## One, Tooth, Ree, …

Were a state assessment item writer to get his hands on the puzzle that I discuss below, I believe that this is what the problem might look like:

Fifteen congruent segments labeled A-O are arranged horizontally and vertically to form five squares, with five of the segments shared by two squares each. Which three segments can be removed to leave three congruent squares?

- {A, B, C}
- {A, C, E}
- {B, K, N}
- {D, E, F}
- {G, I, N}
- {H, A, L}
- {I, B, M}
- {K, L, M}

You’ve likely seen this problem before, but in a simpler and more elegant form.

Remove three toothpicks to leave three squares.

Eli shared this puzzle with us the other night at dinner, after hearing it during his math enrichment class that day. (It’s my sincere hope that every family has similar conversations at mealtime.) This was a great problem to discuss at the dinner table, since a collection of manipulatives was close at hand.

As you might expect, Eli and his brother started by removing toothpicks randomly and seeing what happened. But it turns out that random toothpick removal is not a great strategy. There are **15 toothpicks** in the arrangement, so there are **15C3 = 495 ways** to select three of them.

Of course, some of those selections are obviously wrong, such as selecting the three toothpicks in the middle row, or choosing the three highlighted in pink:

But even ignoring the obviously wrong ones, that still leaves a lot of unobviously wrong combinations to consider. Yet I let my sons randomly select toothpicks for a few minutes without intervention. Why? Because of a primary tenet for effective problem solving:

Get dirty.

You can’t get any closer to a solution if you don’t try *something*. Even when you don’t know what to do, **the worst thing to do is nothing**. So, don’t be afraid to get your hands a little dirty. Go on, try something, and see what happens. You might get lucky; but if not, maybe it’ll shed some light.

After a few minutes, they conceded that random toothpick selection was futile. That’s when they modified their approach, leading to another important aspect of problem solving:

Look at the problem from a different perspective.

“Hmm,” said Eli. “That’s a lotta toothpicks.” He thought for a second. “But there are only five squares.” That’s when you could see it click, leading to the most popular problem-solving principle on the planet:

Solve a simpler problem.

There are **5 squares** in the arrangement, and there are only **5C3 = 10 ways** to select three of them. If you first choose three squares, then you can count the number of toothpicks that would need to be removed. For instance, you’d have to remove five toothpicks to leave these three squares:

That doesn’t quite work. But the problem is a whole lot easier if you focus on “leave three squares” rather than “remove three toothpicks.” There are only 10 possibilities to check.

From there, it was just a hop, skip, and jump to the solution.

So, maybe this problem, as well as the million or so other remove-some-toothpicks problems, isn’t very mathematical. It only contains two mathematical elements — basic **counting** and **geometry**. In the vernacular of Norman Webb’s Depth of Knowledge Levels, the kinds of questions that you could ask about this problem — “How many toothpicks are there?”, “How many are you removing?”, “What is a square?” — would be DOK 1.

But they’re fun, and they’re engaging to kids, and they develop problem-solving skills, such as those mentioned above as well as perseverance.

For more problems like this, check out **Simply Science’s collection of 16 toothpick puzzles**.

Speaking of the number 16, here’s my favorite toothpick problem.

Remove 4 toothpicks to leave 4 small triangles.

And finally, because I know you came here looking for jokes, here is the only joke I know that involves toothpicks. It’s rather disgusting. Continue at your own risk.

One night, as a bartender is closing his bar, he hears a knock at the back door. It’s a math graduate student. “Can I have a toothpick?” he asks. The bartender gives him a toothpick and closes the door.

Five minutes later, another knock. “Can I have another toothpick?” asks the student. The bartender gives him another one and closes the door.

Five minutes later, a third knock. “Can I have a straw?” asks the student.

“Sure,” says the bartender, and hands him a straw. “But what’s going on out there?”

“Some lady threw up in the back,” says the grad student, “but all the good stuff is already gone.”

## You Say It’s Your Birthday

*It’s my birthday, too.*

Okay, not really, but it wouldn’t be surprising if it were. Today (September 17) is the fourth most common date on which to be born. Or at least it had been from 1973 to 1999 in the United States, according to an analysis by Amitabh Chandra, a professor of social policy at Harvard who compiled 28 years of birth date data.

Yesterday (September 16) was the overall most common birth date, and tomorrow (September 18) was the tenth most common.

In fact, all of the top ten most common birth dates occur in September, and even the lowest-ranking date of the month is still in the top third of the most common dates (September 5, #125). Apparently November and December are busy times for gettin’ busy.

It’s a busy birthday time elsewhere, too. Statistics New Zealand issued a warning about the excessive number of birthdays that will occur near the end of September.

How common is your birth date? Take a gander at the image below. Or, for an interactive experience, click around on **Andy Kriebel’s version that uses Tableau**.

This information (which is not new, btw — an initial analysis was completed by Chandra for a paper published in the Journal of Political Economy in 1999; a table with Chandra’s full data appeared in a column by David Leonhardt in 2006; and, various versions of the image above have been floating around the internet for several years) throws a wrench into the famous Birthday Problem, which asks, **“At a party of n people, what is the probability that two of them share a birthday?”**

The solution to the Birthday Problem assumes that birth dates are evenly distributed across the calendar. But clearly they’re not. How much does real data affect the solution, though? As it turns out, not much.

Using real birth date data from 1985-88 pulled from the CDC, Joe Rickert completed an analysis using Revelation R Enterprise 6. The results look like this:

On the other hand, the theoretical probability of two people sharing a birth date in a group of *n* people is given by the formula:

And if you try to graph that formula, you get something like the following:

Admittedly, I was lazy. The scatterplot above only includes *n*-values from 2‑10 and multiples of 5 from 15‑100. That’s because Excel can’t handle factorials larger than 170!, so I had to use a large number calculator online and enter the values into an Excel sheet by hand. Still, it gives a pretty good idea of what the shape of the curve.

A typical benchmark for the Birthday Problem is *n* = 23, when the probability of a pair sharing the same birth date first exceeds 50%. Using real data, P(23) = 0.5087, and using the theoretical model, P(23) = 0.5073. Good enough for government work. By visual inspection, you can see that the other values match up rather well, too.

Finally, here are some math jokes about birthdays.

Statistics show that those who celebrate the most birthdays live longest.

I only drink twice a year: when it’s my birthday, and when it’s not.

Happy integer number of arbitrary units of time since the day of your birth!

You don’t need calculus to figure out your age.

They don’t make birthday cards for people who are 85 years old. So I almost bought you a card for an 80-year old and a 5-year old. But they I figured no one wants to do math on their birthday.

## Loot™ — Best Game Ever?

If **Loot** isn’t the best game of all time, it’s at least the best game for International Talk Like a Pirate Day (September 19).

For those who don’t know, Loot is a pirate-themed card game in which you can do three things:

- Send a merchant ship out to sea.
- Attack a merchant ship with pirate ships.
- Play a pirate king or admiral to increase your attack strength.

I bought the game on a whim when visiting Powell’s City of Books several months ago, and it’s clearly the best random purchase of my life. (It far exceeds the tattoo of Rene Descartes that mysteriously appeared on my posterior the morning after one helluva night in Hoboken, NJ. Don’t ask.) The game has many appealing qualities:

- Simple to learn. It takes less than 10 minutes.
- Fast to play. A game can be completed in 20 minutes.
- Strategically challenging. The strategy is not obvious. (I’ve played quite a few times, and I’m not even sure that my strategy is effective, let alone optimal.)
- And not least important, the following is a direct quotation from the official rules:

*We find that the game’s even more fun when everyone talks in pirate accents.*

How do you not love a game that gives that kind of directive? When playing in teams, we ask the following questions to elicit a pirate response:

- What’s the circumference of a circle divided by 2π?
- What’s the eighth most common letter in the English language?
- What is
*d*÷*t*? - What’s the 18th letter of the alphabet?
- Which set of numbers includes the rationals and irrationals?
- Which letter appears most often in
*refrigerator*? - What electrical property is measured in ohms (Ω)?

Here are some other mathy pirate jokes that might amuse.

What has 12 legs and 12 eyes?

A dozen pirates!What is the pirate alphabet?

ABCCCCCCCDEFGHIJKLMNOPQRSTUVWXYZWhat do pirates and Descartes have in common?

They think, therefore they ARR!How much did the pirate pay for his peg and hook?

An arm and a leg!Did you know that 3.14% of sailors are pi-rates?

Finally — even though it’s not Pi Day — here’s an image you might enjoy, compliments of Illuminations.

## Lighthouses and Math in Oregon

My family and I recently spent a week along the Oregon coast, where we hiked, biked, and — of course — did some math.

While on a hike through Ecola State Park, we had this view of the Tillamook Rock Lighthouse, a structure 62 feet tall that stands atop an offshore island:

Never one to resist the opportunity to pose a math question, I asked my sons, “How far do you think it is **from here to the lighthouse**?” My sons — never ones to resist reading the informational placard at a trailhead — already knew the answer and quickly responded, “About 1.2 miles.”

Ah, hell no. I’m itching for a mathy conversation, damn it, and you’re not getting off that easy! So I then asked, “How far do you think it is **from here to the horizon**?” Luckily, the answer to that question hadn’t been included in the display at the bottom of the hill.

Eli used proportional reasoning. He extended his arm and measured the distance from the shore to the lighthouse between his thumb and index finger; then, he used that same unit to measure the distance from the lighthouse to the horizon. “I’d guess about 2 miles,” he said. “It’s farther from here to the lighthouse than from the lighthouse to the horizon.”

Alex used intuition. “I think it’s a little farther,” he said. “Maybe three miles?”

“Actually,” I said, “it’s much farther than that. I won’t tell you how *much* farther, but I will help you figure it out.” (Typing that last sentence makes me realize that having a constructivist father is not an easy burden to bear. My answer to most questions is usually another question.)

After much discussion, we finally concluded that we were about **90 feet above sea level**. Estimating vertical distance, it turns out, is not an easy task, especially for eight-year-olds. But once we agreed, the rest of the math was more or less straightforward. The image below provides a reasonable model for the situation:

*h*= height above sea level of the observer*R*= radius of the Earth*x*= the straight-line distance to the horizon from the observer

The sight line of the observer is a tangent line to the circle, and the point of tangency is the horizon that the observer sees. Consequently, a right angle is formed by the tangent line and radius. We now have a right triangle with which to work, so we can lean on the Pythagorean theorem. But first we’ll need some numbers.

At the Tillamook Rock Lighthouse, we had the following numbers:

*h*= 90 feet (estimated)*R*= 4,000 miles (give or take)

And *x*, of course, is the unknown distance to the horizon that we are trying to find.

This leads to the following:

A rough estimate of the Earth’s radius *in feet* is 4,000 × 5,000 = 20,000,000, which is why that number appears in the equations above. Two things should be noted here. First, using 5,000 for the number of feet in a mile instead of the more accurate but less friendly 5,280 allows for easier calculations throughout. Second, we’re only looking for an **estimate** of the distance to the horizon, so we should feel free to take some liberties with the numbers if it’ll help with computation.

Using 20,000,000 instead of 4,000 allows feet to be used as the units throughout, so the answer obtained at the end (*x* = 62,000) is also in feet. To find the distance to the horizon *in miles*, we’ll need to divide by 5,000, which gives:

62,000 ÷ 5,000 = **12.4 miles**

Now that’s all well and good, and the boys were amazed that the horizon is so far away. But what if you want to generalize to find the distance to the horizon from *any height* above sea level?

The answer depends on the math, boating, or physics book you choose to reference.

One possibility is to use this graph, which we discovered a few days later while visiting the Yaquina Head Lighthouse:

It shows the distance to the horizon on the horizontal axis and the feet above sea level on the vertical axis, which I suppose makes sense from a visual perspective; but it’s more conventional to put the independent variable (feet) on the horizontal axis and the dependent variable (miles) on the vertical axis. Nonetheless, the information provided by the graph is still useful, and it nicely shows that the relationship between height and distance to the horizon is clearly not linear.

On the other hand, if you search for “distance to horizon formula” online, you’ll find any of the following formulas:

The differences are due to whether miles or nautical miles are used, as well as whether you’d like to account for the refraction of light, which greatly influences the distance an observer would be able to see. Personally, I prefer the first formula for its simplicity, but you’re free to use whichever one you like.

Sadly, I don’t know a single joke involving the horizon. (Do you? Post in the comments.) So I’ll leave you instead with a quote about the horizon:

Make the horizon your goal. It will always be ahead of you.

— William Makepeace Thackeray

## MJ4MF and Math Clocks

How stoked was I to see that @WeAreTeachers is giving away a Swag Bag that includes a copy of **Math Jokes 4 Mathy Folks**? In a word — inordinately!

My only concern is that the bag of goodies also includes this clock:

This is the second time in less than two weeks that I’ve seen this clock. I was disheartened to see it hanging in a middle school that I visited recently. I believe that no self-respecting math teacher should be willing to display it. Cute? Yes. Mathematically accurate? No. To wit:

**Painful:**The equation 50/2 = 100/*x*yields the solution*x*= 4. But an hour after 3 o’clock is 4 o’clock, not “*x*= 4 o’clock.” This equation should be replaced by the expression 2 × 100 ÷ 50. Similarly, the equation -8 = 2 –*x*should be replaced by -8 – (-2).**Just Plain Wrong:**So, maybe you have no problem with the use of a equation on this clock, and you’re okay that the solution of the equation is meant to represent the missing number. Okay, fine. But there can be no way that you have no problem with the equation at 7 o’clock. It’s*quadratic*, so there are*two*solutions. C’mon! Anyone who solves that equation won’t know if it’s 7 o’clock or -6 o’clock!**Egregious:**When the big hand of this clock points north and the little hand points west, the time will be 9.00477796077… o’clock (or thereabouts). Teachers spend a good part of their lives trying to help students unlearn that π = 3.14. It’s not; that’s only an approximation. Yet this clock perpetuates that misconception.

I would like to suggest that the folks at We Are Teachers consider including a different clock in their Swag Bag. I’m a much bigger fan of the Math Expressions Wall Clock:

Or perhaps one of these Math Clocks would be more appropriate:

## 144 Quadrillion Reasons to Never Attend a Baseball Game with Me

Baseball fans do lots of dumb things. Wear a foam finger. Act like they’re smart because they correctly chose the attendance from the four choices on the big board. Throw trash at the lower levels. Streak the field. Root for the Yankees. (And while we’re at it, the 1980’s called. They want “The Wave” back.)

Yet few things are as dumb as the following game, which some friends and I play at Nationals Park.

- To start, everyone puts in a $1 ante.
- The first gambler then adds another $1 to the cup. If the player batting gets a hit, that gambler gets all the money in the cup. If the player batting does not get a hit, that gambler passes the cup to the second person.
- The second gambler then adds $2 to the cup. Again, that gambler gets the money if the player batting gets a hit, or passes the cup to the next gambler if the player batting does not get a hit.
- This continues, with the next gambler adding double what the previous gambler added, until there’s a hit.

Because of exponential growth, it doesn’t take long for this game to exceed most people’s comfort level. At a recent game, we had gone through five batters without a hit when the cup was passed to Dave. He gladly added $32, but then passed the cup to Joe when the batter struck out.

Joe was reluctant to add $64, but like the rest of us, he was certain that the paucity of hits couldn’t continue. It did, however, and he passed the cup to me.

Now that the cup was requesting a three-digit donation, this was getting serious. Most of my friends don’t carry $128 to a baseball game, and even fewer of them are willing to put it in a betting cup. Dave looked at me. “You in?” he asked.

“Sure am,” I replied as I put $100 in the cup. “You’ll have to trust that I’m good for the other $28,” I said. “That’s the last of my cash.”

It occurred to me that **52.9% of the money now in the cup had come from my wallet**. That made it extra hard to pass the cup to Adam when the batter struck out.

But Adam just looked at it. “Don’t even think about handing that to me,” he said. “I’m out.”

So the cup went back to Dave, begging for $256 more. “If you’re in, I’m in,” he said to me. “Trust me that I’m good for it?” he asked. I nodded my head to let him know that I did.

When Dave handed the cup to Joe after the next batter flew out, it was **Dave’s money that now accounted for 56.3% of the pot**. Joe just passed the cup directly to me. “I’m out, too,” he said.

Judgment is the better part of valor. But no one has ever accused me of being valiant. *If I put in $512 and there’s a hit*, I thought, *I’ll win $377. But if not, I’ll be passing a cup to Dave that has $649 of my money.*

“I’m in,” I said, surprising even myself. Dave looked at me with equal parts disbelief and dismay… amazed that I had pulled the trigger, unsure what he would do if the cup came to him, stunned that we had gotten to this point.

He’d need to deal with the question sooner than he thought. The batter hit a routine ground ball to the second baseman on the first pitch. As the ball popped in the first baseman’s mitt, my gulp was audible. Folks in our section eyed Dave suspiciously as he cheered an out for the home team. I passed the cup to him and asked, “To you for $1,024, sir. Are you willing to sacrifice your kids’ college fund for this?”

Dave looked at the cup. He looked at me. He looked back at the cup. After several long seconds, he looked at me again. “Split it?” he asked.

I paused. “What?”

“Yeah… what?” said Joe.

Dave explained. “You and I split it. Joe and Adam are out.”

“No!” protested Joe. “That’s not fair! You set us up!”

I ignored Joe, because I realized I now had an exit strategy. “Are you suggesting 50/50?” I asked Dave.

“That’s what I was thinking.”

“But I put two-thirds of the money in that cup.”

“Okay, sure,” Dave said. “We’ll split it 2:1.”

So in the end, I won $33, Dave won $51, and Adam and Joe were disconsolate over what had just transpired.

There had been a streak of 10 batters without a hit (one of the batters had walked, but that doesn’t count as a hit, so our silly game continued), and the pot had grown to $1,027. It would have climbed to $2,051 if Dave had continued. That might only be a drop in the bucket for the Koch brothers, but it’s a sizable amount for a math educator.

The hitless streak made us wonder what game in baseball history would have made us dig deepest into our pockets. The all-time winner would have been a game on May 2, 1917, between the Cincinnati Reds and the Chicago Cubs, when **Fred Toney and Hippo Vaughn each threw no-hitters through nine innings**. In the top of the 10th, Vaughn got one out before giving up the first hit of the game… so that’s at least 55 batters without a hit. It’s hard to find play-by-play data for games from 1917, but the box score shows that there were 2 walks in the game. If both of them happened before the first hit, that’d increase the pot to a staggering 2^{57} + 3 = $144,115,188,075,855,874. That’s **144 quadrillion**, for those of you who, like me, go cross-eyed trying to read such large numbers.

Why do you dislike the number 144?

Because it’s gross.Why do you dislike the number 144 quadrillion?

Because it’sverygross!

Perhaps most interesting to me, though, was **the percent of the cup contributed by the person currently holding the cup**. A tabulation of the possibilities is shown below. This scenario assumes four gamblers with no one dropping out. The rows highlighted in yellow (rounds 4*n* – 3) indicate when the first gambler is holding the cup. Likewise, the second gambler’s turns occur in rows 4*n* – 2, the third gambler’s turns occur in rows 4*n* – 1, and the fourth gambler’s turns occur in rows 4*n*.

Round |
Contribution |
Gambler Total |
Cup Value |
Percent |

1 | 1 | 2 | 5 | 40.00% |

2 | 2 | 3 | 7 | 42.86% |

3 | 4 | 5 | 11 | 45.45% |

4 | 8 | 9 | 19 | 51.43% |

5 | 16 | 18 | 35 | 52.24% |

6 | 32 | 35 | 67 | 52.67% |

7 | 64 | 69 | 131 | 52.90% |

8 | 128 | 137 | 259 | 53.20% |

9 | 256 | 274 | 515 | 53.26% |

10 | 512 | 547 | 1,027 | 53.29% |

11 | 1,024 | 1,093 | 2,051 | 53.31% |

12 | 2,048 | 2,185 | 4,099 | 53.33% |

13 | 4,096 | 4,370 | 8,195 | 53.33% |

14 | 8,192 | 8,739 | 16,387 | 53.33% |

15 | 16,384 | 17,477 | 32,771 | 53.33% |

16 | 32,768 | 34,593 | 65,539 | 53.33% |

17 | 65,536 | 69,906 | 131,075 | 53.33% |

Here’s how to read the table:

- The
**Contribution**is the amount that a gambler adds to the cup in that round. - The
**Gambler Total**is the combined amount contributed by that gambler so far. For instance, the “Gambler Total” in Row 13 is $4,370, because the gambler has contributed 4,096 + 256 + 16 + 1 on his previous four turns, plus the $1 ante at the beginning of the game. - The
**Cup Value**is the amount in the cup, which includes the ante plus all previous contributions. - The
**Percent**then shows the percent of the money in the cup that was contributed by the gambler holding the cup.

In Round *n*, the Cup Value is given by the wonderfully simplistic formula

because the sum of the first *n* powers of 2 is 2^{n} – 1, and the initial ante contributes $4 more.

More difficult is the formula for the combined contribution shown in the Gambler Total column. For Gambler 1, the formula is:

The Gambler Total formula for gamblers 2, 3, and 4 are similarly complex. (You can have fun figuring those out on your own.) But the purpose of determining those formulas was to investigate the percent of the Cup Value represented by the Gambler Total. And as you can see in the final column, **the percent of the cup contributed by the gambler holding the cup tends toward 53.33%**.

So here’s the way to think about this. If you pass the cup and the next gambler wins, 53.33% of the money in the cup is his; **26.66% of the money in the cup was yours**; and, the remaining 20.00% came from the other two gamblers. Which is to say, **you’ve contributed 26.66/46.66 = 57.14% of the winner’s profit**.

Ouch.

Taking all this into account, and realizing that I could bankrupt my family if there were ever a streak of just 20 batters without a hit, some additional rules became necessary:

- A person can choose to leave the game at any time. Play continues with those remaining. Obviously, if only one gambler remains, he or she wins.
- Anyone can declare a “cap,” wherein the amount added to the cup continues at the current rate without increasing. For instance, if I added $32 to the cup on my turn and called a “cap,” then every person thereafter would simply need to add $32 on their turn, instead of doubling.

This latter rule seems like a good idea; it transfers the game from exponential growth to linear growth. Still, the pot grows quite rapidly, as shown below:

After 10 consecutive batters without a hit, the pot would still grow to $160, even if a “cap” had been implemented at the $32 mark.

We’ve also switched to **the Fibonacci sequence**, because that grows less quickly over time.

At a recent Nationals-Cubs game, four of us played this game with the amended rules. Everyone anted $1 to start, and then the amounts added to the cup were:

1, 2, **3**, 5, 8, 13, **21** (*cap*), 21, 21, 21, **21**

Dave was third, and the amounts he added to the cup are bold in the sequence above. Both Tanner Roark and Tsuyoshi Wada started strong, and the first 10 batters were retired without a hit or a walk. But the 11th batter of the game got a hit in the bottom of the second inning. Dave had contributed $46 of the $141 in the cup, so he won $95 on that beautiful swing by Wilson Ramos of the Nationals. (It wouldn’t have been “beautiful” if it had been made by a Cub or if it had cost me more than $27.)

The adjustment of using the Fibonacci sequence was a good one. The seventh donation to the cup was a reasonable $21, compared to a $64 contribution using the doubling scheme. In addition, instituting the “cap” permitted folks to continue playing long after they otherwise would have dropped out.

We continued playing through the end of the game. Which brings me to one final rule:

- At the end of the game, the gambler holding the cup does not win. Instead, the money in the cup goes to the next gambler.

This only makes sense. You wouldn’t want a gambler winning if he was holding the cup when an out was recorded.

Last Friday’s game ended with an astonishing 18 hits. There were no other streaks of 10+ batters without a hit, so none of the wins were as large as Dave’s first. On the other hand, I won more than I lost the rest of the game, and I was only down about $10 by game’s end.

Still, I’m concerned that I’ll go broke by the end of the season. One solution, of course, would be to stop playing, but that would require willpower and a higher intellect. Instead, I’m thinking that perhaps I should try to earn some extra money. If you’d like to contribute to this worthy cause, **I’m available for tutoring, stand-up comedy, and blog post writing**.

And if you have a need for analyzing the dumb games that boys play while watching a baseball game, well, there’s data to suggest that I’m pretty good at that, too.

—

*I’d like to thank Marjan Hong for her help in analyzing this game. I’d also like to thank Dave Barnes for teaching me this game, for wasting countless hours discussing the rules and amendments, and for taking my money.*