Category Archives: Mathematics

Penny is for Penalize

The elementary school where my boys attend is hosting a fund raising contest where students bring their change to donate. Instead of accumulating donations to win for their own grade, they are putting their change in other grades canisters. The grade with the lowest weight in donated change at the end of the fund-raising wins. Pretty innovative, I think. The kids seem motivated, so fortunately for the school they are not colluding to all bring nothing which would keep their weights down. Instead they are piling in the change especially those with siblings in other grades.

Strolling through the halls, I overheard a conversation where teachers were wondering if anyone had found out which coins weighs the most so they could give more of those.  Of course, that got me thinking.  The obvious answer would be that the larger coins like the half-dollar or presidential dollar would be the heaviest.

Although, don’t you really want to have the most weight for your money?  Sure, I could put in 10 presidential dollar coins, but which would weigh more, 10 dollars in half dollars, 10 dollars in quarters, or 10 dollars in dimes?

So, of course, I had to know the answer.  Checking out the U.S. Mint, I learned the following weights for each of the coins:

1 penny = 2.5 g  which means $1 in pennies is 250 g.
1 nickel = 5 g which means $1 in nickels is 100 g.
1 dime = 2.268 g which means $1 in dimes is 22.68 g.
1 quarter = 5.670 g which means $1 in quarters is 22.68 g.
1 half-dollar = 11.340 g which means $1 in half-dollars is 22.68 g.
1 dollar (coin) = 8.1 g which means $1 in a single coin is 8.1 g.

In spite of the larger weights for the larger coins, you are still much better off dumping in those pennies.  I did learn a pretty interesting fact though: A dollar in dimes weighs the same as a dollar in quarters which also weighs the same as a dollar in half-dollars.  Pretty cool!

Second Chance Exam

imageHave you ever heard of a “second chance exam”?  I came across the concept for the first time in an article at Faculty Focus, Revisiting Extra Credit Policies.

Here’s how the author explains it:

The instructor attaches a blank piece of paper to the back of every exam. Students may write on that sheet any exam questions they couldn’t answer or weren’t sure they answered correctly. Students then take this piece of paper with them and look up the correct answers. They can use any resource at their disposal short of asking the instructor. At the start of the next class session, they turn in their set of corrected answers which the instructor re-attaches to their original exam. Both sets of answers are graded. If students missed the question on the exam but answered it correctly on the attached sheet, half the credit lost for the wrong answer is recovered.

I currently have a standing policy in all of my classes that allow students to correct missed problems on a test after it has been graded.  They’ll receive a bonus point on their exam grade for every correctly revised problem.  Instead of a flat bonus, this gives the most reward to students who put in the most work in the corrections. 

I used to do a flat 10 point curve for corrections.  At one point I was having students hand a test notebook at the end of the term.  The notebook contained corrected versions of their tests and they were rewarded with 3 bonus points on the their final average.

I’m considering trying this new approach, the “second chance exam” because it requires students to assess what they know, put in the work of correcting a problem and it also reduces the amount of time it takes to get a final grade into the grade book.  Right now, students take a test, then get it back the next class, then turn in corrections after that, and then I eventually return their corrections.  This new way, I collect the second chance exam the next class after the exam and then return the fully graded exam after that.

Of course, a sizeable percentage of fellow faculty would probably argue that extra credit only encourages laziness and procrastination on the part of the students but if the opportunities can be manipulated into a learning experience, isn’t that better than not learning at all?

CAMT 2011 – Our Nerd Vacation


My wife, Lori, and I have come to Grapevine, TX to attend one of the largest state-level mathematics conferences in the U.S.  We are at the Conference for the Advancement of Mathematics Teaching (CAMT) which is a joint conference held by the NCTM, MAA and TASM.  (Google them if you really have to know what those acronyms stand for)

The conference is primarily for K-12 math teachers but there are few like me here that participate or supervise in the area of teacher education.  I’m here to learn two things, better techniques to teach our up and coming teachers and changes coming due to STAAR and the EOCs.  (The tests replacing the current TAKS.)

The venue is the Gaylord Texan Resort and Convention Center in Grapevine, Texas.


We have a beautiful room with an excellent view of the atrium.

IMG_1180 Stitch


Now this is more for me as a reminder than for the readers but here are the talks I have attended so far:


Day 1 (Monday, July 18)

First Timer’s Session (8:00 – 9:00)

Algebra I Activities With The TI-NspireTM Handheld – Andi Parr, Region 13 ESC (9:15 – 10:15)

The Role of Inquiry Teaching Methods in Secondary Mathematics Classrooms – Mark Daniels, University of Texas at Austin (10:30 – 11:30)

Ignite Session! – Tim Pope, Key Curriculum Press (11:45 – 12:45)
Note: This session had a unique format.  Nine speakers each had 5 minutes to present.  They each had 20 powerpoint slides that would advance ever 15 seconds automatically.  This year’s speakrs included; Pam Harris, Paula Moeller, Michelle King, James Epperson, Amber Branch, Emma Trevino, Cindy Schneider and Cindy Schimek.

Exploring AP Caluclus Activities with the TI-NspireTM – Noe Medrano (3:30 – 4:30)


Day 2 (Tuesday, July 19)

Small Group Instruction in the Secondary Classroom – Richard Yoes, Joda Mendoza, Pasadena ISD (8:00 – 9:30)

The New TI-Nspire Navigator SystemTM – Holly Larson, McKinney ISD (9:15 – 10:15)

Big Gains from Small Struggles – Cathy Seely, Charles A. Dana Center (10:30 – 11:30)

Math Curriculum Makeover – Dan Meyer, Author (11:30 – 1:00)


…more to come later

Arcs in a Square (or Snakes on Plane)

image In a recent MAA publication, Shai Simonson, attempts to bring the joys and excitement of the world of mathematics to the non-technical reader.  In Rediscovering Mathematics: You Do the Math, Simonson covers a wide array of topics ranging from number theory to the application of probability in sports, casinos and gambling.  I have added the book to my reading list and you might want to take a look at the article that begins his text, “How to Read Mathematics.”

One of the problems from the book was posted over at Math Mama Writes… and when a puzzle like this piques my interest, I’m at its mercy until I figure it out.  Thanks to a recent illustration I made in calculus last week and a cartoon that reinforced my perspective, I have a new motto for next year’s courses:

Math problems aren’t solved, they are conquered!

Well, this problem below was one that I had to defeat.  I went to battle with it and after losing a few skirmishes (i.e., trying approaches that failed) I finally beat it into submission.  From now on, when I see that feisty integral that won’t behave or a simple number puzzle whose pattern defies identification, I’ll strap on my armor (or sweater vest), grab my sword (or calculator) and wage full-out war on that problem.  No problem is safe!

Arcs in a Square (or Snakes on a Plane)

Given the square ABCD, with side length 4 and circular arcs centered at each vertex, find the area of the region at the center – without using calculus.

And by the way, the Snakes on a Plane reference is my own and I’ve not ever heard this problem referred to in this way but it sounded good to me (arcs [tex]approx[/tex] snakes, square [tex]approx[/tex] plane).  However, I am a strange duck.

I’ll post a couple of different approaches that conquered this problem in later posts.

The Locker Problems Solved

imageA few days ago, I posted a couple of problems that I had stumbled across over at Math Jokes for Mathy Folks.  If you haven’t already, take a moment to read the puzzles:  The Locker Problems.

Problem 1

To start with, we can simply begin walking through process by hand.  Locker 1 will begin closed, then the 1st person will come in an open it.  After that, no one touches the 1st locker so we know it stays open.

Locker 2 is opened by the 1st person, closed by the 2nd and it stays closed.  Locker 3 is opened by the 1st person, left alone by the 2nd and then closed by the 3rd.  This is going to get tedious if I keep explaining in words.  Lets use a table.  If a locker is open, I’ll put a “o”, if it’s closed I’ll put a “c”.



This is also going to get tedious if I want to do this for 1000 lockers and 1000 students.  We definitely need to find a pattern. 

My next step was put together a lazy little Matlab script to do the same exact thing as my table. (See here.)  Using the “spy” function in MATLAB allows me to see the sparsity pattern of the matrix.  The open lockers are represented by 1’s and the closed ones by 0’s.  Here’s what it looks like for 100 lockers and 100 students.


There is definitely a pattern!  Notice those open lockers form the bands you see in the picture.  So why do some end up open and some end up closed?

The easiest way to see why is to consider what happens to a single locker.  For example, think about locker 24.  When does it change state?  Obviously, person 1 opens the locker, person 2 closes it, person 3 opens it, person 4 closes, person 5 does nothing, etc.  Notice that if the locker number, 24, is divisible by the person number, then the state changes:


The number 24 has 8 factors, that is, 8 numbers that divide evenly into it.  So the key is that any locker number with an even number of factors will end up with closed and any with an odd number will end up open. So what numbers have an odd number of factors?

At first thought, it seems that all numbers should have an even number of divisors since they always occur in pairs: 

24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6

Consider a number like 36, though.

36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 6 x 6 <—- Ah ha!!

Perfect squares are the only numbers that can have an odd number of divisors.  So locker numbers which are perfect squares (i.e., 1, 4, 9, 16, 25, …) will remain open while the rest stay closed!! That matches exactly with the table above and the graph from MATLAB.

All we need to know is how many perfect squares are less than 1000. 

[tex]sqrt{1000} approx 31.6222[/tex]

ANSWER:  There will be 31 open lockers

Problem 2

In the second version of the problem, we have only 30 students and some are absent.  In fact, we need to determine which ones were absent so that all the lockers except the first on remain closed.

Starting in a similar fashion as before, we could build a table by hand and determine who would have to be sick.  For example, the 1st student is clearly present since the first locker remains open.  After that, the 2nd and 3rd are both present since those lockers end up closed.  The 4th person must be absent since if they were present then the 4th locker would be open. 

So any locker number that has an odd number of factors (only counting factors if that person is present) must not be there.  For example, consider the 8th locker.  The number 8 has 4 factors but one of them is 4 so it actually only has 3 factors when counting those present.  So 8 must be out sick. 

Following this pattern, we eliminate all multiples of 4.  Then when we get to the 9th locker, they will also absent.  So will all multiples of 9.  Continue this pattern until you reach 30.


The numbers that are not shaded are called squarefree since they have no repeated factors when you they are prime factored.  (Read more about squarefree numbers.) The absent folk are the "non-squarefree” numbers, also known as squareful.

ANSWER: There are 11 squareful numbers less than 30 so 11 students are absent.

That was fun!

The Locker Problems

image Problem 1

Every day, 1000 students enter a school that has 1000 lockers. All of the lockers are closed when they arrive. Student 1 opens every locker. Student 2 closes every other locker. Student 3 then “changes the state” of every third locker – that is, he opens it if it’s closed, and he closes it if it’s open. Student 4 then changes the state of every fourth locker, Student 5 changes the state of every fifth locker, and so on, so that Student n changes the state of everynth locker.

Which lockers are open after all 1000 students have finished opening and closing lockers?


Problem 2

Every day, 30 students enter a room with 30 lockers. All of the lockers are closed when they arrive. Student 1 opens every locker. Student 2 closes every locker. Student 3 then “changes the state” of every third locker — that is, he opens it if it’s closed, and he closes it if it’s open. Student 4 then changes the state of every fourth locker, Student 5 changes the state of every fifth locker, and so on, so that Student n changes the state of every nth locker.

One day, some students are out sick. Regardless, those present repeat the process and just skip the students who are absent — for instance, if Student 3 were absent, then no one would change the state of every third locker.

When they finish, only Locker #1 is open, and the other 29 lockers are all closed. How many students were absent?


HT: Math Jokes for Mathy People

Solutions have been posted:  The Locker Problems Solved

Hunting for Simpson’s Paradox (part 2)

(If you haven’t already, you should read part 1 of this article which was posted on April 13, 2011)

Baseball & Glove Over WhiteSimpson’s paradox illustrates an intuitive error in reasoning that’s hard to accept without credible examples.  It occurs when a correlation or trend that is present in groups is reversed when the groups are combined.  The example we’ve been working with deals with batting averages.  A hypothetical situation was presented in the first part of the article in which we stated that one batter had a higher batting average against left-handed pitchers and against right-handed pitchers, separately.  However, when the overall batting average was calculated, that batter had a lower average.

Here’s the table once more just for reference:


Certainly it was an interesting example but it was entirely made-up.  I just tweaked the numbers a bit to get the table to work.  I set my sights on finding some of my own examples of Simpson’s paradox based on REAL data. 

Attempt 1

Yahoo Sports provides some easily accessible baseball team split stats, e.g., home vs. road, pre-all star break vs. post-all star break, turf vs. grass, indoor vs. outdoor (  I started by pulling the overall team batting averages and then their averages split by home games versus road games (for the 2010 season).  By pasting into Excel and sorting, then manually checking for a good 15 minutes, I was able to establish that there was no example of Simpson’s paradox in this data set. Bummer.

Attempt 2

This process had to be automated because there was no way I was going to spend 15+ minutes checking every split, every season, until I found an example.  While this could be fairly easily done with a VBA script (e.g. macro) in Excel, I decided to move over to MATLAB simply because I’m more well-versed in scripting there.  Algorithmically speaking, it really is a piece of cake to automate.

NOTE: If the details of the script don’t interest you, you can skip down to the results.  And yes, I found some examples.  WOOHOOO!

First, you need your data read into the program.  In Excel, I had a table with the first column as the team names, the second column was the overall batting average and the third and fourth column were the split stats (i.e., team batting averages for road vs. home).  To move this into MATLAB, I created a cell array called teams which would contain team names.

teams = {}

Then, after opening the teams variable in the variable editor window (double-click it in the Workspace), I pasted in the team names (right-click, “Paste Excel Data”).  Next, I created a variable x and pasted in the three columns of batting average data.

At this point a very simple little script (simphunt.m) tests every pair of teams for Simpson’s paradox.  The main step of the script is to compare the overall averages for each pair and the split stats for each pair.  If the relation of the split is reversed in the overall, then we have an example.

if (x(i,1)>x(j,1) && x(i,2)<x(j,2) && x(i,3)<x(j,3)) || …
   (x(i,1)<x(j,1) && x(i,2)>x(j,2) && x(i,3)>x(j,3))


The first split I tested was, again, home vs. road (to which I already knew the answer):

There are no cases of Simpson’s Paradox

The next was day vs. night.  Disappointed again.

There are no cases of Simpson’s Paradox

Fortunately, it was only taking a minute or so to fetch the data and move it into MATLAB.  Nevertheless, I was still feeling a little disheartened.  I was beginning to expect to run through dozens of splits, maybe from multiple seasons, or even moving on to player vs. player stats instead of teams.

Then I went with indoor vs. outdoor.  JACKPOT!!!

There are 5 cases of Simpson’s Paradox

‘ Chicago Cubs’          ‘ Oakland Athletics’
‘ Cleveland Indians’     ‘ Tampa Bay Rays’  
‘ Colorado Rockies’      ‘ Milwaukee Brewers’
‘ Los Angeles Angels’    ‘ Tampa Bay Rays’  
‘ New York Mets’         ‘ Toronto Blue Jays’

Five Cases!!!  Looking back at the data we can confirm:


I could clearly keep hunting but I’ve satisfied my desire for finding my own examples.  Next time I bring up Simpson’s paradox in any of my classes, I’ll start with my manufactured example but quickly move on to these REAL examples.

Read More…
Find out more about Simpson’s Paradox at the following links:

Simpson’s paradox – Wikipedia, the free encyclopedia
When Combined Data Reveal the Flaw of Averages

Unboxing the Casio EX-ZR100

Here at the School of Math and Sciences at Wayland Baptist University, we just purchase a new point-and-shoot camera for integration into our curriculum.  As part of a student project that I’m directing this term, we needed to capture some video filmed at a relatively high frame-rate.  After much research on available options within our price range, we decided on the Casio EX-ZR100 model camera which provides not only 12.1 megapixels and a 12.5x optical zoom, it can also film lower resolution videos at 240, 480 or even 1000 frames per second.

I’ll take time later on this blog to document the project for which we used the camera, but in the meantime you might enjoy watching this.  Actually, you should be forewarned, this may be the most disturbing footage of a mathematics professor anywhere on the internet:

Hunting for Simpson’s Paradox, part 1

Baseball & Glove Over WhiteLet’s say we  happen to know the batting average for two baseball players.  Overall, player 1 has a higher average that player 2.  However, if you consider only how each player hits against left-handed pitchers we find that player 2 actually has a better average player 1.  In this hypothetical scenario, it also turns out that player 2 also has a better average against right-handed pitchers.  How is that possible?

Doesn’t it make intuitive sense that if player 2 is better than player 1 against left and right handed pitchers separately, that he must be better than player 1 against all pitchers?  While that may be what our intuition tells us, it turns out that it’s not necessarily true.

Consider the following table.  Note that batting average is simply the ratio of a players number of hits over the number of at-bats.



This table presents exactly the hypothetical scenario described above.  Separately, player 2 had a higher average than player 1 against left and right handed pitchers, but over all player 1 has a higher average than player 2.

This phenomenon is commonly known in statistics as Simpson’s paradox.  It demonstrates how our intuition can get us into trouble.  Briefly stated, Simpson’s paradox occurs when a correlation or trend that is present in groups is reversed when the groups are combined.

I was recently reminded of Simpson’s paradox when @Math_Bits posted a link on twitter to an article, “Instances of Simpson’s Paradox” by Thomas R. Knapp.  It got me thinking.  Sure I can manufacture an example and I’ve seen a few examples in papers, texts and even wikipedia.  But I want to find my own examples.  And of course, manufactured examples like the table above don’t count.  I need real data.

I figured the best place to start for real data that’s easy to find is in sports, say baseball, while we’re thinking of it.  I started simple and pulled up the first split data set I found.  Over on Yahoo sports, I pulled up team statistics for the full 2010 season, see  I pulled the data over into excel and began (manually, ug…) hunting for an example of Simpson’s paradox.  Of course, I would start the hardest way possible.  I looked at batting averages (overall, home and on the road).  I made three lists, one for each category: overall, home and road.  Then, I sorted the teams from highest to lowest and began looking one-by-one for pairs of teams where one had a higher average overall but lower both at home and on the road.

… to no avail.  I even reversed the process by sorting from lowest to highest.

I’m beginning to believe that Knapp was right when he claimed that examples of Simpson’s paradox are extremely rare.

The next step was to automate this process.  As a programmer, I began devising a simple code that will take an overall list and lists for each group and identifies all those pairs that satisfy Simpson’s paradox.

In the next post, I’ll walk through the progress I made using Matlab to do my dirty work.

Read More…
Find out more about Simpson’s Paradox at the following links:

Simpson’s paradox – Wikipedia, the free encyclopedia
When Combined Data Reveal the Flaw of Averages