## 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.

### P(Winning Lottery) > 0… but Just a Little

I know a fair bit about the probability of winning the lottery.

State-run lotteries are a tax on the mathematically challenged.

Given the odds of winning, then you might wonder why I occasionally buy scratch-off lottery tickets. Lord knows, my wife often wonders aloud about it. Believe it or not, there are three reasons that I buy these tickets:

- First, I’m from a rural town in the-middle-of-nowhere Pennsylvania. The rural poor are infamous consumers of lottery tickets. Consequently, I believe that buying lottery tickets is part of my genetic code.

- Second, it’s a guilty pleasure that is easier to indulge than buying PowerBall or Daily Number tickets. When you buy one of those, there is a human interaction, and I imagine that the clerk selling me the ticket is thinking, “Loser! Don’t you know how low your odds of winning are?” For the scratch-off tickets, you insert your money in a vending machine, and the tickets are dispensed. Sure, you may get a disapproving eye from a passer-by, but at least there’s no formal exchange with another human.

- Third, and most importantly, I know a bit about probability, but I also know a little about the intersection of math and psychology. As it relates to the lottery, the idea is fairly simple — make every third or fourth ticket a winner, and people who buy scratch-off tickets will win often enough that they’ll keep coming back for more. Truth is, the winning tickets usually have a prize equal to the price of the ticket or twice the price of the ticket. For instance, if the tickets cost $5, then the winning tickets usually have a pay-out of $5 or $10. When four tickets are sold for $5 each, the state collects $20 and only pays out $5 or $10. Good work if you can get it, eh?
This last point is actually the one that hooks me in. If I buy four tickets at a time, I can almost guarantee that one of them will be a winner. Consequently, I’ll only be giving $10 or $15 to the state instead of $20. (What a bargain, right? I walked into the store with $20, and I get to leave with $5 or $10. Who could pass that up?) But on the off chance that there are two winners in this group of four, or if one of the tickets is a big winner with a prize of more than double the price, well, then, this could work out all right for me.

Yes, I am fully aware that my argument is irrational and that I am slightly delusional. Recognizing my irrationality and delusion, I don’t buy scratch-off tickets very often; but, I do buy them occasionally.

So, why am I telling you all this? Because this morning, I bought four scratch-off tickets at the local supermarket.

**First Ticket:** It had a “5 Times” logo next to $10. That means I won $50.

“Wow!” I thought. “I’m already ahead $30.” And then, of course, I realized how unlikely it was that the other three would be winners.

**Second Ticket:** I matched not one, not two, but *three* of my numbers to the winning numbers for $5 each. That means $15 in winnings on the second ticket.

“Holy schnikeys!” I said out loud, though probably too soft for anyone else to hear. (I hope.)

**Third Ticket:** I matched two numbers for $5 each. That means another $10.

**Fourth Ticket:** Nada.

But, whatever. I was up $55, so who cares about that stupid fourth ticket?

I collected my winnings, and I walked across the street to Panera and ordered a chai tea latte and a bagel. I handed my MyPanera card to the clerk — indicentally, I hate the recent trend of naming something as MySomething, because then it’s really awkward when I want to refer to the MySomething that belongs to me by calling it my MySomething; but, I digress — and he told me that I had earned a free bagel. “You can have this one for free, if you want,” he said. Well, hell yeah!

A few hours later, I went to lunch with a new professional acquaintance. Even though *I had asked her* if she wanted to meet for lunch, *she* picked up the tab!

Can you believe it? Fifty-five dollars in lottery winnings, a free bagel, and a free lunch. Financially speaking, this could have been the luckiest day of my life. (Well, except for the day when I learned that an essay I’d written had won a honeymoon in Oaxaca for my wife and me. But that’s a story for another day.)

Yep, the chances of one good thing happening in a day are low. But the chances of three good things happening in a day? Infinitessimal! Guess I’m just blessed.

My former boss, Jim Rubillo, knows a thing or two about probability and statistics, too. Somehow, his favorite line seems appropriate for this post.

If you don’t believe in the power of random sampling, then the next time your doctor requests a blood sample, tell her to take it all!

### Kicking Off Could Become Flipping Off in NFL

One of the most exciting plays in the history of professional (American) football was the opening play of the second half of Super Bowl XLIV, when the New Orleans Saints recovered an onside kick. They then scored to take a 13–10 lead, and eventually won the game 31–17.

But onside kicks could be a thing of the past. Yesterday, New York Giants’ co-owner John Mara suggested that kickoffs might someday be eliminated from the NFL. This caused a lot of sports pundits to react, saying that it would inherently change the game. On the *Mike and Mike Show*, analyst Mark Schlereth responded with these rhetorical questions:

What’re you gonna do, flip a coin three times in a row? You gotta get heads three times in a row to get an onside kick?

Once again, probability was placed front-and-center in recent football discussions. While I like Schlereth’s new, less violent, and more mathematical approach to onside kicks, I just wish he had gotten the math right.

If you flip three coins, the probability of getting three heads is 12.5%. That’s not enough. Data shows that onside kicks in the NFL are successful 26% of the time. So the following would be a reasonable modification to Schlereth’s proposal:

Flip two coins. Two heads results in a successful onside kick.

Then the probability would be 25%, closer to the current reality.

Unfortunately, that’s not exactly right, either — it’s based on a misleading statistic. The success rate of onside kicks is highly dependent on whether the team receiving the kickoff is expecting it or not. When teams are expecting it, the success rate hovers around 20%; when teams aren’t expecting it, however, the success rate jumps to 60%. Considering that data, the process might be modified as follows:

- Kicking team indicates to referee that they will try an onside kick.
- Of course, this must be done secretly, so as not to arouse the suspision of the receiving team. I propose that one referee be assigned to each team; the team would encode the message using RSA encryption, and the assigned referee would be given the corresponding RSA numbers. A message can then be passed without fear of interception by the receiving team. To ensure that this procedure does not signficantly delay the game, messages stating “we WILL try an onside kick” and “we WILL NOT try an onside kick” could be prepared in advance, and unemployed math PhD’s could be hired as NFL referees to decode the messages.

- The receiving team must similarly indicate whether or not they suspect an onside kick.
- Again, use RSA encryption.

- If the kicking team chooses an onside kick, and the receiving team suspects an onside kick, then:
- Flip 9 coins. If 9, 8, 3, or 1 of them land heads, the onside kick is successful.
- P(9, 8, 3, or 1 head with 9 coins) = 20.1%

- If the kicking team chooses an onside kick, but the receiving team does not suspect it, then:
- Flip 9 coins. If 9, 8, 5, 4, or 2 of them land heads, the onside kick is successful.
- P(9, 8, 5, 4, or 2 heads with 9 coins) = 60.0%

- If the kicking team does not choose an onside kick, then:
- Flip 9 coins, just so the receiving team is unaware of what the kicking team decided to do, which will allow for the element of surprise with future kicks.

If the NFL decides to accept Mark Schlereth’s suggestion for using coins to determine onside kicks, I am hopeful that they will give my proposal serious consideration. If necessary, I have an Excel spreadsheet that I would be willing to share with them.

### Random Number Generation

While playing Nurikabe, my sons completed the following puzzle:

The puzzle itself isn’t very interesting, but did you notice the Puzzle ID? Exactly 1,000,000. The boys thought this was pretty cool, and I did, too. Yeah, yeah, I know, the occurrence of 1,000,000 shouldn’t impress me more than the appearance of, say, 8,398,176 or 3,763,985. But there are just under 10,000,000 unique 5 × 5 puzzles on the site, and only nine of them contain six 0’s. How lucky were we to get that random number?

Generating random numbers can be a difficult proposition, especially for a computer. This article from WIRED magazine — which describes a pattern that inadvertently appeared on lottery tickets, making it possible to predict winning tickets before they were scratched — shows how difficult it can be to generate numbers that appear to be random. (The article really is worth a read, especially for math geeks. Truth be known, WIRED is the only magazine that I read cover-to-cover every month.)

Robert Coveyou, a mathematician who worked on the Manhattan project, was an expert in pseudo-random number generators. He is most famously remembered for the following quote:

The generation of random numbers is too important to be left to chance.

Of course, Randall Munroe at xkcd has a foolproof method for generating a random number:

I would hate for you to need a random number and then have difficulty generating one. I’m here to help, so I present the…

**MJ4MF Random Number Generator (PDF)**

Creating the MJ4MF RNG is quite simple. Just follow these steps:

- Download and print the PDF from the link above.
- Cut out all six squares, one for each number 1-6.
- For each square, make two folds: first, fold the paper to the center vertically; then, fold the paper to the center horizontally. The result of these two folds is shown, below left.
- When all six pieces are folded, interlace them to form a cube. This is shown, below middle. The assembled cube is shown, below right.

Finally, a joke about random numbers.

A student is asked for the probability that a random number chosen between 0 and 1 will be greater than 2/3. The student answers 1/3. The teacher says, “Great! Can you explain to the class how you arrived at your answer?” The student says, “There are three possibilities: the number is either less than, equal to, or greater than 2/3, so the probability is 1/3!”