been possible to predict this result with a more sophisticated mathematical model? Let’s suppose that in addition to the *n* people vying for the car, there’s a certain “success factor” *p*, which represents the probability that the person with the winning key will be able to start the engine. In other words, 1 – *p* represents the probability that someone will screw up with the winning key (either due to nerves, unfamiliarity with the concept of keys, or whatever other reason you can think of). Of course, everyone’s success factor may be different, but for simplicity let’s assume that *p* is a universal constant.

How does this impact our model? Well, now the probability that the first person will win the car is *p*/*n*, since the first person must both (a) have the winning key and (b) not screw things up when called upon to turn the car on.

The general situation is more complex, but the result is quite nice. To write down a rigorous argument takes some work^{1}; if you’re more interested in a little convincing as to why the result holds, think about what would happen if you ran this experiment — letting *n* people try to win a car by giving them a key — a large number of times. If you put pen to paper, you’ll see that in this case, given that the first *k* people don’t win the car, the probability that the (*k + *1*)*^{st} person wins is equal to

As ought to be the case, when *p *= 1 this gives us our simpler model from before.

Although the formula is relatively simple, it has some interesting consequences compared to our earlier model. For example, with our earlier model we knew that if we made it to the last person, that person was guaranteed to win; in other words, when *p* = 1 and *k* = *n - *1, the probability above is equal to 1. But for general *p* this is no longer true; instead, if we make it to the last person in this new model, the last person will win with probability

Notice that regardless of *p*, as long as *p* < 1, the above quantity goes to 0 as *n* grows. In other words, if you’re competing against a large pool of people, even if the probability of not screwing things up is quite high, it’s probably more likely that someone screwed up than that you’re the winner. We can actually quantify this a bit more: with a little work, you can show that if the contest makes it to the last person, the probability that the last person actually has the winning key is equal to

So, what’s more likely: that the last person will win, or that someone before the last person screwed up? Well, it’s more likely that someone else screwed up precisely when

After a little bit of algebraic rearrangement we arrive at the simple condition that *p* < 1 – 1/*n*. In other words, if you’re the last competitor from a pool of 100 people, even if people don’t screw up 98.9% of the time, if you get a chance to turn the key, you shouldn’t get your hopes up; it’s more likely that someone else screwed up than it is that you’ll drive out of the rink a winner.

^{1} You’ll need to be savvy with your conditional probabilities in order to make this argument fully rigorous. Try the case of the second person being the winner first. If we let *X _{i}* = 1 when the

Most of my thoughts on the subject are encapsulated in the Mathalicious response. (Both articles come highly recommended, and what I say below may not make much sense if you haven’t read them first.) The conversation got me thinking, though, and so I’d like to offer my own personal aside/addendum.

When I began writing in this corner of the internet in the summer of . . . → Read More: Keeping it Real: An Addendum]]>

Most of my thoughts on the subject are encapsulated in the Mathalicious response. (Both articles come highly recommended, and what I say below may not make much sense if you haven’t read them first.) The conversation got me thinking, though, and so I’d like to offer my own personal aside/addendum.

When I began writing in this corner of the internet in the summer of 2008, my goal was simply to talk about mathematical ideas in a way that was accessible for a general audience (and in particular, an audience that didn’t necessarily think of itself as mathematically inclined). I believed that talking about math through the lens of popular culture would help with this, because, in my experience, most people (a) prefer pop culture to math, and (b) don’t believe that the intersection between the two is all that rich. I probably haven’t always been successful in hitting the sweet spot in the intersection between pop culture and mathematics, but those successes have brought with them a not-insignificant sense of pride.

I’ve measured that success over the last several years, in part, by how accessible I’ve made some bit of mathematics to someone who wouldn’t have otherwise been interested. (I realize this is perhaps one of the least useful metrics ever conceived; I don’t even know what the proper units of measurement would be.) To put it another way, my goal has been to make math more “real” to people both in and out of the classroom. When I say “make math real,” I mean something like “make the beauty of mathematics more transparent,” though of course this simply replaces one squishy word (“real”) with another one (“beauty”). I guess what it boils down to is trying to imbue people with some of the joy for mathematics that I feel (on most days, anyway).

The world external to mathematics, if done properly, is frequently a resource when working towards this goal, since people have a variety of interests, not all of which are (obviously) mathematical in nature. This is why the work I do, both here and at Mathalicious, so frequently begins with something that is about the world at large, rather than about prime numbers or the Riemann Hypothesis (both of which are also things that I love dearly). Having said that, most of this work, for me, is in service of having great conversations around interesting problems, and strictly speaking there’s no requirement that the world external to mathematics make an appearance. Indeed, I don’t believe such externality is necessary in order for a task to be “real.” Beauty doesn’t discriminate when it comes to the origin of interesting questions. Sometimes, great mathematical questions are inspired by something external to mathematics (the search for these questions occupies a nontrivial amount of my time during the day). Sometimes, great mathematical questions are inspired by mathematics alone.

And sometimes, mathematics inspires “real-world” applications unbeknownst to the original authors. Elliptic curves were studied for hundreds of years before possible applications to cryptography were discovered by Neal Koblitz and Victor Miller in 1985. It’s conjectured that the zeros of the Riemann zeta function satisfy a property which was first discovered in the context of the quantum mechanics. In other words, there’s precedent for deep, pure mathematics to have unexpected connections to the “real” world around us. Or, to put it another way, the intersection between the external and internal worlds of mathematics is likely larger than any of us anticipate.

This is why, for me, keeping a task “real” in the sense I’ve described above is more important than keeping it “real” in the sense of drawing from something external to mathematics. A “fake” task, in my view, is one which doesn’t fit properly in either the internal or the external world (e.g. version C in the problem described by Dan). To the extent that I have no idea how deeply the internal and external worlds of mathematics intertwine, I can only do what I think is best: get equally psyched for both. And on days when I get other people psyched as well, all feels right the world. (Or worlds. Depending on your interpretation.)

]]>In one of the campaign’s more recent ads, however, I was disappointed to see a teachable moment go to waste. I suppose this is what happens when you have a cell phone company spokesman in a room full of children instead of an actual teacher. (Though to be fair, the math involved isn’t really suitable for elementary school.)

Here’s the ad:

In . . . → Read More: It’s Not Complicated. Or is it?]]>

In one of the campaign’s more recent ads, however, I was disappointed to see a teachable moment go to waste. I suppose this is what happens when you have a cell phone company spokesman in a room full of children instead of an actual teacher. (Though to be fair, the math involved isn’t really suitable for elementary school.)

Here’s the ad:

In case you don’t have time to watch cell phone commercials, here’s a transcript of their conversation.

AT&T Guy: What’s the biggest number you can think of?

Girl 1: A trillion billion zillion!

AT&T Guy: That’s pretty big. How about you?

Boy 1: Ten.

AT&T Guy: …Okay. How about you?

Boy 2: Infinity!

AT&T Guy: Can you top that?

Girl 2: Infinity and one!

AT&T Guy: Actually, we were looking for infinity plus infinity, sorry.

Girl 1: What about infinity times infinity?

(AT&T Guy’s mind is blown.)

There are a number of objections one could raise to the above conversation. I’m not interested in all of them – in particular, I’m not interested in discussing the question of whether or not infinity is a number. What interests me most is whether or not this AT&T Guy gave a fair shake to all of the responses he received.

Before diving in, we’ll need to agree on some terms. Specifically, what do these children mean when they throw around the word “infinity”? Since these children are relatively young, it’s safe to assume they have no notion of the different cardinalities of infinity, and instead think of infinity as something akin to a size larger than the largest number. Let’s assume that “infinite” in this setting means the size of the set of counting numbers. (In other words, we’ll take infinite to mean countably infinite.)

Given this working definition, is there really a distinction between infinity, infinity and one, infinity plus infinity, and infinity times infinity? To answer this question, we need some way to compare sets with infinitely many elements. Thankfully, mathematicians have already thought through these issues. We say two sets are the same size (more precisely, they have the same cardinality) if (1) each element from the first set can be mapped to a *unique* element from the second set, and (2) every element in the second set gets mapped to by something in the first set. The first condition makes the mapping an injection (also known as 1-to-1), while the second makes the mapping a surjection (also known as onto). Mappings which are both injections and surjections are called bijections. (In other words, a bijection is invertible and vice-versa.)

The jargon can be a little hard to wrap one’s head around, so here are some pictures that may help. The first mapping is an injection, but not a surjection. The second is a surjection, but not an injection. Only the third mapping is both.

Two sets are said to have the same cardinality if there’s a bijection between them. For finite sets, this is nothing more than the statement that a set of size *n* is as large as a set of size *m* if and only if *n* = *m*. For infinite sets it can be a little trickier, but as long as we can find a bijection between two infinite sets, we can say that they have the same size.

So what does this mean for the AT&T commercial? Let’s first compare “infinity” to “infinity plus one.” To make things concrete, we’ll compare two explicit sets: the set of counting numbers {1, 2, 3, … } to the set of whole numbers {0, 1, 2, 3, … }. Both sets have a (countably) infinite number of elements, but the set of whole numbers clearly has one element that the set of counting numbers does not: the number zero. So one could argue that if the set of counting numbers is of size infinity, the set of whole numbers is of size infinity plus one.

But these two sizes are really no different! Indeed, we can define a mapping *f* from the natural numbers to the whole numbers via the formula *f*(*n*) = *n* – 1. This is a bijection because no two counting numbers are mapped to the same whole number (*f* is injective), and no whole number gets overlooked by this mapping (*f* is surjective: if *m* is a whole number, then *f*(*m* + 1) = *m*). In other words, there is a bijection between these two sets, and so they have the exact same cardinality.

What about infinity plus infinity? For this, let’s compare the counting numbers {1, 2, 3, … } to the set of nonzero integers {1, -1, 2, -2, 3, -3, …}. Since the second set is essentially two copies of the first (the counting numbers and their opposites), it’s not entirely crazy to say that if the first set has size infinity, the second has size infinity plus infinity. But once again, these sets actually have the same cardinality! Here’s one example of a bijection in this case:

(I’ll leave the verification that this is a bijection up to you.)

In other words, the AT&T guy shouldn’t have been so quick to brush off the answer of “infinity and one” in favor of “infinity plus infinity” – these two sizes are really the same.

But what about infinity times infinity? That certainly seems larger, right? For an example of such a set, we could consider all the ordered pairs (*m*, *n*) where *m* and *n* are whole numbers. Since there are infinitely many options for the first coordinate, and infinitely many options for the second, there are, in some sense, infinity times infinity ordered pairs in this set.

Once again, though, this set actually has the same cardinality as the other infinite sets explored here. Writing down an explicit bijection may not be as straightforward in this case, but there are a number of ways to do it. The picture below provides insight into one possible approach:

The conclusion is that all of the infinite sets described in the ad really have the same cardinality. It may not be the most complicated thing in the world, but it’s certainly more complicated than AT&T would have us believe.

]]>As some of you may know, in general I don’t hold our country’s voting methods in very high regard. Think about the way we vote for president, for instance. Aside from not asking voters to state any preferences at all, it’s difficult to do worse than our current system: we can only show our support for a single candidate, when in fact our preferences may be more nuanced. Moreover, since we can only vote for a single candidate, there’s little incentive to vote for our favorite one, unless our favorite happens to be a front-runner. This is known all across the universe, as evidenced by the Presidential runs of Kang and Kodos:

Even worse, a third party candidate who garners a decent amount of support may end up hurting his own party and parties more closely aligned to it by acting . . . → Read More: Down with Plurality!]]>

As some of you may know, in general I don’t hold our country’s voting methods in very high regard. Think about the way we vote for president, for instance. Aside from not asking voters to state any preferences at all, it’s difficult to do worse than our current system: we can only show our support for a single candidate, when in fact our preferences may be more nuanced. Moreover, since we can only vote for a single candidate, there’s little incentive to vote for our favorite one, unless our favorite happens to be a front-runner. This is known all across the universe, as evidenced by the Presidential runs of Kang and Kodos:

Even worse, a third party candidate who garners a decent amount of support may end up hurting his own party and parties more closely aligned to it by acting as a “spoiler.” Of course, the most well-known example of this is Ralph Nader, who many people believe cost Al Gore the 2000 election (for more on the spoiler effect, see here).

For all these reasons and more, our current system (known as Plurality Voting, or first-past-the-post) seems woefully inadequate. The good news is that there are lots of alternatives. The bad news is that because there are lots of alternatives, it can be difficult for someone with little background in election theory to be able to recognize what separates a great voting system from a poor one.

This is where the Center for Election Science comes in. As part of their goal to “educate the general public and advocate election systems that most benefit the public good,” they have recently launched an Indiegogo campaign to raise money for a video highlighting the awfulness of Plurality Voting, and explain a simple alternative known as Approval Voting. Approval Voting is like Plurality Voting, but with one change to the rules: instead of being forced to vote for a single candidate, you can vote for as many as you like. The person with the most votes wins.

I’ve written about Approval Voting before, and the Center for Election Science does a good job explaining the system (and comparing it to other systems) on their website. But very briefly, here are some of the perks:

- It never hurts to vote for your favorite candidate.
- Approval Voting eliminates the spoiler effect.
- From a practical standpoint, Approval Voting is easy to understand and would be straightforward to implement using current technology.

There are other advantages too, but I’ll let you dig deeper if you’re curious. The main takeaway is that the Center for Election Science is doing important work, and if you’ve got some coin to toss their way, I’d encourage it.

This weekend they’ve lowered the bar for contributions down to a single dollar. Compared to what our forefathers sacrificed in the name of democracy, that’s a pretty good deal.

Here’s all the info, if you want to learn more:

]]>One of the more recent trends in the world of Simpsons memorabilia is the advent of the Mini-Figure collections, produced by Kidrobot. Each series (there have been two so far) consists of around 25 small Simpsons figures, each with his or her own accessories. The figures cost around $10 each ($9.95, to be precise), so an avid collector would need to spend something like $250 to complete each of the two collections, right?

Well, not quite. When you buy one of these figures, you have no idea which one you’ll get, because the box containing the figure doesn’t indicate what’s inside. All you know are . . . → Read More: Mathalicious Post: Most Expensive. Collectibles. Ever.]]>
*Simpsons *themed cautionary tale for collectors on a budget. Here’s a sample:

One of the more recent trends in the world of Simpsons memorabilia is the advent of the Mini-Figure collections, produced by Kidrobot. Each series (there have been two so far) consists of around 25 small Simpsons figures, each with his or her own accessories. The figures cost around $10 each ($9.95, to be precise), so an avid collector would need to spend something like $250 to complete each of the two collections, right?

Well, not quite. When you buy one of these figures, you have no idea which one you’ll get, because the box containing the figure doesn’t indicate what’s inside. All you know are the probabilities for each figure, and even those are sometimes missing…

Given this information, here’s a natural question: how many of these boxes should you expect to buy if you want to complete the set, and how much will it cost you?

Peep the link to read the full story!

]]>Of course, as you move closer to the present, the equations get a little more sophisticated. Even so, Dr. Mackenzie does his best to ground the equations to something relatable to a wide . . . → Read More: Math in Books: The Universe in Zero Words]]>

Of course, as you move closer to the present, the equations get a little more sophisticated. Even so, Dr. Mackenzie does his best to ground the equations to something relatable to a wide audience (and by and large, he’s quite successful). For instance, he uses whales as a way to talk about non-Euclidean geometry: you can read more about this example here, and can download the relevant chapter, too. While the book won’t turn you into a mathematical genius, it will teach you the history surrounding some of the subject’s most important equations, and will give you a reasonable idea as to what the equations are communicating. Words can’t always do justice to the economical beauty of an equation (after all, that’s what makes equations so appealing in the first place!), but if you suffer from math anxiety, a book like this may help to alleviate some of your most severe symptoms.

Once I finished the book, Dr. Mackenzie was kind enough to let me pick his brain a little bit. Here’s a little Q and A to give you a better impression of what the book is about and the thought that went into writing it. I’ve added some links to more math where appropriate. Thanks to Dr. Mackenzie for taking the time to chat!

Are there any equations you wanted to include but ended up having to strike?Absolutely. The first draft of the table of contents had something like 48 equations. On the second draft I pared it down to 30, and as you might expect, the last six cuts were the hardest. Two areas that seem underrepresented in my book are statistics and mathematical biology.

From statistics, I would have really liked to include the formula for the normal distribution and Bayes’ Theorem. However, the 1700s and 1800s, which those formulas came from, were already extremely well-represented, and I don’t know what I would cut to make room for them. Also, Bayes’ Theorem poses a bit of a problem — discovered in the 1700s, but really only appreciated in the second half of the twentieth century. So which century do you put it in?

As for mathematical biology, the problem I faced was a lack of equations that are really central to the entire subject. In mathematical physics, everybody agrees on the importance of Newton’s laws or Maxwell’s equations. But biology is more fragmented, and it seems to deal less in universal laws and more in reasonable models or rough approximations. The closest thing to a universal law in biology is natural selection, but Darwin did not express that in mathematical form.

Having said that, though, I think that a very strong candidate for my book would have been the exponential law of population growth when resources are unlimited, and perhaps the logistic law when resources are limited. Both of these equations have such huge ramifications. It was a tough call leaving them out — the law of exponential growth was on my top 30 list — but again it was a case where there was such strong competition for the six equations from the 19th century.

Do you have any personal favorites among the equations included?Well, of course to some extent they are all favorites! But yes, there are some that are more closely related to the kind of mathematics I did when I was still actively involved in research. Hamilton’s quaternions are a special favorite, because they came up twice in a VERY unexpected way in my research on minimal surfaces (surfaces of least area). Once may be an accident, but twice really gets you thinking. The quaternions and their cousin, the octonions, are closely related to all sorts of “exceptional” phenomena in algebra and geometry, even including the dimension of space.

In the book I didn’t even mention the connection of quaternions to my own research, because there are too many other interesting things about them. They are closely related to spinors, and as I say in my book, they are the absolute best way to mathematically represent anything that spins or rotates. That includes electrons and really all subatomic particles (except the Higgs particle!), and so it created a direct link between the chapter on quaternions and the chapter on Dirac’s equation for the electron. This makes a nice segue also to your next question…

One of the things I enjoyed about your list of equations is that they are not treated in isolation, but instead weave together a nice narrative on the history of mathematics. Did working within this narrative framework present any challenges when trying to decide on which equations to include?This is one of the reasons for the beauty and the incredible power of mathematics. It’s almost impossible to start writing about a sufficiently deep formula or theorem and NOT start finding connections to all of the rest of mathematics. I did not have to go looking for these connections; I also didn’t really plan on them when I worked out the table of contents. They just showed up by themselves without any effort on my part. All I had to do was point them out.

Some of the connections are even meta-mathematical. For instance, when I was working on part 2 of the book, which covers roughly the years from 1500 to 1800, I couldn’t help noticing the conflict between publicity and secrecy that kept playing itself out in different ways in different times. Some mathematicians tried very hard to keep their ideas proprietary. Others understood that the best way to advance the subject and to advance your own reputation is to share your discoveries freely. I think it was Euler, who was perhaps the most prolific mathematician of all time, who really led by example and turned mathematical scholarship decisively toward the mode of sharing. Nevertheless, this is a battle that continues to be fought in every century and every generation. Now we have mathematics done for the military or intelligence agencies, and we have mathematics done for private companies (e.g., every investment firm has its own version of the Black-Scholes equation). Will these discoveries be shared, or secreted away? This isn’t a big theme of my book, but it’s something I noticed and pointed out where appropriate.

As you travel from the past to the present, the equations necessarily get more complicated. Did you find your approach to writing about an equation vary depending on the mathematical sophistication required to be familiar with the equation? Was it harder to write about Black-Scholes compared to something like the Pythagorean Theorem?Yes, this was definitely part of the challenge of writing a mathematics book. One thing I’ve noticed about a lot of what I call “Big Honking Histories” of mathematics is that they stop around the early twentieth century. Modern mathematics seems to be just a bridge too far for these books. I was determined to make this book different and cover mathematical developments right up to the present. I did not want to give the message that math was all completed in the 1800s, or give the message that laymen should not even try to understand anything after 1900. That would be a terrible message. Every other science has its popularizers who are trying to convey the science of TODAY, and mathematics needs to do the same.

But it’s hard. In some cases the material was hard even for me to understand, let alone for the reader. Nevertheless, in every case I think the formula must express some idea that is simple and far-reaching; otherwise it could not be a great formula. In the case of Black-Scholes, the heart of the matter is that the movement of stock prices is a diffusion process, just like the movement of molecules in the atmosphere. If I can uncover that basic idea, then I can write something that is accessible to all readers, even if they don’t understand the mechanics of how you turn the central idea into an equation.

Even so, I might not have been completely successful. I have seen one or two reviews that say the last part of the book is more challenging than the first part, where I was writing about formulas like 1+1 = 2 and approximations to pi. But even if it does require a little bit more effort from the reader to read the last part, I think that the effort is worth making. Dirac’s equation, for instance, is too important for us not to at least try to understand what it says. I may not succeed in making it completely understandable, but I’ll get you closer and I will at least put it on your intellectual radar screen.

If you’d like to know more, pick up a copy of Mackenzie’s book! I think you’ll enjoy it. If you’re feeling too lazy to navigate away from this page, you can order the book directly from the widget below, or from the “Shop!” tab up at the top. Happy reading!

“You always double down on 11, baby.” Sage advice from Vince Vaughn’s character in the 1996 film Swingers. At one point in the film, Trent (played by Vaughn) and Mike (played by Jon Favreau) make an impromptu trip to Las Vegas, and Mike ends up completely out of his depths at a high-stakes blackjack table…Mike receives a six and a five, giving him a total of eleven. Trent urges him to double down, and indeed, this seems like good advice. After all, in a deck of 52 cards, 16 of them have a value of 10 – that’s over 30%! Always doubling down on eleven is also consistent with the basic blackjack strategy popularized by Edward O. Thorp in his book Beat the Dealer. . . . → Read More: Mathalicious Post: Doubling Down]]>

“You always double down on 11, baby.” Sage advice from Vince Vaughn’s character in the 1996 film Swingers. At one point in the film, Trent (played by Vaughn) and Mike (played by Jon Favreau) make an impromptu trip to Las Vegas, and Mike ends up completely out of his depths at a high-stakes blackjack table…Mike receives a six and a five, giving him a total of eleven. Trent urges him to double down, and indeed, this seems like good advice. After all, in a deck of 52 cards, 16 of them have a value of 10 – that’s over 30%! Always doubling down on eleven is also consistent with the basic blackjack strategy popularized by Edward O. Thorp in his book Beat the Dealer. From a mathematical standpoint, Trent is right. You should always double down on eleven.

Interested in the rest of the story? Click here!

]]>How Math Goes Pop! will be changing is the subject for another post. One thing’s for sure, though: I’ll be contributing to the Mathalicious blog regularly. My first post, on whether or not it makes sense to foul the opposing team at the buzzer in a close basketball game, went live last week. Here’s a small sample:

A three point shot by Sundiata Gaines turned a two-point loss for the Jazz into a one-point win. No doubt that’s a tough defeat for Cavs fans and players alike, but in such a situation, there’s really nothing the defense could’ve done to change the outcome.

Or is there? What . . . → Read More: Mathalicious Post: To Foul Or Not To Foul]]>

How Math Goes Pop! will be changing is the subject for another post. One thing’s for sure, though: I’ll be contributing to the Mathalicious blog regularly. My first post, on whether or not it makes sense to foul the opposing team at the buzzer in a close basketball game, went live last week. Here’s a small sample:

A three point shot by Sundiata Gaines turned a two-point loss for the Jazz into a one-point win. No doubt that’s a tough defeat for Cavs fans and players alike, but in such a situation, there’s really nothing the defense could’ve done to change the outcome.

Or is there? What if, instead of letting Gaines take the shot, the defense had fouled him? Could that have increased the Cavs’ likelihood of maintaining their lead? If Gaines had been fouled he would’ve been given three free throws, but would’ve had to make all three in order to win. Making three shots certainly sounds harder than making one shot, even if a shot from the line is easier to make than a three-pointer. Though

ethicallymurky, is fouling a sound strategymathematically?

Click here for the full story!

]]>In keeping with my summertime theme of mathematics and food (see e.g. here and here), I’d like to share with you a story about a recent dinner I shared with my better half. After a day spent apartment hunting, we decided to treat ourselves to a dinner out.

Everything we learned about treating ourselves we learned from Parks and Recreation.

In keeping with the theme of treating ourselves, we ordered two desserts at the end of the night, and both looked quite delicious. We agreed to each eat half of one dessert and then trade for the second half. One was in the general pie family of desserts.

Given a slice of pie, the most . . . → Read More: Pi(e) Mathematics]]>

In keeping with my summertime theme of mathematics and food (see e.g. here and here), I’d like to share with you a story about a recent dinner I shared with my better half. After a day spent apartment hunting, we decided to treat ourselves to a dinner out.

Everything we learned about treating ourselves we learned from Parks and Recreation.

In keeping with the theme of treating ourselves, we ordered two desserts at the end of the night, and both looked quite delicious. We agreed to each eat half of one dessert and then trade for the second half. One was in the general pie family of desserts.

Given a slice of pie, the most natural way to divide it in half is to bisect the angle formed at the end of the slice.

To cut a slice of pie in this manner, one generally requires an appropriate tool, such as a knife. However, our desserts were accompanied only by spoons. Rather than try to bisect the angle using such a crude instrument, we decided the person who started with the pie should simply use his or her best judgement, and stop eating when roughly half the slice had been consumed.

In other words, rather than bisecting the area by slicing *vertically*, we decided to bisect the area by slicing *horizontally*, since one typically eats pie by starting at the bottom corner and working up. Of course, this raises an interesting question: how high up must the horizontal slice be in order for it to cut the piece of pie in half? Little did I know that at the same time, Dan Meyer was discussing the same problem over at his blog. Commenters there seemed to be split over whether or not this question is contrived; the present discussion either provides evidence that it is not, or evidence that we should have simply asked for a knife.

Let’s return to the question of where we ought to place our cut. Certainly halfway up isn’t right, because then piece below the line will have a smaller volume:

To find the horizontal line which cuts the volume in half, we’ll need to do a bit of calculation. Let’s introduce some notation: treat a slice of pie as a sector of a cylinder of radius r and height *h*. Suppose also that the angle formed by the sides of the slice of pie is denoted ; if the pie consists of *n* equal slices, then . The suitability of this model will be discussed at the end, but for now, let’s take it as a given. Here’s a diagram of the situation:

Because the volume of a cylinder with radius *r* and height *h* is , and the pie is cut into *n* equally sized pieces, the volume of a slice must equal

Now suppose we cut a line through the slice of pie as in our pictures above, cutting it into two pieces. Let’s call the distance between this cut and the tip of the slice *c*. The two pieces are shown here from a top down view; note the blue slice corresponds to a triangular prism.

Because the volume of a triangular prism is equal to the area of the triangular base times the height of the prism, we can compute the volume of the blue portion of the slice. Using some trig and the fact that *c* bisects the angle , we see the area of the triangle must equal , and therefore the blue piece has volume equal to

In order for this to equal half of the total volume, the following equality must hold:

Notice we can cancel the factors of *h* on both sides, and after simplification we see that the ratio of *c* to *r *must satisfy

Now, for a typical pie there are only a few possibilities for the value of *n*. By considering different values, we can determine how large the ratio *c*/*r* needs to be; in other words, we can determine *c* as a percentage of *r*.

We can also ask how much one person gets screwed if the pie is sliced by simply taking *c* = *r*/2. This is a natural cut to make since it slices the *length* of the pie in half, as discussed above. In this case, the blue portion of the slice will have volume equal to

meaning that the ratio of the blue portion’s volume to the total volume equals

Here’s a table of these values (the ratio of *c* to *r* and the ratio of volumes) for *n* = 4, 6, 8, 10, 12, or 16 slices.

n | c/r (≈) | Ratio of Volumes (≈) |
---|---|---|

4 | 0.626 | 0.318 |

6 | 0.673 | 0.276 |

8 | 0.689 | 0.264 |

10 | 0.695 | 0.259 |

12 | 0.699 | 0.256 |

16 | 0.702 | 0.253 |

As you can see, when the number of slices increases, *c* needs to be a larger fraction of *r* in order to cut any one slice in half (Exercise! Show that as *n* goes to infinity, *c*/*r* goes to ). Similarly, the fraction of volume represented by the blue portion in the picture above gets smaller and smaller as the number of slices increases if a slice is cut in half lengthwise (Exercise! Show that as *n* goes to infinity, this ratio goes to 1/4).

Here’s a picture of the different slices represented in the table along with where you should make the cut to divide each slice into pieces of equal volume. The slices are superimposed on one another.

Similarly, here’s what it looks like if you take *c* = *r*/2 in each case listed in the table.

I don’t know the value of *n* for the slice of pie we shared that evening. Lesson learned: never go out to eat without your protractor. I did, however, take a picture of our estimate. How do you think we did?

Hungry for more? Chew on this: how accurate is the model used here? Is a slice of pie really best modeled by a sector of a cylinder? Or is it really more of a sector of a frustum? Does this significantly alter the results? Also, what’s different if the pie is cut into three slices (i.e. what happens when *n* = 3)? And who gets more crust with this approach? How much more?

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As might be expected, the piece has furrowed quite a few brows. A few friends have asked me for my opinion, but I’m a little late to the game, and there are a number of people who have expressed my views in their own words quite well. I’ll briefly add my own to cents, peppered with links throughout.

. . . → Read More: Asking the right questions]]>As might be expected, the piece has furrowed quite a few brows. A few friends have asked me for my opinion, but I’m a little late to the game, and there are a number of people who have expressed my views in their own words quite well. I’ll briefly add my own to cents, peppered with links throughout.

First, I agree with Dan Meyer that the question “Is Algebra Necessary?” is not the right question. In the strictest sense, I suppose it isn’t; certainly one can go through one’s entire professional life without using a lick of algebra (though I can’t say I’d recommend it). But the purpose of education isn’t to supply people with only the information they will need in their career. By this measure, nearly all of what students learn is not necessary As Rob Knop points out, “liberal arts education is to make people into good citizens, not into good workers.”

A more important question, though no less trivial, is the question “Is Algebra Valuable?” I don’t think there’s much room for debate here either. If you don’t think algebra has any value, that’s probably because you don’t understand algebra. This may be through no fault of your own – maybe you had a terrible teacher, or a terrible textbook, or a home life that made it difficult to concentrate on your studies. Whatever the cause, once people feel slighted by mathematics, many of them decide there are better things they could be doing with their time. But the benefits to understanding algebra (or more generally, to building critical thinking and reasoning skills) exist and are quantifiable. As Daniel Willingham notes (see his links for more info), “Economists have shown that cognitive skills–especially math and science–are robust predictors of individual income, of a country’s economic growth, and of the distribution of income within a country.”

A better question may be something like “How can we convince students of algebra’s value?” It’s no secret that math has kind of a PR problem. Textbooks can be dry, and the questions students are tasked to answer frequently seem contrived and completely disjointed from their everyday world. But this is not a problem inherent to algebra, only the way it is presented. Good teachers know this, and are able to make mathematics relevant to their students. Once the material no longer seems arbitrary, it is easier to understand. It can also be, dare I say it, fun.

But even for those who enjoy math, it can still sometimes be difficult. Difficulty alone, however, is insufficient reason for changing the curriculum, especially when the US trails so many other countries in the mathematics ability of its students. Hacker is undoubtedly well-intentioned, but I don’t think his argument stands up under scrutiny.

I’ll stop now, because other people have refuted the op ed better than I could. Feel free to check out the links I’ve already mentioned, or recent posts by Ilana Horn, Andy Soffer, and this superb roundup by Damon Hedman.

I’ll be back to my usual irreverence next time, I promise!

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