## Posts tagged ‘probability’

### Shoestring Probability

March 15 is Shoe the World Day. And April 5 is One Day Without Shoes Day.

Shoes and math have a lot in common.

A shoe salesman consults a mathematician on what size shoes to keep in stock. The mathematician tells him, “There is a simple equation for that,” and shows him the Gaussian normal distribution.

The shoe salesman stares at the equation for a while, then asks, “What’s that symbol?”

“That’s the Greek letter π.”

“What is π?”

“The ratio between the circumference and the diameter of a circle.”

The shoe salesperson thinks for a minute. “What the hell does a circle have to do with shoes?”

As it turns out, there are at least 43 different ways to arrange the laces on your shoes. My favorite is the hexagram method:

But there are some fun things to do with your shoelaces other than lacing up your kicks. Here’s one.

Take the shoelaces out of your shoes. Fold the shoelaces in half and hold them in one hand so that the four aglets are exposed but the rest of the shoelaces are hidden in your palm. Like this:

Have a friend select two of the aglets and tie those ends together (I recommend a square knot). Then, have your friend tie the other two ends together. Finally, offer your friend the following wager:

You give me $1 if you formed one large loop.

I’ll give you $1 if you didn’t.

Is it a fair bet?

Too easy? Then try this. Take your shoelaces *and* your friend’s shoelaces, fold them in half, and then expose the eight aglets. Choose two at a time and tie them together. The wager remains the same.

Now is it fair?

Is it possible to create a fair wager with any number of shoelaces? If so, how many?

### 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.

### What’s Your Problem?

Problems in the MathCounts School Handbook are presented “shotgun style,” that is, a geometry problem precedes a logic puzzle and follows a probability question. (I worked for MathCounts for seven years and then served as a writer and chair of their Question Writing Committee, so I’m not unbiased.)

By comparison, textbooks often present 50 exercises on the same topic, each one only minimally different from the previous one. That tips the hand to students, methinks, and makes them realize, “Oh, I just need to do the same thing.” I prefer the MathCounts approach, where students have to dig into their bag of tricks to find a viable solution strategy.

With that in mind, here are a few problems I’ve encountered recently, each one not like the others.

**Problem 1.** The simple polygon is made from 73 squares, connected at their sides. What is the perimeter of the figure?

**Problem 2.** What is the expected number of times that a six-sided die must be rolled to get each number 1–6?

**Problem 3.** A wall is to be constructed from 2 x 1 bricks (that is, bricks that are twice as long in one direction as the other). A strong wall must have no **fault lines**; that is, it should have no horizontal or vertical lines that cut entirely through a configuration, dividing it into two pieces. What is the minimum size of a wall with no fault lines? The figure below shows a 3 × 4 wall that has both horizontal and vertical fault lines.

**Please share great problems you’ve recently encountered in the Comments.**

No answers, but here are some hints.

*Problem 1. *Look for a pattern.

*Problem 2.* Check out this simulation for the Cereal Box problem.

*Problem 3.* The smallest arrangement without a fault line is larger than 3 × 4 and smaller than 10 × 10.

### Go to Vegas, Saul

Saul is a statistician. He leads a comfortable life — he has tenure at a respected university, an impressive list of publications to his credit, and the admiration of his colleagues. Less than a year from retirement, he hears a voice from above. “Saul, quit your job,” the voice says.

He ignores it.

The next day, the voice returns. “Saul, quit your job.” And the next day. And the day after that. And it becomes more frequent, occupying most of his waking hours as well as his dreams. “Saul, quit your job.”

It continues relentlessly for months. “Enough already!” Saul shouts when he can take no more. He delivers a letter of resignation to his dean that morning.

“Saul, take your life savings out of the bank.”

*I’m not taking out my money*, Saul thinks. But the voice continues relentlessly. “Saul, take your life savings out of the bank.”

After several sleepless nights, he finally gives in. “Now what?” he asks.

“Saul, go to Vegas.”

He buys a ticket to Vegas. When he arrives, the voice tells him, “Saul, go to the blackjack table.”

He obeys.

“Saul, bet all of your money on one hand.”

“That’s insane!” he shouts.

“Saul, bet all of your money on one hand.”

He knows that the voice will continue if he doesn’t listen, so he does it.

He’s dealt an 8 and a king. 18. The dealer is showing a 6.

“Saul, take a card.”

“But the dealer has…”

“Saul, take a card.”

“But the laws of probability…”

“Saul, take a card!”

He takes a card reluctantly. It’s an ace. 19. He sighs relief.

“Saul, take another card.”

“C’mon!”

“Saul, take another card!”

He takes another card. Another ace. 20.

“Saul, take another card.”

“But I have 20!” he shouts.

“Saul, take another card!”

He shakes his head. “Hit me,” he says sheepishly. A third ace. 21.

And the voice booms, “Un-fucking-believable!”

### The Twelve Days of Crisp Math – Day 8

For the **Eighth Day of Crisp Math**, here’s a problem for you. Best of luck solving it before Day 9…

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

A. 25%

B. 50%

C. 60%

D. 25%

### Kindergarten Math

Alex and Eli started kindergarten on Tuesday.

At a “Meet the Teacher” event last week, we were told that this is the most kindergarten classes they’ve had at the school. “We had to add another class this year, so we now have eight,” one of the teachers said.

“How many students are in each class?” I asked.

“Twenty-one,” she said, “so it’s really good that we added that eighth class — or else there’d be, like, 26 students in each class!”

I was too polite to tell her that 8 × 21 ≠ 7 × 26.

Today, we had a parent-teacher conference. On the bulletin board in her class was the following chart with student names:

There are 19 students in the class, and all of them have a first name that that begins with a letter in the first half of the alphabet. There are 3 A’s, 1 B, 3 C’s, 1 D, 2 E’s, 1 I, 3 K’s, 3 L’s, and 1 M.

I mentioned this to the teacher. “I know!” she said. “Isn’t that an amazing distribution!”

*Well, yeah*, I thought. *It’s quite amazing, in fact.*

If you assume that names are evenly distributed across the alphabet, the probability that all 19 students would have a first name in the first half of the alphabet is an astounding (1/2)^{19} = 0.00000002%.

But of course, names are not evenly distributed across the alphabet. I don’t know how they’re distributed, but the first letters of English words are distributed as follows:

That means that 52.005% of all English words start with a letter in the first half of the alphabet. If you assume that names follow the same distribution, then the probability doubles to 0.00000004%.

Yup. Still pretty low.