Football Pools, Part 3

This is the third in a series of posts about pools used for betting on the outcome of football games (part one can be found here, and part two here).  Let me briefly recall the setting, which is probably familiar to anyone who has been to a Super Bowl party.  Typically, one bets on the outcome of a football game using a 10 x 10 grid.  People can buy any number of the 100 squares on the grid, and when all the squares have been purchased, each row and each column is assigned a random digit from 0 to 9.

Suppose, for example, that you buy four squares, and after the rows and columns have been labeled, you find that you own square 3-7, square 2-5, square 9-0, and square 6-6.  You will win money if, at the end of any one of the four quarters, the last digit in each team’s score matches your pair.  For example, if the score after the 3rd quarter is 13-27, you will win some money, since the last two digits are 3 and 7, and you own square 3-7.  There are variants of this: some pools only pay out every half, not every quarter, and usually the payouts vary by quarter, so that having the right square at the end of the game wins you more money than having the right square at the end of the first quarter.

Here's an example of a football pool which has been tagged in the four squares mentioned above.

In the first part of this discussion, we introduced a new way to conduct the pool: rather than looking at the last digit of a team’s score, we looked instead at the digital root of the team’s score.  Recall that the digital root of a team’s score is obtained by adding the digits in their score.  If that sum is between 1 and 9, we stop – if it is larger than 9, we compute the digital root again, until we get a digit between 1 and 9.  For example, the digital root of 14 is 1 + 4 = 5, while the digital root of 38 is 2, since 3 + 8 = 11, and 1 + 1 = 2.  We then analyzed the distribution of scores, and found that the digital root of a team’s score is more evenly distributed between 1 and 9 than the last digit of a team’s score is evenly distributed between 0 and 9 (this is subject to the convention that we assign 0 a digital root of 9, since 0 is the only number with digital root equal to 0).

In the second part of the discussion, we tackled questions of independence.  Namely, we asked whether the last digit in one team’s score is independent of the last digit of the other team’s score, and similarly we asked whether the digital root in one team’s score is independent of the digital root of the other team’s score.  In both cases, we found the answer to be negative.

The subject of this article is based on the following observation: when you have wagered in a traditional football pool, it’s not uncommon for a small number of squares to be hit with high frequency during the course of a game.  For example, suppose you watch a game in which one team scores 7 points, then 3, then 7, then 3, while the opposing team never scores.  This means that the game’s score will go from 0-0, to 7-0, to 10-0, and then to 17-0.  So, while there are four unique scores in the game, with the usual football pool, only two squares will be hit: the 0-0 square, and the 7-0 square.  However, with the digital root pool, four squares will be hit: again using the convention that we assign 0 a digital root of 9, the squares will be 9-9, 7-9, 1-9, and 8-9.

The reason the digital root pool hits more squares in this case is because whenever one team increases its score by 10, the last digit of their score will return to a previous value.  However, with the digital root method, if a team increases its score by 10, the digital root increases by 1.  Because a score increase of 10 is a relatively common occurrence in football (all one needs is a touchdown, extra point, and field goal), one may therefore guess that using the digital root pool, more squares should be hit during the course of the game.

Whether one would want more squares to be hit or not is up for debate, but I see certain benefits.  For example, if more squares are hit during the game, then more people will have something invested in the game as it airs.  If you are sitting on the square that represents the current score, you want the score to remain the same through the end of the quarter so that you can reap the rewards – but if the winning squares keep bouncing around between a small number of people, there may be fewer people actively invested in the score as the game progresses.  This is especially true in Super Bowl parties, when many of the attendees are less interested in the game than they should be.

In other words, I’m of the belief that if more squares are hit, it’s a good thing.  It therefore becomes natural to ask whether or not the digital root pool actually does hit more unique squares than the traditional pool.  Thankfully, we have a wealth of data which we can use to answer this question.

I looked at all the games from this current season, and counted the number of boxes that would have been hit in each game using the traditional pool and the digital root pool.  Averaged over 331 games (this includes preseason and postseason), the number of squares hit using the traditional pool is approximately 6.84.  By comparison, the number of squares hit using the digital root pool is 8.43 – an increase of 1.59 boxes, or an increase of about 23%.  This effect is amplified when one considers the fact that the digital root pool uses only 81 squares, as opposed to the traditional pool’s 100.  This means that as a proportion of the total number of squares, the traditional pool hits about 6.84% of its squares, while the digital root pool hits 10.4% – here we have an increase of over 50%!

This is strong evidence that the digital root pool hits more squares than the tradition pool.  In fact, the data shows that an average game will have a change in the score approximately 8.73 times, which is only a bit higher than the average number of boxes hit by the digital root pool.  This makes sense when we slice the data another way: of the 331 games analyzed, in 252 of them the number of squares hit with the digital root pool was equal to the number of changes of score, meaning that no square got hit more than once.  The same cannot be said of the traditional pool – in this case, the number of games in which no square got hit more than once was only 62.

The data has convinced me that the digital root pool may be better suited for festive gathering, where wagering on football will be but one of many activities designed to induce merriment.  At the very least, it’s hard to argue that the traditional pool will hit as many squares as the digital root pool. Some may balk at a break from football pool tradition, but that’s ok.  I won’t watch football games with them anyway.

January 31, 2010 | Tags: , , , , | No Comments »

Meta-post

Hi everyone,

This is just a quick note to welcome you to the new Math Goes Pop!  We are still tweaking the look of the site, but hope you enjoy the changes.

If you haven’t already done so, I’d encourage you to subscribe to the RSS feed.  If you’re already subscribed, please check your feed URL, as it has now changed from the blogspot address.  The new URL is http://www.mathgoespop.com/feed.  Of course, you may find it easier to just subscribe via the link up top.  I encourage you to do so!

January 27, 2010 | No Comments »

e day?

If you come here regularly, you know of my complaints regarding so-called “math holidays” that get plenty of press, but rarely have anything to do with actual mathematics. The most well known is pi day, celebrated here in the states on March 14th, also known here as 3/14.

Aside from the mathematical arguments one can make for or against this holiday, there is a larger problem. It’s all well and good to celebrate pi day on the date representing the first three digits of pi, but this is only possible if we write dates in the MM/DD format. Most of the world, however, uses the (more logical) DD/MM format, therefore depriving them of such a delicious play on numbers.  Many loyal international fans of this holiday no doubt decry the fact that April has only 30 days, for otherwise they could simply celebrate pi day on 31/4. As it is, they are left with two options: Celebrate on 3/14 like those of us in the states, or enjoy a neutered version of this play on numbers by celebrating on 3/1.

Today I would like to propose an alternative to those for whom the DD/MM notation is standard. Rather than trying to work with imperfect solutions to the pi day problem, take a different number and celebrate it in your own way: the number e.

While e may not be as popular as its irrational sibling pi, it is no less important. No doubt many would argue that it is more important. It is certainly not as well-known in popular discourse, and so highlighting it, I would argue, is more important than highlighting the attention-whore known as pi.  Moreover, since the decimal expansion of e begins with 2.71828183…, countries that use the DD/MM format could celebrate e day today, January 27th.  Sadly, since February does not have 71 days, and since there are not 27 months in a year, people in America would be unable to celebrate in quite the same way – but given all the press that pi day has received over the past few years, I think that’s fair.

Of course, in order to celebrate the holiday properly, one needs activities.  Topics could include the ways in which this fantastic number arises naturally, or a discussion of exponential growth (and orders of magnitude in general).  One could also prove that e is irrational, a fact which follows quite easily from the Taylor series expansion of the exponential function ex at x = 1.  Perhaps I’m being overly optimistic though – such a holiday would probably include less exciting activities, such as a recitation of the decimal expansion of e to a certain number of digits (a mind numbing activity which is practiced without fail every pi day).

Special consideration needs to be given to a replacement for the act of eating pie, which seems like a suitable activity to do on pi day, but not on e day (especially since the surfaces of pies are circular).  I’m not sure what natural analogue exists – there is one thing that comes to mind when one wants to celebrate a day called “e day,” but I don’t want to promote drug use.  Perhaps instead one could eat foods that start with the letter e, like eclairs, eggplants, and elephants.  But these foods don’t work on a higher level, in that they don’t really relate to the number e in the way that the circular shape of a pie can be related to the number pi itself.

Eggs for e day?

There are obstacles to overcome, that much is certain.  But if we’re going to celebrate holidays related to math, we may as well do a halfway decent job of it.  So happy e day to you – don’t do anything I wouldn’t do.

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Judge v. Justices

Just as you can’t judge a book by its cover, it is not always easy to determine a person’s mathematical background based on his or her occupation. Sure, a burger flipper at McDonald’s may not look like the next Einstein, but how can you be sure she’s not just working a summer job to afford university? Conversely, just because someone is highly educated doesn’t mean he knows the difference between a prime and a composite number (although I’d argue that it should).
Case in point: Supreme Court justices may or may not know the meaning of the word orthogonal. Here’s a snippet from the oral arguments in the case of Briscoe v. Virginia (courtesy of blog The Volokh Conspiracy):

MR. FRIEDMAN: I think that issue is entirely orthogonal to the issue here because the Commonwealth is acknowledging -
CHIEF JUSTICE ROBERTS: I’m sorry. Entirely what?
MR. FRIEDMAN: Orthogonal. Right angle. Unrelated. Irrelevant.
CHIEF JUSTICE ROBERTS: Oh.
JUSTICE SCALIA: What was that adjective? I liked that.
MR. FRIEDMAN: Orthogonal.
CHIEF JUSTICE ROBERTS: Orthogonal.
MR. FRIEDMAN: Right, right.
JUSTICE SCALIA: Orthogonal, ooh.
(Laughter.)
JUSTICE KENNEDY: I knew this case presented us a problem.
(Laughter.)
MR. FRIEDMAN: I should have — I probably should have said -
JUSTICE SCALIA: I think we should use that in the opinion.
(Laughter.)
MR. FRIEDMAN: I thought — I thought I had seen it before.
JUSTICE SCALIA: Or the dissent.
(Laughter.)
MR. FRIEDMAN: That is a bit of professorship creeping in, I suppose.

While Friedman uses “orthogonal” in a bit of a metaphorical sense, this use is far from unprecedented – indeed, this use is even documented in the venerable internet database ubrandictionary.com, which defines orthogonal as a term that is “used to describe two things that are independent of one another. One does not imply the other.” Claiming that this usage is just a “bit of professorship” sounds a bit like a cop out. I wish Friedman had embraced it more completely.

In any event, the mathematical definition of orthogonal should be given in any halfway decent high school geometry course, if only as a synonym for perpendicular. The fact that Scalia and Roberts seem so unfamiliar with the concept is, at the very least, a little disappointing.

But all is not lost. On the other hand, last weekend Fox aired a special commemorating 20 years of The Simpsons, appropriately titled The Simpsons Anniversary Special: In 3-D! On Ice!. Several people contributed interviews to the special, including Mike Judge, creater of Beavis and Butthead and King of the Hill, among other comedic gems. Watch the clip below for a bombshell revelation:

That’s right – without The Simpsons, Judge believes he would be a math teacher. In fact, after doing some research online, I discovered that Judge didn’t begin playing with animation until the age of 26, while he was doing graduate studies in mathematics in the hopes of becoming a teacher.

Does this mean that Beavis and Butthead are smarter than Roberts and Scalia? Of course, some may cry out that this is an unfair comparison, but I think I can provide a fair answer.

Yes.
January 18, 2010 | Tags: , , , | No Comments »

A Mathematical New Years Game

First, let me begin by wishing a happy 2010 to you all. If you celebrate the holidays the way I do, then the past few weeks have seen you spending time with friends and family. And if you really celebrate the holidays the way I do, then some of that time with friends and family will have been spent with mathematical puzzles.

Very recently I was with a group of friends, discussing all that would come to pass in this new year. One friend, whose anonymity I will preserve by referring to him only as “Smith,” was in the enviable position of being the only one among us whose age divided the current year (I won’t embarrass him by revealing his age, but given that it’s a divisor of 2010, this certainly restricts the possibilities). Once we realized this, it became natural to ask how common an occurrence this should be. In other words, how often can you expect your age to divide the current year? Of course, implicit in this is a choice of calendar – for our purposes, we will stick to commonly used Gregorian calendar, although the results would be equally valid under a different choice (e.g. the Hebrew calendar or Islamic calendar). For example, if you were 1, 7, 41, or 49 last year, your age divided the year (of 2009). Next year, only the one year olds will win out, since 2011 is a prime number.

Depending on the year you were born, you may find that this happens quite frequently, or not very frequently at all. For example, if you were born in the year 0, you’re in luck, because your age will divide the current year for at least part of the year for every subsequent year. The phrase “part of the year” is important, because in a given year you will be two different ages – the age before your birthday, and the age on and after your birthday. Of course, this isn’t an issue if you were born on January 1st or December 31st, but we will ignore this (simpler) case.

Let’s take a more detailed example. Suppose you were born in 1982. In 1983, after your first birthday, your age will divide the year (since 1 divides everything). Similarly, in 1984, your age will divide the year after your 2nd birthday, since 1984 is even. And in 1986 your age will divide the year until your 4th birthday, since 1986 ÷ 3 = 662. Unfortunately, you will be too young to appreciate this arithmetic coincidence at any of these opportunities, and unless you live to be 661, you’ll never again be able to say that your age divides the year.

However, if you were born just a few years earlier, in 1979, you’ll find that your age divides the year quite frequently. In fact, by the year 2000, the only years in which your age wouldn’t have divided the year at all would have been 1987, 1988, 1993, 1994, 1996, 1997, and 1999.

Why is it that some years allow for one’s age to be divisible by the year quite frequently, while other years do not? The answer is quite simple. Suppose we let b denote the birth year, and we let a denote a person’s age. That person will be a years of age from their birthday in year b + a until their birthday in year b + a + 1. Therefore, your age will divide the year from your birthday until the end of the year if a divides b + a, or from the first of the year until your birthday if divides b + a + 1. So, the question becomes: when does a divide b + a, and when does it divide b + a + 1?

In the first case, since a always divides a, we know that a divides b + a if and only if a divides b. By the second same argument, we see that a divides b + a + 1 if and only if a divides b + 1. In other words, we conclude the following:

Your age will divide the current year if, and only if, either (i) it is between January 1st and your birthday, and your age divides the year after you were born, or (ii) it is between your birthday and December 31st, and your age divides the year you were born. To put it more simply, your age will divide the year for at least part of the time you are at that age if and only if that age divides the year of your birth or the year after your birth.

With this knowledge, it’s easy to see why people born in 1979 will have their age divide the current year more frequently than people born in 1982. In the former case, determining the set of ages which will divide the current year is equivalent to finding the divisors of 1979 and 1980. 1979 is a prime number, so it will never be the case that your age will divide the year between your birthday and December 31st (except after your 1st birthday); on the other hand, 1980 has a prime factorization of 2 x 2 x 3 x 3 x 5 x 11, which gives it a large number of small factors, and consequently a large number of solutions to the problem.

By contrast, if you were born in 1982, you won’t get many factors either way: 1982 factors as 2 x 991, and 1983 factors as 3 x 661. This is why, if you are born in 1982, your age won’t divide the current year after you’re 3.

While it’s not often that mathematics comes up when I’m with my friends at home, I certainly relish every opportunity. I hope that this may serve as an example to all of you who would like to make mathematics more of a part of your everyday life, especially in social circles into which math rarely intrudes. Single guys looking for first date conversation material are especially urged to keep this sentiment in mind.

January 10, 2010 | Tags: , , | 2 Comments »