Math and the NBA All-Star Game

How awesome was Anthony Davis last night? In a word: very. He set a new All-Star Game scoring record with 52 points, adding 10 rebounds and 2 steals.

(Disclosure: I’ve liked A.D. since he was one-and-done at Kentucky. But it wasn’t until last night that I bought an Anthony Davis jersey:

Anthony Davis JerseyBand. Wagon.)

But as much as I like A.D., I couldn’t help thinking that there needs to be an asterisk next to this new All-Star scoring record. If you only look at points, sure, 52 > 42, so Davis scored more points last night than Wilt Chamberlain scored in the 1962 All-Star Game. But that’s only a part of the mathematical story.

Wilt ChamberlainFirst, let’s talk scoring percentage. In 1962, the final score of the game was 150‑130, meaning that Chamberlain accounted for 15.0% of all scoring. The final score of last night’s game was 192‑182, meaning that Davis accounted for 13.9% of all scoring. Chamberlain gets the nod, but only slightly, and I’ll admit it’s not insignificant that Davis only played 31 minutes last night, while Chamberlain played 37 minutes in 1962. So, maybe this is a push.

But let’s consider shooting percentage. Last night, both teams combined for 55.5% shooting, whereas in 1962, they managed just 43.8% shooting. Perhaps the all-stars from 50 years ago just didn’t shoot as well as players today? Actually, that’s somewhat true: The league FG% for 1961‑62 was 42.6%, the league FG% for 2016‑17 (so far) is 45.6%. But the all-stars last night were 9.9% above the league average, whereas the all-stars in 1962 were just 1.2% above their league average, suggesting that the defense in New Orleans was negligible at best. Which brings me to my next point…

Let’s talk defense. Maybe the combined 374 points that were Anthony Davisscored last night doesn’t convince you that defense was nonexistent. Then how about this: In the 1962 game, there were 62 personal fouls. Last night, there were only 16. Even more stark, though: In 1962, all-stars shot 95 free throws during the game; last night, they only shot 8. That’s not a typo, and it’s a pretty clear indication that no one was making much effort to contest shots.

Davis played a great game, but it doesn’t feel right that he unseats Chamberlain, given the circumstances. Not to mention, Chamberlain played a more complete game — shooting  73% from the field, grabbing 24 rebounds, and adding 1 assist.

This brings me to my final point, proportions. The teams last night scored 1/3 more points than their 1962 counterparts, and if you take away that extra third from Davis, he’d have ended the night with 39 points. So if an asterisk is good enough for Maris’s 61 and Flo-Jo’s 10.49, then it ought to be just fine for Davis’s 52, too.

But it is what it is. Congratulations, Anthony Davis.

Looking at the math of basketball is something I get to do quite a bit these days. Discovery Education has formed a partnership with the NBA, and we’re creating a collection of “problems worth solving” using NBA stats and highlight videos. Wanna see some of what we’ve done? Check out www.discoveryeducation.com/NBAMath.

February 20, 2017 at 9:01 pm Leave a comment

2 Good 2 Be True

I was eating a bowl of shepherd’s pie at the Irish pub in our neighborhood. A man walks up to my table and asks, “What’s your favorite number?”

“Uh, 153,” I respond.

“And 153 × 2 is 306,” he says, then hurriedly scurries away.

He approaches another table, asks another patron for her favorite number, and again multiplies it by 2. He does this over and over, popping from table to table, annoying customer after customer. Eventually, the manager notices this eccentric behavior and approaches the man.

“Sir,” says the manager, “You can’t keep interrupting people’s dinners by asking them for a number and then multiplying by 2.”

“What can I say,” he responds. “I love Dublin!”

A little while later, the gentleman at the table next to me says to his companion, “I know a sure-fire way to double your money.”

This piqued my interest, so I leaned over to eavesdrop on his advice.

“Fold it in half,” he said.

Dismayed that you’ve read this far and have only heard two terrible jokes? Well, buck up, because your fortune is about to change. I can’t help you double your money, but I can help you get twice as much for it.

Perhaps you’ve been wanting a copy of Math Jokes 4 Mathy Folks, but just haven’t pulled the trigger yet. Well, now’s the time. Robert D. Reed Publishers is offering a BOGO special for MJ4MF, so now you can buy a copy for yourself at regular price and get another for the special math geek in your life at no charge!

http://rdrpublishers.com/blogs/news/yes-math-is-fun

And check this out.

  • If you buy 2 copies, you’ll get 2 additional copies absolutely free!
  • If you buy 3 copies, you’ll get 3 more at no cost!
  • Buy 4 copies, and 8 copies will be delivered to your door!
  • And if you buy 50 copies? Why, you’ll have 100 copies arrive to your home, office, or post office box for the exact same price!
  • If you want n copies, you’ll only pay for n/2 of them!

Folks, this is a linear relationship that you’d be foolish to ignore!

January 16, 2017 at 10:56 am Leave a comment

Math Problem for 2017

Happy New Year! Welcome to 2017.

Here are some interesting facts about the number 2017:

  • It’s prime. (Okay, so that fact isn’t very interesting. But just you wait…)
  • Insert a 7 between any two digits of 2017, and the result  is still a prime number. That is, 27,017, 20,717, and 20,177 are all prime. (See? Told you it was gonna get better.)
  • The cube root of 2017 is approximately 12.63480759, which uses all ten digits 0‑9, and 2017 is the least positive integer that has this property. (Mind blown yet?)
  • The decimal expansion of 20172017 has 6,666 digits.
  • 2017 = 442 + 92
  • 2017 = 123 + 63 + 43 + 23 + 13 = 103 + 93 + 63 + 43 + 23

If you need some more, check out Matt Parker’s video.

Sorry, no video from me. But in honor of our newly minted prime year, I have created a problem for you to solve.

In the area model below (not to scale), the area of the five blue regions is indicated by the number inside the rectangle. What is the area of the yellow region with the question mark inside?

2017puzzle

Sorry, I don’t give answers. Feel free to have at it in the comments.

January 3, 2017 at 10:11 pm 4 comments

Predicting Your Child’s Height, and the Eventual Demise of Our Species

I’m between 6’1” and 6’4”, depending on which convenience store I’m leaving. My wife, on the other hand, doesn’t frequent convenience stores, because she can barely see over the counter, not to mention her extreme disdain for Slurpees.

slurpee

The discrepancy between our heights has its pros and cons. On the plus side, I can reach the pasta pot on the top shelf in the kitchen, and she barely has to bend over to retrieve the Tupperware containers from the bottom shelf. On the down side, we both strain our necks when kissing unless she’s standing on the second step.

Moreover, our sons were measured at 33” tall at their two-year check-ups, which puts them in the 20th percentile. This is when I realized that marrying a woman only 5’2” tall makes it unlikely that either of my boys will earn a spot on the varsity basketball team.

growth-birth-36-boys

Growth Chart, Boys: Birth to 36 Months

This was a distressing thought. If my sons continued to track at the 20th percentile, they’d reach a height of only 5’7” as adults. When I mentioned this to my mother-in-law, she attempted to allay my fears. “Kids always grow up to be twice their height at age 2,” she told me. That formula predicted their adult height to be 5’6”, which did not reassure me in the least, despite the fact that I didn’t believe it for a second. (Recently, I learned that the Mayo Clinic actually considers this a reasonable formula for predicting height, although they claim that it only works for boys. For girls, they suggest doubling the height at age 18 months.)

Growing up, the shortest person in my family was my oldest sister, who is 5’10” tall. Thinking that my sons might not even reach 5’8” was a major concern for me. As a result, I began to do some research. Although I knew I could not alter the course of their physical development, I had to at least know if they stood a chance at being above average in height.

As it turns out, predicting the adult height of a child is very difficult, and not just for kids whose growth has been stunted by wrestling or gymnastics. I found no fewer than a dozen formulas in widespread use, yet none of them were very reliable. Which is to say, none of them gave the results I was looking for. I spoke with multiple pediatricians who said the best predictor of a child’s eventual height is the mean of the mother’s and father’s heights. That didn’t sit well with me, though, partially because it just seemed too easy, but mostly because it said my sons would be 5’8” tall at adulthood. Still shorter than I’d hoped, so I looked some more.

The Mayo Clinic suggests the following method to predict boys’ height:

  • Add the mother’s height and the father’s height.
  • Add 5 inches to that sum.
  • Divide by 2.

This formula may not be superior to the “double height at age 2” formula above, but at least it felt more mathematically rigorous. Sadly, it predicted that my sons would reach a height of 5’8½“, which is better but still not what I was looking for. So, my search continued.

After several months of research, I finally stumbled on a formula that I thought I could trust. It began like many others, taking the mean of the mother’s and father’s height. But then it compensated for height differences due to gender. For boys, multiply the mean by 13/12; for girls, multiply the mean by 12/13. That is,

Boys: \frac{13}{12} \times \frac{m + f}{2}

Girls: \frac{12}{13} \times \frac{m + f}{2}

where m is the mother’s height and f is the father’s height. It roughly means that a boy has an extra inch added to the prediction for every foot in the mean, whereas a girl has an inch subtracted.

Finally, I had stumbled on a formula that predicted my sons would be above average, stretching the tape to just shy of 6’2” tall. This was a win-win, too — this formula predicted that they would be tall, but not quite as tall as their dad.

What’s interesting about this formula is its long-term prognostication. It implies that the human population will continue to increase in size indefinitely. Let me explain.

Google says that the average man is 5’6” tall and the average woman is 5’2” tall. This means that the average human is approximately 5’4”, or 64”, tall. Then an average man and average woman would have an average son who is 69.33” tall and an average daughter who is 59.07” tall — according to the formulas above — and the average human would be 64.20” tall. When this next generation has children, the sons would be 69.55” tall and the daughters would be 59.26” tall, and the average human height would be 64.41” tall. Do you see what’s happening? The average height is increasing slightly with each generation.

It’s not hard to see why this happens. The average of 13/12 and 12/13 is

\frac{1}{2} \cdot (\, \frac{13}{12} + \frac{12}{13}) \, = \frac{313}{312} \approx 1.003

which means that the average height will increase by 0.3% with each generation. Admittedly, that’s not much, but it portends almost certain doom! By Y3K, the average human will be almost 6’3” tall; and by Y4K, the average human will be nearly 7’3” tall.

There is some historical data to suggest that human height has been increasing for quite some time. In the early 18th century, researchers found that the average English male was 5’5” tall; today, the average English male is 5’9” tall. That’s an increase of about 0.6% per generation. Further, archaeological research shows that the average Greek woman was about 153 cm tall, and the average Greek woman today is closer to 165 cm, an increase of about 0.07% per generation.

HelenaPaparizou.pngShould we expect humans to someday stretch the tape to 8’ or more? Probably not. The paragraph above is a little lesson on how to lie with statistics. While it may be the case that the modern Greek beauty Helena Paparizou (5’7”) is taller than ancient beauty Helen of Troy (5’6”), it hasn’t necessarily been a steady increase. The average female height of a Greek woman in the 10th century was over 162 cm; just two centuries later, the average height dipped below 160 cm.

While the formulas above may serve to make reasonable predictions now, they may not work so well in the future. So if you’re 5’4” tall, don’t be bummed if your great-great-great-great-great-great-grandchild still isn’t able to dunk.

Alex and Eli are now 9 years old, and both of them are 4’3” tall. I strongly suspect that neither of them will reach 6’0”, and that’s just fine. My wife may not be tall, but she is kind, smart, funny, and compassionate, traits that she has generously shared with her children. As a result, my sons are well above average in ways that actually matter.

December 17, 2016 at 4:36 pm 2 comments

It’s Not What’s on the Outside…

Through the Academic and Creative Endeavors (ACE) program at their school, my sons participate in the Math Olympiad for Elementary and Middle School (MOEMS). While passing the door to the ACE room yesterday, I noticed a sign with the names of those who scored a perfect 5 out of 5 on the most recent contest — and my sons’ names were conspicuously absent. Last night at dinner, I asked Alex and Eli what happened, and they told me about the problem that they both missed. (What? Like we’re the only family in America that discusses math problems at the dinner table.) Here’s how they explained it to me:

Nine 1-cm by 1-cm squares are arranged to form a 3 × 3 square, as shown below. The 3 × 3 square is divided into two pieces by cutting along gridlines only. What is the greatest total (combined) perimeter for the two shapes?

3x3 Grid

The answer to this problem appears below. Pause here if you’d like to solve it before reaching the spoiler.

[Ed. Note: I didn’t see the actual exam, so the presentation of the problem above is based entirely on my sons’ description. Apologies to MOEMS for any substantive differences.]

This problem epitomizes what I love about math competitions.

  • The answer to the problem is not obvious. This is the case with many competition problems, unlike the majority of problems that appear in a traditional textbook.
  • The solution does not rely on rote mechanics. Again, this differentiates it from a standard textbook problem or — shudder! — from the problems that often populate the databases of many skill-based online programs.
  • Students have to get messy. That is, they’ll need to try something, see what happens, then decide if they can improve the result.
  • Students have to convince themselves when to stop. Or more precisely, they’ll need to convince themselves that they’ve found the correct answer. For instance, let’s say a student divides the square into a 3 × 2 and a 3 × 1 rectangle. The combined perimeter is 18 units. Is that good enough, or can you do better? This is different from, say, a typical algebra problem, for which students are taught how to check their answer.  3x2-and-3x1-rectangles

The problem also epitomizes what many people hate about math competitions.

  • There’s a time limit. Students have 26 minutes to solve 5 problems. Which means that if students spend more than 5 minutes on this one, they may not have time to finish the other four. (There was a student of John Benson who, when asked about his goal for an upcoming math competition, replied, “I hope to solve half the problems during the competition and all of them by the end of the week.” That’s the way mathematicians work.)
  • It’s naked math. Sorry, nothing real-world about this one. (But maybe that’s okay, because real may not be better.)
  • The problem is presented as a neat little bundle. This is rarely how mathematics actually works. True problems often don’t present themselves all at once; it’s through investigation and research that the constraints become known and the nuances are revealed.

All that said, I believe that the pros far outweigh the cons. Benjamin Franklin Finkel said, “Many dormant minds have been aroused into activity through the mastery of a single problem.” I don’t remember the last time a mind was aroused by the solution to x + 7 = -3, but I’ve witnessed awakenings when students solve problems like the one above.

And here’s the tragedy in all of this: Many teachers believe that only kids who participate in math competitions can handle — or appreciate — math competition questions. No! Quite the opposite, in fact. Students who have tuned out have done so because they’ve never been challenged and, worse, have never felt the thrill of solving a problem on their own.

What I really love about the problem, though, is it made me think about other questions that could be asked:

  • How many ways are there to divide a 3 × 3 square into two pieces that will yield the maximum total perimeter?
  • What is the maximum total perimeter if a 3 × 3 square is divided into three pieces?
  • What is the maximum total perimeter if a 4 × 4 square is divided into two pieces? …a 5 × 5 square? …a 6 × 6 square?
    • The answer to the problem about the 4 × 4 square appears below. Pause here if you’d like to solve it before reaching the spoiler.
  • (wait for it) What is the maximum total perimeter if an n × n square is divided into two pieces?

It was that last question that really got the blood pumping.

Here’s a solution for how to divide a 3 × 3 square to yield the greatest total perimeter:

3x3solution

And here’s a solution for how to divide a 4 × 4 square to yield the greatest total perimeter:

4x4solution

What’s the solution for an n × n square? That’s left as an exercise for the reader.

December 5, 2016 at 5:32 am Leave a comment

NBA, Discovery, and the Math of Basketball

Last week, Discovery Education and the National Basketball Association (NBA) announced a partnership in which real-time data from stats.nba.com will flow into Math Techbook, and students will use that data to solve problems.

How cool is that?

Eighty students from John Hayden Johnson Middle School in Washington, DC, participated in the event, which was emceed by Hall-of-Famer “Big” Bob Lanier and made silly by Washington Wizards mascot G-Wiz.

assembly_jrnba_discovery

G-Wiz and students at John Hayden Johnson Middle School in Washington, DC, at a joint event of the NBA and Discovery Education.

The event received a lot of press coverage, and as you may have heard, there’s no such thing as bad publicity. But one of the articles quoted me as saying:

It’s not like a beautiful, traditional math problem.

That is not what I said. I am absolutely certain that I have never used the words beautiful and traditional in the same sentence. Well, perhaps when referring to a wedding dress or an Irish cottage, maybe, but certainly not when referring to a math problem.

I was also quoted as saying:

It’s going to be messy, for sure.

That is, in fact, one of the things I did say. Because by definition, good math problems are messy. For this project, our writing team created problems that don’t have one right answer. For instance, one problem asks students to generate a formula to predict which players should be on the All‑NBA 1st Team. Should they use points and rebounds as part of their formula? If so, how much weight should they give to each? And should there be a deduction for the number of turnovers a player has? All of that is up to the student, and it’s certainly possible that more than one formula would give reasonable results. (If you don’t believe me, do a search for NBA Efficiency, TENDEX, Thibodeau, VORP, or New SPM to get a sense of some formulas currently used by professional statisticians.)

A microsite with a four sample problems is available at www.discoveryeducation.com/nbamath. To see all 16 problems and to experience the NBA Math Tool, you’ll need to login to Math Techbook; sign up for a 60‑day trial at www.discoveryeducation.com/mathtechbook.

I’m ecstatic about the problems that our writing team — which includes folks who love both math and basketball, like Brenan Bardige, Ellen Clay, Chris Shore, Shauna Hedgepeth, Katie Rhee, Jen Silverman, and Jason Slowbe — has created. One of the simpler problems they’ve written, meant for middle school students, is to determine which player should take a technical free throw. It’s not a hard problem, but students get to choose which team(s) to examine and how to use free-throw data to make their choice. With the NBA Math Tool that we’ve created, which includes FTM and FTA but not FT%, one possible formula is =ROUND(100*FTM/FTA,1), which will display the free-throw percentage to one decimal place of accuracy — though there are certainly less sophisticated formulas that will get the job done, too, and students could bypass formulas entirely by using equivalent fractions.

But a different article said that the “questions may look something like” this:

Andrew Wiggins is making 49.1% of his two-point shots and 52.3% of his threes. Which shot is he more likely to make?

Actually, we would never ask a question like this in Math Techbook, either as part of this NBA project or otherwise. By the time students start working with percentages in middle school (6.RPA.3.C), we expect that they already understand how to compare decimals (4.NF.C.7). Though basic exercises are included in the service, most problems — and especially those based on NBA data — exist at a greater depth of knowledge.

But what we might do is ask students to use proportions to make a prediction.

As you know, basketball announcers and sportswriters make predictions all the time. They talk about players being “on pace” to score some number of points or to grab a certain number of rebounds. In fact, the Washington Post recently prophesied that Steph Curry will hold the NBA’s all-time three-point record before the next presidential election.

During the first part of the event at Johnson Middle School, the students set out to make a prediction:

How many assists will John Wall finish the season with?

John Wall recently set the Wizards franchise record for assists, so the context was timely.

To solve this problem, students explored the NBA Math Tool, which now resides inside Math Techbook. This tool allows students to analyze both NBA and WNBA stats. Students considered data for the Washington Wizards:

Wizards Stats - NBA Tool

Row 6 shows that John Wall had 98 assists through 11 games. Good information, to be sure, but it led to more questions from students than answers:

  • Some players on the Wizards have played 13 games. How many games have the Wizards played so far this year?
  • How many games will John Wall play this year?
  • How many games are in an NBA season?

Looking at team data in the NBA Math Tool, students learned that the Wizards have played 13 games so far this year. And one student knew that every NBA team plays 82 games in a season. Good info… but now what?

One approach is to set up a proportion with the equation

\frac{11}{13} = \frac{g}{82}

which yields the number of games (g) that we can expect John to play this year (69), and then the equation

\frac{11}{69} = \frac{98}{a}

can be used to find the number of assists (a) that we can expect John to record (602).

But the eighty students in the gym were sixth- and seventh-graders, and they weren’t ready for algebraic equations. Instead, they attacked the problem by noting that Wall had 98 assists through the first 13 games, so they estimated:

  • He should have about 200 assists through 25 games.
  • He’d have about 400 assists through 50 games.
  • He’d have about 600 assists through 75 games.
  • That’s 7 games shy of a full 82‑game season, and Wall should have about 50 assists in 7 games.
  • So, we can expect him to finish the season with about 650 assists.

My role at the event was to lead students through the solution as a group-problem solving activity; and then, to work with them in the media center on the free-throw problem described above. It was an incredible day! I got to co-teach with Ivory Latta, point guard for the Washington Mystics:

Latta Vennebush Math

I got to meet some incredible people, including current players, former players, and NBA executives:

NBA Launch Event

(L to R) Etan Thomas; Felipe Lopez; Patrick Vennebush, MJ4MF and Discovery Education; Bill Goodwyn, President and CEO of Discovery Education; Ivory Latta, Washington Mystics point guard; Elizabeth Lipscomb, Discovery Education; Todd Jacobson, Sr. VP of Social Responsibility, NBA; and “Big” Bob Lanier, Hall-of-Fame player for the Detroit Pistons and Milwaukee Bucks.

But most importantly, I was finally able to let the world know about this amazing project, which my team has been working on for a year.

The NBA slogan is, This is why we play. But today I say, this is why we work: to develop rich curriculum resources that are fun, relevant, and powerful in teaching kids math.

#mathslamdunk

November 26, 2016 at 6:24 am Leave a comment

Fractional Fun

It’s well known that 5 out of 4 people have trouble with fractions, but even the mathematically advanced may have a little trouble with this puzzle. Your challenge is simple…

Find the sum of all items in the following table.

A hint is below the table, and the answers are below that. Good luck!

fishbird
woodenfortwoodenfort
whiskeybottlewhiskeybottlewhiskeybottlewhiskeybottle
sith
wraithwraithwraithwraith
tenttent

 

Hmm… it seems that you scrolled down here a little too quickly for the hint. Try harder. To put some distance between you and the hint, here are some fraction jokes:

How is sex like a fraction?
It’s improper for the larger one to be on top.

Which king invented fractions?
Henry the Eighth.

There’s a fine line between the numerator and denominator.
(And it’s called a vinculum.)

Okay, you’ve waited long enough. Here’s your hint. The items in the table are a fird (fish + bird), wooden forts, bottles of whiskey, the Sith lord Darth Maul, wraiths, and tents. (By the way, thanks to www.HikingArtist.com for the cool drawing of the fird!) Hope that helps.

To put some space between the hint and the answer, here are some more fraction jokes:

A student once told me, “To prove to you that I understand equivalent fractions, I only did three-sevenths of my homework.”

I was scared half to death… twice.

What is one-fifth of a foot?
A toe.

Okay, you’ve waited (and endured) enough. Without further adieu, the answer is 2.5. The images in the table are:

  • one fird
  • two forts
  • four fifths
  • one Sith
  • four wraiths
  • two tents

the sum of which is

\frac{1}{3} + \frac{2}{4} + \frac{4}{5} + \frac{1}{6} + \frac{4}{8} + \frac{2}{10} = \frac{300}{120} = 2\frac{1}{2}.

You’re welcome.

Thanks for stopping by. Have a great day!

November 11, 2016 at 11:11 am Leave a comment

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About MJ4MF

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.

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