Death By Coconut

In “I Know What You Did Last Summer of the Shark”, then Daily Show correspondent Stephen Colbert states that falling coconuts kill 150 people each year. You might assume that “death by coconut” was a purely random occurrence, but that might just be what the coconuts want you to believe. What's are the chances that the coconuts are out to get you? 

According to Wikipedia, 54 million tonnes of coconuts were produced in 2009. From this, we know that if each coconut weighs 10 lbs, then roughly 1.0×10¹⁰ coconuts were produced. Now, we could quibble about the actual number. Some grow in the wild which might make the actual number larger. Some coconuts are picked instead of falling, so that might make the actual number smaller. You can argue either way, so let’s stick with this figure just to keep the problem relatively simple.

There are 6.7×10⁹ people in the world, each of which has about 1.5 ft² of area that the coconut could land on giving a total area of about 1.0×10¹⁰ ft².  According to Wikipedia, the total land area in the world is about 1.5×10⁸ km².  If people are randomly distributed across the land area of Earth, then the probability of being hit by a coconut is equal to the fraction of land area that people take up at any given time.  Using Google’s calculator, we get

From this, we suspect that each year roughly 6 out of every one million people get bonked by a coconut.
It’s difficult to say how many people actually get killed by coconuts. From our calculation above and the world population, we can estimate the number of people that get hit by coconuts each year, but not everyone who gets hit from a falling coconut will be killed by it. Assuming all hits were fatal, we could calculate the total number of deaths by multiplying “hits per person” times the “total number of people”, to get

(6×10^-6 hits per person) × (6.7×10⁹ people) = 40,000 fatal hits.

If hits are only fatal 1% of the time, then 400 people will die from coconuts falling. However, this 1% statistic may be off substantially—possibly much more than an order of magnitude—so our estimate is very rough.

It’s difficult to tell how many coconut-induced fatalities will occur, but our estimate of “40,000 coconut hits” suggests that the number of deaths could be substantially higher than 150. If this were the case, then coconuts are certainly not out to get you since they kill fewer people than they would by chance.

We’ve assumed the following:

· Coconuts are 10 lbs on average.
· The mass of all the falling coconuts in the world each year is equivalent to the mass of coconuts produced each year.     
· The probability of a falling coconut hitting a person is equal to the fraction of land area taken up by people.
· Only 1% of coconut hits are fatal.

These assumptions seem reasonable, but that does not mean they are necessarily correct. Coconuts are certainly between 1 and 100 lbs, so the first assumption seems decent. It’s possible the number of falling coconuts each year is off for the reasons stated above. Likewise, the percentage of fatalities could be off by several orders of magnitude. Equating the probability of a falling coconut hitting a person to the fraction of land area taken up by people is a reasonable first guess, but there are factors that might throw this assumption off. For example, perhaps more people live near coconut trees because people like to live in tropical climates. 
 
While our estimate doesn’t have enough precision to answer this question conclusively, this example does illustrate an important point. As Weinstein and Adams describe in their book Guesstimation, estimates generally break up into three “Goldilocks” catagories: too big, too small, and just right. In this case, being “just right” means your estimated result is too close to call. When this happens, you need to put more effort into refining your estimate if you want to draw any conclusions. Refining the coconuts estimate to high precision is beyond the scope of what I can do in a silly blog post, but there is still a valuable lesson to be learned: In estimation as in life, there are times when even the best answer leaves a wide degree of uncertainty and it’s important to acknowledge when we don’t have enough information to draw a conclusion. That said, there are many examples where a test produces results that are so unlikely we can conclude they are not due to random chance.

Lucky Numbers

Anna and I went out for Chinese food in Philadelphia today. As I looked at the lucky numbers in my fortune cookie, I couldn't help but wonder, "If everyone who ate Chinese food today played their lucky numbers in the lottery, what are the chances at least one of them would win?"

Both fortune cookies and lottery numbers usually show about 5 numbers that can range from roughly 1 to 50.  The probability of picking the first number correctly is 5 out of 50.  The probability of picking the second number correctly is 4 out of 49.  The probability of picking the third number correctly is ...  Multiplying these probabilities together, we can find the total probability of finding the right sequence of numbers¹,

P = [(5)! · (50-5)!] / 50! = 4.7×10^-7.

That's about one in two million. I generally go out for Chinese food about once per month, which seems like a reasonable amount for most people.  Taking that as the average and using the fact that there are 3.0×10⁸ Americans, we can estimate the number of people that went out for Chinese today,

# of people going for Chinese = (prob. of going out for Chinese) · (total # of people)

= (1 day / 30 days) · (3.0×10⁸ people)

= 1.0×10⁷ people.

The probability that everyone will will pick the right numbers is P^10,000,000.  Likewise, the probability of everyone picking the wrong number is (1-P)^10,000,000. The probability that at least one person will win is then just

1 - (1-P)^10,000,000

= 0.009.

There's about a 1% chance that if everyone played their lucky fortune cookie numbers at leat one would win.

[1] This is the well known binomial distribution.
[2] I'm assuming the fortune cookie's "lucky numbers" are random and uniformly distributed.

Special Guest Natalie Angier

Today we're pleased to have a question from special guest Natalie Angier.  Ms. Angier is a Pulitzer-prize winning science journalist for the New York Times.  She has authored several books, most recently, The Canon: A Whirligig Tour of the Beautiful Basics of Science.  She writes, "How many leaves are raked up nationwide on an average weekend afternoon in October?"


Before I begin, I have leaf-raking riddle for you.  Without adding or rearranging the words, add punctuation to the following sentence to make it grammatically correct: "A boy raking leaves." The answer is below.

If you're going to rake leaves, you need leaves to rake.  Some states like Arizona are desert-y and won't have many leaves to rake, but most places in the U.S. will have trees that shed.  Even if you live in the right climate, you still need a yard with at least one tree in it.  I'll assume that 1 out of 10 people owns a yard with a tree in it, since it's very likely that the actual number is greater than 1 out of 100 and less than 1 out of 1.  Of these people, some will rake, but many will use a leaf blower or just let the leaves lie.  Using a similar order-of-magnitude argument to the previous one, I'll assume 1 out of 10 people who have leaves rake them.  The average leaf raker might rake his/her leaves once in October, and there are about 8 good leaf-raking October weekend afternoons each year.  There are 3.1×10⁸ people in the United States.   Combining these assumptions, we can estimate that,

# of people raking = (3.1×10⁸ people) · (0.1 tree owners per person)
· (0.1 raker per tree owner)· (0.125 chance of raking now)

= 390,000 people raking leaves each weekend afternoon in October.

But the question specifically asked for the number of leaves raked.  This will clearly depend on the number of leaves a person has in his/her backyard.  According to at least one source, a mature tree can have up to 200,000 leaves.  To confirm this, I looked at leaves strewn across Tappan Square in Oberlin.  The mean separation was about 6 inches between leaves.  If you spread them out over a reasonably sized lawn (about 1/5 of an acre), you get about 200,000 leaves.  If each raker rakes this many leaves one a weekend, there will be,

# of leaves = (200,000 leaves per raker) · (390,000 raker)

= 7.8×10¹⁰ leaves raked.

That's 78 billion leaves raked nationwide each afternoon in October.  Thanks for the great question, Ms. Angier! 

For those wondering about my earlier grammatical riddle, the correct answer is "A boy, raking, leaves."

A Relatively Good Calculation

In 1905, Einstein published his special theory of relativity.  The most well-known part of this theory is almost certainly the famous E=mc² equation that predicted a future with nuclear bombs and atomic energy, but this is not the only surprising prediction.  The theory also predicted that objects shrink when they move really fast.¹  After hearing a professor describe this strange and fascinating phenomenon, I wondered two things.   First, what was Einstein smoking?  Second, if I ran really fast, would I be able to see atoms?  How fast does a person have to run to be atom sized?

According to special relativity, the length of a moving object is equal to its original length times an extra factor

L' = L [1 – (v/c)²]^0.5.

Here, c is the speed of light, L' is the length of the object when it's moving, L is the length of the object when it's not moving, and v is the velocity at which it's moving.  Our new length L' will be about 10^-10 m or roughly the size of an atom.  Our original length will be about 1.5 m.  We can solve for v

v = c [1 – (L'/L)² ]^0.5

 = (3.0×10⁸ m/s)[1 – (10^-10 m / 1.5 m)² ]^0.5

= 0.9999999999999999999977778 c.

You would need to move very close to the speed of light to be atom sized.

[1] This phenomenon is called "length contraction".

Lieutenant Commander Banana Clip

I saw this on Totally Looks Like the other day, and it certainly brought back a few repressed memories.  When I was a kid, I definitely wore a banana clip pretending to be Geordi LaForge, and my family wasn't even big Star Trek fans.  How many people in the U.S. wore banana clips over their eyes pretending to be Lieutenant Commander Geordi LaForge?

Apparently, I was not alone.  It strikes me that role playing as Geordi takes three things: a certain level of maturity, Star Trek knowledge, and banana clips.  I chose the phrase "a certain level of maturity" carefully.  I'm pretty sure most 2-year olds and 90-year olds didn't do this.  You need just the right amount of maturity.  With too much, you just think it's silly.  With too little, you can't see how genius the banana clip Geordi visor really is.  I suspect about 1/3 of the population has the appropriate maturity level. 

After maturity, you need Star Trek knowledge.  As I said earlier, I wasn't even that big of a Star Trek fan, but I still rocked the Geordi visor.  That said, I suspect at least 1 out of 10 people had my level of Star Trek knowledge. 

The final ingredient is, of course, banana clips.  These were fairly ubiquitous in the late 80s and early 90s when Star Trek the Next Generation was popular.  As a reasonable guess, I would say at least at least 1 out of 10 people had access to banana clips at this time.

Using the assumptions above, we can estimate that about 1 out of every 300 Americans, or about one-million people pulled off the Geordi look.  LeVar Burton, you deserve a commission from the banana clip companies.

[1] Doh!  "Jordi LaForge" should be spelled "Geordi La Forge."  Thank you, to my student Kara Kundert for the correction!

Safety in Numbers

It’s commonly said that airplanes are the safest mode of transportation. It’s true that more people die in car crashes than plane crashes each year, but most people also drive more often than fly.  On a "per trip" basis, which mode of transportation is safer?

The question did not specify whether we were considering just crashes in the U.S. or in the entire world.  Assuming crashes are equally likely in all parts of the world, the “fatalities to trips” ratio should be about the same in both cases.  Just make sure you are consistent (i.e. don’t divide the number of U.S. plane crash victims by the total number of flights in the world.)

According to “Ask a Scientist” ¹, in the U.S. there are

“…about 40,000 deaths per year in automobile accidents vs. about 200 in air transport.”

You can check these numbers against other references to make sure they are accurate.  To answer the question, we need to estimate how often a person flies and how often a person drives.  On average, we might guess that an American flies about once per year.  This is reasonable since you’d certainly expect people to fly more than once every 10 years and less than once per month.  In contrast, most of us drive (or ride in) a car about twice per day, even if it’s just to get to and from work or school.  Since there are 3.1×10⁸ Americans, this means there are

# flights per year = (1 flight per American per year) × (3.1×10⁸ Americans)

= 3.1×10⁸ flights per year,

and

# car trips per year= (2×365 car trips per Amer. per year) × (3.1×10⁸ Americans)

= 2.2×10¹¹ car trips per year.

The fraction of deaths is then just,

Fraction = (# deaths per year) / (# trips per year)

= (200 deaths per year) / (3.1×10⁸ plane trips per year)

6.5×10^-7 deaths per trip

or

= (40,000 deaths per year) / (2.2×10¹¹ car trips per year)

1.8×10^-6 deaths per trip.

You are 2-3 times more likely to die in a car trip than a plane trip, so according to our numbers plane travel is still safer.  Given the precision of the estimation, it’s possible that other reasonable assumptions would come up with the opposite result, since both figures are within an order of magnitude of each other.

[1] You can find this stat and a nice discussion of the topic at here.  Their numbers are different partly because they’re talking about fatalities per mile not fatalities per trip.

I Can Haz Estimation?

If you’re like me, there’s nothing you like better after a long work day than kicking back and looking at some grammatically incorrect cats.  And who better to provide all your LOLcat needs than The Cheezburger Network?  In honor of my favorite intertubes diversion, how many pictures of cats are on the entire Internet?

This is a difficult problem.  That being the case, we expect our estimate will likely be very different from the actual number.  In cases like these, it is helpful to determine what the wrong answers are.  To do this, you need to calculate upper and lower bounds.  To start, let’s estimate how many people are cat owners.  You might guess that 10% of people own cats.  Is this reasonable?  Well the actual number is certainly less than 100% of people, and very likely greater than 1% of people.  If you asked a 100 of your friends and family, at least one of them probably has a cat.  The next question you could reasonably ask is how many cats do cat owners typically have.  Some people have 10 cats, while others only have one.  A reasonable guess is 2 cats per owner. 

Now comes the tricky part.  What fraction of people put pictures of their cat on the web?  Many of you are tech savvy and have accounts on Facebook, Flickr, etc.  If you have a cat, you very likely have at least one picture of it on the web somewhere.  But what about your parents, grandparents, and friends that may not be as tech savvy as you?  What about people in developing nations that may not have Internet access?  If you average over everyone including people in other countries, what percentage of cat owners will post pictures on the web?  To be safe, lets say 10% again.  Is this reasonable?  As before, the actual number will certainly be less than 100%.  Will it be great than 1%?  Possibly, but there are a lot of people that don’t like putting their information up on the Internet and the ones that do might not put their cat on the Internet.  To be safe, let’s put the lower bound at 0.1%.  We can summarize what we know in a chart like the one below:

	Upper	Guess	Lower
% of people that own cats	100%	10%	1%
Cats per owner	10	2	1
% of owners that put cat on web	100%	10%	0.1%
Pictures per cat	100	10	1
World population	6.7x109	6.7x109	6.7x109
# of cat pictures	6.7x1013	1.3x109	6.0x104

Our upper and lower bounds are very far apart (i.e. 9 orders of magnitude).  The answer certainly lies between them, but is there any way we can tighten these bounds?

Let’s try calculating the answer in a different way.  According to Netcraft, there are 2.5x10¹⁰ web pages as of a few years ago. As an upper bound, we might say that at most, every page has 10 cat pictures on it.  Even in this extreme case, there would only be 2.5x10¹¹ cat pictures, so we’ve effectively reduced our upper bound by a factor of 100. 

Now lets see if we can adjust our lower bound.  A quick Google image search for the word “cat” pulls up about 1.5x10⁸ results.  Looking at the pictures on the first page, we can see that the vast majority of the images are of cats, but occasionally you get one that’s not a cat.  This is very rare.  In fact, it seems much more likely that Google is missing cat pictures since it doesn’t have access to private photos on sites like Facebook and Flickr.    For this reason, we could probably take the Google result as a lower bound.  Just to be extra careful, let’s say the lower bound is 1/10^th the number of “cat” images found by Google.

Our final results:

Upper Bound -- 2.5x10¹¹ cat pictures

Actual Estimate – 1.3x10⁹

Lower Bound -- 1.5x10⁷