## Solution to Super Bowl XLIX Problem

In the post When Super Bowl XLIX Starts to Bore You on January 29, you were asked to consider the following problem:

Imagine that the NFL has eliminated divisions, and there are just two conferences with 16 teams each. To simplify things, every team plays the other 15 teams in their conference exactly once each. At end of the regular season, what is probability that every team within a conference has a different record? (Assuming, of course, that each team is equally likely to win on any given Sunday, and assuming that ties are not possible.)

[UPDATE: Upon further review, it seems unlikely that you would have been bored during last night’s game. Unless you hate football and watched just for the commercials, which were terrible; or during the halftime show, which could be described as no better than fine; or during the post-game interviews, although if you’re like me, you were too busy retching over the Patriots’ victory to be solving a math problem just then.]

This is not a trivial problem, and a little exploration is good before you dive into a theoretical solution. Let’s start with a familiar heuristic, Solve a Simpler Problem.

If there were just two teams in the conference, one team would win the game they played. The winning team’s record would be 1‑0, and the losing team’s record would be 0‑1. That is, the records of the two teams would always be different, so P(different, 2 teams) = 1.

If there were three teams in the conference, it gets a little more complex, but not too bad. We can enumerate the outcomes of a season:

 A > B B > C A > C A > B B > C C > A A > B C > B A > C A > B C > B C > A B > A B > C A > C B > A B > C C > A B > A C > B A > C B > A C > B C > A A: 2-0 B: 1-1 C: 0-1 A: 1-1 B: 1-1 C: 1-1 A: 2-0 B: 0-2 C: 1-1 A: 1-1 B: 0-2 C: 2-0 A: 1-1 B: 2-0 C: 0-2 A: 0-2 B: 2-0 C: 1-1 A: 1-1 B: 1-1 C: 1-1 A: 0-2 B: 1-1 C: 2-0

Of the eight possible outcomes, six result in the three teams having different records. That gives P(different, 3 teams) = 3/4.

Unfortunately, the number of possible outcomes jumps pretty quickly. With four teams, there are 64  possible outcomes, and I don’t have the time or energy to list them all here. You’ll either have to trust me, or you can write them out yourself. Of those 64, it turns out that 24 of them result in different records for all the teams. So, P(different, 4 teams) = 24/64 = 3/8.

By this point, you may have noticed a pattern. Or maybe not; it’s subtle.

So let’s start thinking theoretically.

Pick one of the teams. The probability that the first team you pick wins all its games is (1/2)15.

Then pick a second team. This team has one loss guaranteed, since the first team was undefeated. So, the probability that the second team wins its 14 remaining games is (1/2)14.

Then pick a third team. This team has two losses guaranteed, and the probability it will win its 13 remaining games is (1/2)13.

And so on.

The next-to-last team has a probability of (1/2)1 of winning its one remaining game.

And the last team has a probability of (1/2)0 of winning its zero remaining games. Which is a little weird, admittedly, but I include it here for completeness.

Now the kicker: There are 16! ways to choose the order of the teams.

So the probability of 16 teams each finishing with a different record is

$16! \cdot \bigl(\frac{1}{2}\bigr)^{15 + 14 + 13 + 1 + \cdots + 0} = \frac{16!}{2^{210}} \approx 1.27 \times 10^{-50}$

Which is to say, it’s not impossible… but nearly so.

## When Super Bowl XLIX Starts to Bore You…

I appreciate that the Super Bowl unites all Americans in their inability to read Roman numerals.

(Caution — oxymoron ahead!) If you’re a smart football fan, then you’ve invited folks to your house to watch the game, so you don’t have to drive home drunk on a cold Sunday night in February.

If this year’s Super Bowl is like last year’s, your guests will be bored by halftime, with the Seahawks leading 22‑0.

Or if deflated balls are used, the Patriots could be leading 38‑7 at the end of the third quarter, like when they played the Colts two weeks ago.

In either case, you’ll need something to keep your guests entertained between commercials and after the Katy Perry halftime show. I suggest the following problem.

Imagine that the NFL has eliminated divisions, and there are just two conferences with 16 teams each. To simplify things, every team plays the other 15 teams in their conference exactly once each. At the end of the regular season, what is probability that every team within a conference has a different record? (Assuming, of course, that each team is equally likely to win on any given Sunday, and assuming that ties are not possible.)

It’s not a trivial problem, so I’ll give you some time to think about it. The answer will be revealed on Monday, February 2, 2015; that is, after the game.

## Popularizing Mathematics: ICME-13

In my professional career, I’ve had the pleasure to:

To say that I’ve been blessed in my professional career would be like saying that Leonhard Euler was pretty good at math.

I was blessed again recently, when former NCTM President Johnny Lott asked if I would lead a topic study group at the 13th International Congress for Math Education (ICME-13), which will be held July 24‑31, 2016, at Universität Hamburg, Germany.

Christian Mercat of University Claude Bernard Lyon 1 is my co-chair for Topic Study Group 7: Popularization of Mathematics.

I would like to invite you to be my guest at ICME‑13 and to attend our topic study group. In addition, I welcome any suggestions you have of mathematicians or math educators who popularize mathematics. We are seeking papers and presentations from a diverse, global panel, and any and all recommendations are greatly welcomed. Please post your suggestions in the comments, or email me at patrick@mathjokes4mathyfolks.com.

## Combinations in Deal a Story

Robert D. Reed Publishers — the publishing geniuses behind Math Jokes 4 Mathy Folks — recently published Deal a Story by Sue Viders, a card game in which writers deal themselves a “hand” to use as the basis for fiction writing. Think Richard Simmons’ Deal-A-Meal, but for fiction writers instead of fatties (for those of you old enough to remember).

The cover of the box boasts that there are “101 cards and 1,000,001 story ideas” contained within Deal a Story. But the math in that statement is typical marketing math — in the words of Cleone L. Reed, the promotional materials designer at RDR Publishers, “it sounds good, but I wonder how many combinations are really possible.”

So she asked if I could help her figure it out.

Here’s how Deal a Story works:

There are 101 cards:

• 16 hero cards
• 16 heroine cards
• 16 villain cards
• 16 character flaw cards
• 16 plot cards
• 16 genre cards
• 5 “wild” cards

Play happens as follows:

• A “player” (writer) chooses three character cards: 1 hero, 1 heroine, and 1 villain.
• For each character, a player chooses 1 character flaw card (so there is a flaw associated with each character).
• A player then chooses 1 plot and 1 genre card.
• And then finally, a player can choose 0‑5 wild cards, just to spice things up.

I think it’s a nice combinatorics problem, figuring out the number of different scenarios that result from dealing the cards. You’re probably not surprised to hear that there are more than 1,000,001 combinations — but how many more? I’ll leave that to you.

And a combinatorial joke for you:

A mathematician walks into McDonald’s and declares, “I’m so hungry, I could eat 5 burgers!”

His physicist friend says, “Oh, yeah, well I’m so hungry, I bet I could eat 10!”

The mathematician responds, “That’s absurd. There’s no way you could eat 3,628,800.”

## Making Heads or Tails of Randomness

Which of the four graphs below have a collection of truly random dots?

To put a little distance between the question and the spoiler below, let me interject with a couple jokes about randomness and probability.

From the ANZSTAT listserv:

If you choose an answer to this question at random,
what is the chance you will be correct?

1. 25%
2. 50%
3. 60%
4. 25%

Had enough? Okay, then, on with the show!

The 25 dots shown in each graph were randomly selected, but four different methods were employed:

• Completely Random: The coordinates of each dot were selected completely at random, independent of the other dots.
• Random-ish: The area was divided into a 5 × 5 grid of squares, and one point was then randomly selected within each square.
• Dartboard: I placed a 5″ × 5″ square on a dartboard and threw darts at it until I hit it 13 times. I then turned the square 90°; and threw darts until I hit it 12 more times. (I’m a reasonable dart player, but yes, there were quite a few darts that landed outside the square.)
• Eli’s Choice: I gave my seven-year-old son a 5″ × 5″ square and asked him to draw 25 random points.

Perhaps those descriptions will better help you determine which graph contains truly random dots. Care to take another look?

No? Okay.

Graph C was the most fun; it was Dartboard. I despise dumb-ass probability problems like, “What’s the probability that a randomly thrown dart will land inside the square and outside the circle?” What kind of person throws darts randomly? That’s just stupid, not to mention irresponsible. You’ll put someone’s eye out! Yet throw them randomly I did. And I’m happy to inform you that none of the walls around my dartboard were harmed in the making of this graph.

Graph A was Random-ish. Although there is some white space in the upper right and two dots are somewhat close in the bottom middle, generally the dots are well spaced.

Graph D was Eli’s Choice. Like Random-ish, Eli was rather careful to make sure that none of the dots were too close to one another. I divided the square into 25 unit squares after he drew his dots, and there was a single dot in 23 of them. There are some definite linear tendencies.

By process of elimination, you now know that Graph B was Completely Random. The function =5*RAND() was used to choose each coordinate for all 25 points in an Excel spreadsheet, and then they were graphed as a scatterplot.

So, what’s the point?

Graph D was generated by a seven-year-old, but it’s not terribly different from what would be generated by an average adult. What most people think of as random is usually anything but. Truly random data usually has clusters within it. For instance, flip a coin 200 times, and there is an 80% chance that there will be a run of at least 6 consecutive heads or 6 consecutive tails within your data.

For graph B, the clustering that happens in the upper left and with the two dots that touch in the lower right is completely typical when data is generated at random. As humans, we often try to ascribe some special meaning to these occurrences. For instance, when there are ten homicides in one week in a city that normally only has two homicides per week, headlines will declare “Murder On The Rise,” and the mayor will hold a press conference on what the police are doing to address the problem. But there’s nothing to be done. This kind of grouping is normal. A few weeks later, when there are zero homicides in one week — which is also typical — there will be hardly a notice.

If you didn’t correctly choose graph B, then perhaps you’re random-averse.

Or, more likely, you’re just normal.

## Number Words and Learned Helplessness

How about some number word puzzles? Here’s a well-known puzzle that you’ve likely seen before:

What is the first positive integer that, when spelled out, contains the letter a?

And here’s a modification of that puzzle that you may find a little more difficult:

What is the first positive integer that, when spelled out, contains the letter c?

And taking it one step further:

What letters are never used in the spelling of any positive integer?

Who says that math isn’t useful in English class?

One more problem in a similar vein:

Pick any positive integer you like, and count the letters when that number is spelled out. Now count the letters when the resulting number is spelled out. Continue ad infinitum. What do you get?

Maybe those weren’t your cup of tea. Perhaps anagrams are more to your liking, so here are two (related) puzzles for you.

Try to make an anagram for each of the following three words.

• whirl
• slapstick
• cinerama

Too tough? Then try these three words instead.

• bat
• lemon
• cinerama

If you had trouble with the first set, you’re in good company. There are no anagrams for the words whirl or slapstick.

These two sets of words were used by Charisse Nixon, a pyschologist at Penn State–Erie, who gave the first set of words to half her class and the second set of words to the other half. She instructed them to find an anagram of the first word on their list; those students who had received the second set were successful. Nixon then instructed them to find an anagram of the second word on their list; again, those students who had received the second set were successful. When she then instructed them to find an anagram of the third word on their list — of which there is exactly one, American — those who hadn’t found anagrams for the first two words were less successful than their peers, even though the final challenge was identical.

Afterwards, students who received the first set of words admitted to feeling confused, rushed, frustrated, and stupid.

Nixon was studying learned helplessness, a condition in which a person suffers from a sense of powerlessness, often arising from persistent failure.

This has implications the math classroom. Students who perform at a fourth-grade level but are asked to participate in an eighth-grade class are surely as confused and frustrated as the subjects in Nixon’s experiment. Students need to occasionally feel success, or else they’ll shut down. If you’re a teacher, you don’t need me or a psychological research study to tell you that. So the question is, how can you get students to feel success? That is, what can you do to prevent learned helplessness?

My suggestion is to look for acceptable and accessible entry points.

Consider the following problem, which might be seen in a middle school classroom:

What is the maximum possible product of a set of positive integers whose sum is 20?

As written, that problem contains three words — maximum, product, and integers — that may confound some students. For middle school students who do understand the terminology, finding an appropriate strategy might be daunting.

In my opinion, the following is a better way to present this problem so that all students have an entry point:

Find some numbers with a sum of 20. Now, multiply those numbers together. Compare your result with a partner. Whose result was greater? Can the two of you work together to find a product that’s greater still?

Even a struggling middle school student could start this activity. Surely he could find some numbers with a sum of 20. Certainly, he could multiply them without a problem.

Why is this a better presentation? The wording is simplified. There is encouragement to work with a partner. It feels more like a collaborative game than a traditional math problem. It sounds — dare I say it? — like fun.

When a struggling student is able to get into a problem, and they’re able to make some strides in the right direction, and they’re rewarded by your positive encouragement, they attain some level of success. Maybe they won’t solve the problem entirely, but who cares? For many students, trying is progress.

And for students who are having trouble finding any success, perhaps the following words of encouragement will help.

If at first you don’t succeed, call it version 1.0.

If at first you don’t succeed, destroy all evidence that you ever tried.

If at first you don’t succeed, blame someone else and seek counseling.

If at first you don’t succeed, then skydiving is not for you.

If at first you don’t succeed, get new batteries.

If at first you don’t succeed, try two more times so your failure is statistically significant.

## What I’ve Been Up To

To all of my friends who haven’t heard from me for nearly two years: I apologize. Sorry. I owe you a beer.

But I have a good excuse. I’ve been very busy trying to do something revolutionary.

In fact, I have a sign on my office window that explains what we’re trying to do.

I work for a great company.

We’re developing an incredible curriculum.

And I work with the most incredible math team ever assembled.

• Marjan Hong, who knows more about Common Core and effective teaching than anyone I’ve ever met.
• Peg Hartwig, who never met a piece of technology she didn’t love, and who just may be the greatest Algebra II teacher in the history of ever.
• Shelley Rosen, who’s a master of math representations, as evidenced by the Tallies, Ten Frames, and Baseball Games resource that she created for Illuminations.
• Brenan Bardige, who keeps it real by developing astounding, authentic, real-world problems for kids to solve (and he should know; his graduate research looked at assessment problems that kids would solve with a method other than the one the problem was trying to assess).
• Sia Robinson, who understands how kids learn math, and who reminds me daily how important it is for kids to learn math effectively.

I’d love to tell you all about Math Techbook… about all the amazing math tools we’ve built into the web-based curriculum… about the 200+ interactives we’ve woven through six courses… about the teacher dashboard and the inquiry-based approach we’re using and all the other good stuff… but there’s just no time! Tomorrow, we’re having a big release party. Really big. I mean, MASSIVE. But good news! You’re invited! Lots of cool people will be there, like Danica McKellar, who played Winnie Cooper on The Wonder Years but then got a Ph.D. in math from UCLA; Skip Fennell, past president of NCTM; Portia Wu, assistant secretary at the U.S. Department of Labor; Mark Edwards, superintendent in Mooresville, NC; and, Michele Weslander-Quaid, innovation evangelist at Google (and a very powerful woman).

As well as hundreds of our closest friends from districts all over the country. And thousands more will be joining us on the web.

Register Here – Math Techbook Launch Event

If you can’t make the launch event, I hope you’ll still check out what we’ve been up to.

And if you’re one of the people to whom I owe a beer, gimme a call on January 9.

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

## MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.