Archive for September, 2015

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.

Most Common Birthdays

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:

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

\textrm{P} = 1 - \frac{365!}{(365-n)! \cdot 365^n}

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

Birthday - Theoretical

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 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 then I figured no one wants to do math on their birthday.

September 17, 2015 at 3:51 pm 1 comment

Loot™ — Best Game Ever?

Loot

Loot™ by Gamewright

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?
ABCCCCCCCDEFGHIJKLMNOPQRSTUVWXYZ

What 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.

Loot

September 14, 2015 at 12:40 am Leave a comment

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:

Tillamook Lighthouse

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:

Distance to HorizonIn that model,

  • 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:

\begin{array}{rcl}    x^2 + R^2 & = &(R + h)^2 \\    x^2 + 20\,000\,000^2 & = &(20\,000\,000 + 100)^2 \\    x^2 & = &40\,000\,010\,000 \\    x & \approx &62\,000 \, \textrm{feet} \end{array}

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:

Graph - Distance to Horizon

Graph at the Yaquina Head Lighthouse, Newport, OR

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:

\begin{array}{rcl}    d & = &\sqrt{1.5h} \\    d & = &\sqrt{\frac{7h}{4}} \\    d & = & 1.17\sqrt{h} \\    d & = &R\times\arccos\bigl({\frac{R}{R+h}\bigr)}    \end{array}

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

September 9, 2015 at 7:42 am 3 comments


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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