Tuesday, November 16, 2010

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.

Sunday, October 24, 2010

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 P10,000,000.  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.

Sunday, October 17, 2010

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."

Thursday, October 14, 2010

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".

Tuesday, October 12, 2010

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!

Thursday, September 30, 2010

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,


# 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


= (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.

Sunday, September 26, 2010

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:

% of people that own cats
Cats per owner
% of owners that put cat on web
Pictures per cat
World population
# of cat pictures

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 1/10th 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

Tuesday, September 21, 2010

My Eyes Are Buggin’

I had an interesting conversation with one of my physics students recently.  She had a dance class right before she came to lab.

Her: I hate dancing with glasses.

Me: Because they fall off when you’re doing the turn-y things?

Her: Yes, it’s so much easier with contacts.

Me: You know, you can calculate how fast you’d have to turn to get the contacts to pop out.

How fast would a dancer have to spin to get her contacts to pop out?

Contact lenses weigh about 0.05 g and are about 1.0 cm2 in area.  They are held on mostly by suction.  Suction occurs because there is a partial vacuum created under the contact as it's pulled away from the eye.  To pop out, a contact needs to overcome air pressure, which for our atmosphere is about 100 kPa.  As the dancer spins, her contacts move in a circle with roughly a 10 cm radius.  The force to keep an object in uniform circular motion is given by

force = (mass) · (velocity)2/ (radius).

The maximum this force can be before the contact pops out is determine by atmospheric pressure

force = (pressure) · (area).

Solving for the velocity, we get

velocity = [(pressure) · (area) · (radius) / (mass)]1/2
= [(100 kPa) · (1 cm2) · (10 cm) / (0.05 g)]1/2
= 140 m/s.

A dancer would have to rotate about 1400 per second to get her contacts to pop out.

Smile! You’re the Monocle Smile Contest Winner!

Hello, Ladies.  How are you?  Fantastic.  We have a winner!

The Question: If he worked nonstop for the rest of his life, how many personalized Youtube videos could Mr. Mustafa make?

Born February 11, 1974, Isaiah Mustafa is currently 36 years sexy.  He’s young and he’s in shape, the latter of which makes longevity more likely.  As best I can tell, he only has one major thing going against him: he played NFL football.  However, his career was fairly short and he was a wide receiver, so he’s not as likely to have the same health issues as a 12-year 350 lb lineman.  One simple way to solve this is to subtract his current age from his life expectancy.  According to Wikipedia, Americans are 38th in the world for life expectancy with and average lifespan of about 78.2 years.  Using Wolfram Alpha, we know that Mr. Mustafa was 36.6 years old on the contest deadline.  From this we can compute how many years he has left,

78.2 years – 36.6 years = 41.7 years.

Each personalized video appears to be about 30 seconds in length.  If he worked nonstop1 making 30-second videos for 41.7 years, he would have a total of about 44 million videos.

Congratulations to our winner, Joey Reichert!

[1] Some might argue that we haven’t taken into account sleeping.  True.  If he did sleep, our final number would only be one third as big.  But the original question did say “nonstop.”  “But wouldn’t he die younger if he never slept?” you ask, suspiciously.  Also true.  But we could always film him while he slept, and which of you ladies out there wouldn’t appreciate that?

Tuesday, September 14, 2010

A Cure for Global Warming?

I came across the above photo on Reddit the other day (original link here).  As one Redditor so aptly put it, “The stupid, it burns!!!”  Interestingly, there are multiple layers of stupid here.  Like a giant stupid onion.  Leaving aside the fact that water dumped into a sink would eventually wash back into the ocean, how many buckets of water would each Earthling need to dump to cancel out the rising sea levels?

According to the Wikipedia entry for "Current sea level rise", predicted values for the ocean rise due to global warming range anywhere from 90 to 880 mm.  I'll take 90 mm and just calculate the lower bound.  The total area of the oceans is roughly 3.4 ×1014 m2, meaning that the total volume is about 3.0×1013 m3.   One bucket might hold 50 L and there are 6.7 ×109 people in the world.  Form this, we can compute the total number of buckets needed,
# of buckets = (total volume) / [ (volume per bucket per person) × (number of people)]
  = (3.0×1013 m3) / [ (50 L per bucket per person) × (6.7×109 people)]
= 90,000 buckets.

Even neglecting the fact that the water will just wash back into the ocean, this method would still require every person on the planet to dump 90,000 buckets of sea water down their sink.  As Shakespeare might once have said, "The stupid, it doth burn."   

Friday, September 10, 2010

Bounding Beck

Sarah Palin said there were “hundreds and hundreds of thousands of people.”  Glenn Beck put the estimate at 500,000 people.  Congresswoman Michelle Bachmann stated that she wouldn’t let anyone tell her there were less than 1 million people.  AirPhotosLive.com estimated only 87,000 people based on aerial pictures it took.  How many people were at the Glenn Beck rally?

As you can see from the photos on AirPhotosLive.com, most of the people assembled along the Lincoln Memorial Reflecting Pool.  According to Wikipedia, the pool is approximately 618 m long by 51 m wide.  Judging from the photos, attendees seem to cover an area between 3 and 4 reflecting pools wide, or roughly between 150 m and 200 m.  Multiplying by the reflecting pool length, this means the total area taken up by the attendees is between 92,700 m2 and 125,000 m2.  To be sure we’re not forgetting anyone, we’ll count the attendees that spilled over near the Washington monument and those that got a front row seat.   This looks like no more than 20% of the total number of people, so we’ll say they cover a total area of at least 90,000 mand at most 150,000 m2

Looking at the aerial photo, there’s a whole lot of green grass you can see between people.  This makes sense since most people don’t like to be squished and you need some extra room so that people can move around.  Each person needs at least 1.0 m2 of space to feel comfortable.  Judging from the sparse parts of the crowd, there’s appears to be no more than 10 m2 of space between people on average.  The average density certainly lies between these two extremes.  From these densities, we can use the equation

number of people = (density of people) × (area)

to compute an upper bound,

number of people = (1.0 person per m2) × (150,000 m2)
= 150,000 people

and lower bound,

number of people = (0.1 people per m2) × (90,000 m2)
9,000 people.

As you can see, the “87,000 people” estimate from AirPhotosLive lies right in the middle of the two extremes.  It is certainly a reasonable estimate, and it brings up an important point.  One way to test the accuracy of your estimate is using upper and lower bounds.  By using the largest and smallest realistic numbers in your estimation, you can put a cap on what the actual number must be.  We know there’s certainly more than our lower bound of 9,000 people and certainly less that our upper bound of 150,000 people. 

By using upper and lower bounds, we also see that our politicians and pundits are either lying or possess a certain degree of innumeracy. Of these two options, I sincerely hope it’s lying.  I’ve grown accustomed to politicians bending the truth, but I’m absolutely terrified by the thought that people who have gained this much influence can’t do middle school math. This does beg one to ask, “Why even bother lying about numbers like this?”  For one, false claims like this are easy to debunk even with mediocre math skills.  In addition, 87,000 people at a rally is nothing to sneeze at, so why exaggerate? 

Finally, I should note that I had some hesitation when writing this post.  This is not intended to be a political blog, and I’m not here to spout my personal beliefs.  The fact is that no matter what side of the political spectrum they’re on, rally organizers tend to exaggerate numbers to make it seem like they’re more popular than they really are, and it’s our duty as rationally thinking citizens to call them on it.

Thursday, August 19, 2010

On Birthdays and Monkey Poo

It’s my sister Laurie’s birthday today, and much as it pains me to say it, it’s never easy being the four-years-younger brother of a super genius.  When you’re graduating middle school, she’s graduating high school.  When you’re graduating high school, she’s graduating college.  When you’re getting a BS from MIT, she’s getting a PhD from Harvard.  When you publish your first book, she’s giving a TED talk and getting tenure at Yale.  Even today as she grows ever older and more decrepit, turning an age I can remember our mom and dad being, I can still feel that familiar tingle of our (usually) healthy sibling rivalry.  If there's one thing I take solace in, it's that I've never had to worry about getting monkey poo in my hair.  What’s the probability that at some point in her career as a primate researcher, Dr. Santos has gotten monkey poo in her hair?

Cayo Santiago is one of the places my sister does research on primates.  Located southeast of Punta Santiago, Puerto Rico, the 600 m by 400 m (240,000 m2) island is home to about 950 Rhesus monkeys.  As can be seen from the photo on the right, Cayo is a fairly lush island featuring lots of trees monkeys can climb.1  I’ll assume a typical monkey spends about 50% of its time climbing, meaning that at any moment there are about 475 monkeys in the trees.

Each year, Laurie generally spends about 2-4 weeks on Cayo, and she has done so for about 10 years.  If she works 8 hours per day, she’ll have spent up to 93 days on the island.  Assuming a Rhesus monkey’s digestive schedule is similar to that of a human, each monkey probably “uses the bathroom” at least once per day.  From this and the information above, we can estimate the total number of high altitude monkey poos taken while Laurie has been on the island

(# of poos per day per monkey) · (# of monkeys) · (# of days)
= (1.0 poo per day per monkey) · (475 monkeys) · (93 days)
= 44,175 poos.

From a bird’s—er—monkey’s-eye view, her hair will appear to have about 1.0 ft2 of area.  Our monkey bombardier will have to hit this spot if Laurie is to spend the rest of her night shampooing.  From this we can calculate the probability of a random poo landing in her hair,

P = (area of hair) / (total area)
= (1.0 ft2) / (240,000 m2)
= 3.9×10-7.

The probability of a poo not landing in her hair is, of course, 1 – P.  The probability that none of the 44,175 monkey poos have landed in her hair is (1 – P) 44,175 ~ 0.98.  This means there’s a 2% chance that a monkey has poo-ed on my sister’s hair at least once.

Happy Birthday, Laurie!  You’re as good a science role model and an even better sister than I could reasonably have hoped for. 

[1] For simplicity, I’ll assume that 100% of the island is covered in trees.

Saturday, August 7, 2010

The Monocle Smile Contest

Look at your math.  Now back at mine.  Now back at your math.  Now back to mine.  Sadly, yours isn’t mine.  But if you enter the Diary of Numbers Estimation Contest then your math could seem like it’s mine.  Look down.  Back up.  You’ve won a book featuring the math your math could look like.  Anything is possible when you enter the Diary of Numbers Estimation Contest.  I’m on a blog.

The Question: Isaiah Mustafa (aka the sexy Old Spice man) reached meme status because of his work in a very funny Old Spice commercial.  Recently, he created a series of humorous personalized Youtube videos featuring his character from the original commercial.  If he worked nonstop for the rest of his life, how many personalized Youtube videos could Mr. Mustafa make?

Rules:  You can win a free copy of How Many Licks?  To enter, estimate an answer to the question below and send it to “aaron at aaronsantos period com.” If your answer is closest to my estimate1, I'll mail you a free, signed copy of How Many Licks?  To be eligible, you must submit your entry on or before September 15, 2010.  Don't worry; I won't spam you or share your email with any third parties.

[1] I know, I know.  How do I know my answer is correct?  I don’t.  I make no pretenses that my answer is correct or even close. Your answer may very well be a better estimate than mine. In fact, your estimate may even be exactly right and you still may not win the contest if somebody else's answer is closer to mine. Sorry about that. This is the best way I could come up with to pick a winner and I'm not changing it now. Like any good game, there's an element of luck required even if you do have great skill. With that disclaimer out of the way, good luck and happy calculatings!

Friday, July 30, 2010


Check out my super smart sister giving her TED talk.

Tuesday, July 27, 2010

20,000 Kissing Minutes

A new Dentyne commercial claims we spend 20,000 minutes kissing.  What kind of estimator would I be if I didn't check their math?  How many minutes does the average person spend kissing?

Couples often kiss each other several times per day1.  If we assume the average day consists of 30 seconds of kissing and the average persons spends 60 years in a relationship, then we can estimate the number of minutes spent kissing in a lifetime:

time kissing = (time kissing per day) · (days per year) · (years in relationships)
= (30 s per day) · (365 days per year) · (60 years in relationships)
= 11,000 minutes.

That’s an average of 11,000 minutes spent kissing in a lifetime.  The commercial’s claim is well within what one might reasonably expect.  Good job, Dentyne and “stub”.

[1] It’s can be a lot more if they make whoopee.

Tuesday, July 20, 2010

Special Guest Saul Griffith

This week's question comes from special guest, Dr. Saul Griffith.  In addition to being a 2007 MacArthur Fellow, Saul has received numerous awards for his inventions including the National Inventors Hall of Fame, Collegiate Inventor's award, and the Lemelson-MIT Student prize.  When not inventing, he co-authors children's comic books called HowToons about building your own science and engineering gadgets.  He is also a columnist and contributor to Make and Craft magazines.  Saul writes,

"How many shot glasses of oil per day per person for the average American's personal contribution to the gulf oil spill?"

According to Wikipedia, the Deepwater Horizon oil spill released up to 16,000 m3 of oil per day.  There are roughly 300 million Americans.  One shot glass contains about 60 mL of volume.   From this we can easily compute the contribution per American,

# of shot glasses = (vol. of oil per day) / [ (# of Americans) · (vol. per shot glass)]
= (16,000 m3 per day) / [ (300 million people) · (60 mL)]
= 1.0 glass per day per person.

The oil spill was basically the equivalent of having every American dump a shot glass of oil into the gulf every day.  Thanks for the question, Saul!

Monday, July 19, 2010

Hamster Contest Winner

We have a winner for our “Hamster-Powered Mansions” estimation contest.  The question: How many buff hamsters would it take to completely power a mansion? 

Mansions come in a wide range of sizes, and each will have different energy requirements.  According to Factcheck.org, Al Gore’s mansion used 191,000 kilowatt-hours in 2006.  Dividing by one year, we can compute his average power consumption to be 22,000 W. 

From this MAKE magazine video1, we can see that a hamster is at least powerful enough to light up an LED, but this is only a lower bound since he might be able to power even more LEDs if we connected them in the circuit.  According to Otherpower.com, their hamster Skippy had no trouble lighting up 6 LEDs, and they estimate he should be able to power 200.  Being fairly conservative, I’ll assume our hamsters can power 50 LEDs.  According to Wikipedia, the voltage drop across one LED is anywhere from 1.5-4.5 V and the current should be between 1 and 20 mA.  Assuming the LED acts like an Ohmic resistor,2 we can estimate the electrical power created by a wheel-spinning hamster,

power = (# of resistors) · (current) · (voltage drop)
= (50) · (10 mA) · (3.0 V)
= 1.5 W per hamster.

From this and the power consumption of Al Gore’s mansion given above, we can estimate the number of hamster’s you’d need to power a mansion,

# of hamsters = (power per mansion) / (power per hamster)
= (22,000 W) / (1.5 W per hamster)
= 15,000 hamsters.

You’d need about 15,000 hamsters to power a mansion.  Congratulations to our winner Bryan Merrill.  Bryan will be receiving a free copy of How Many Licks?  Keep reading for our next contest.

[1] Speaking of MAKE magazine, check out Maker Faire in Detroit July 31 and August 1.
[2] I’ve assumed LEDs require 3.0 V and 10 mA.

Tuesday, July 6, 2010

Comma Chameleon

Anyone who reads Diary of Numbers regularly knows I’m far from a grammar nazi.  Despite my best efforts, my grammar is typically atrocious.  There is, however, one thing I’m very particular on.  Even though both of the following are equally correct, I much prefer the latter:

(1) “Dorothy was afraid of lions, tigers and bears.”
(2) “Dorothy was afraid of lions, tigers, and bears.”

Recently, something struck me: since both usages are perfectly correct, I’ve been wasting printer ink all my life.  How many cartridges of ink do you waste in your lifetime by always including the extra comma?

This is a tricky problem because the amount of printer ink a person uses varies considerably depending on his/her career, hobbies, etc.  If you’re a novelist, you’ll probably be printing a lot more commas than a rodeo clown.  Moreover, the problem’s a little wishy-washy because of how I originally phrased it1.  A better way to phrase the question would have been, “How many cartridges of ink would you waste in your lifetime if everyone included the extra comma?”

Some days I print 100 pages, other days I print none.  On average, I probably print about one full page per day2.  Lists appear about once every 10 pages3.  This means I’d be printing one superfluous comma every 10 days.  At this rate, I’d produce 2900 needless commas over an 80-year lifespan.

According to HP’s website, the average HP LaserJet Q2612A Black Print Cartridge yields 2000 pages worth of ink.  A single comma requires about one-fourth the ink of a letter and according to one of my MS-Word documents, there are about 2000 letters per page.  This means one page worth of ink is equivalent to about 8000 commas.  From this and the info above, we can estimate the number of printer cartridges one will waste in a lifetime,

# ink cartridges = (# ink cartridges per page) · (# pages per comma) · (# of commas)
= (1 ink cartridge per 2000 pages) · (1 page per 8000 commas) · (2900 commas)
= 0.00018 ink cartridges.

Over the course of your lifetime, you will waste 1/5,000th of an ink cartridge by including the extra comma.  Even if everyone in America included the extra comma, we’d waste only about 680 printer cartridges each year.

After posting this problem, a good friend, who had previously worked for Yale University Press as a copy editor, told me that Yale never included the extra comma to save ink!

[1] Sorry about that.
[2] This is more than printing one page every ten days and less than printing 10 pages per day, which seems like reasonably good bounds for and order of magnitude estimate.
[3] Again, this estimate is reasonably good if you consider upper and lower bounds.  A list certainly does not appear on every page but you’ll probably see more than one on a hundred pages of text.