# How Do You Solve a Problem Like Fermi? Fun with Fermi Problems

Quick! How many people in the world have exactly the same number of hairs on their head as you do? No fancy analysis, no consulting Google — make a quick estimate off the top of your head! (HAHAHAHA) Pointless question with no possible way to find the correct answer — intriguing or annoying? If you’re currently working out the math on a cocktail napkin, you’re going to enjoy this article.

Maybe you know where I’m going with this, if you’ve heard of Fermi Problems – cool little science and math problems, where the spirit is to do as little real work as possible. Ideally, you come to an accurate answer for oddball question based purely on reasoning and simple approximations. And in my opinion, the weirder the problem, the better. Such as — how many barrels of oil would a typical dinosaur make? Or, did smoking kill more WWII vets than the actual war itself? You can find those at the website that inspired this article, The Straight Dope, where Cecil consistently comes up with the most fascinating examples of Fermi Problems. But while Cecil may have perfected them, he didn’t invent them — that honor goes to Enrico Fermi, esteemed physicist from the prewar era who was famous for coming up with them. The prototypical example is Fermi making an estimate of the number of piano tuners in Chicago, getting to within 20% or so of the right answer with absolutely no data at all. Or (to choose a more ominous topic area), accurately estimating the power of the first atomic bomb test just by observing pieces of paper blown in the blast, a story illustrating how Fermi’s talents became legend even among his fellow physicists. But let’s leave the ominous topic of nuclear fission as abruptly as we approached it, and come up with some of our own Fermi-like problems. In this article, I give a tour of the most interesting and unusual Fermi problems I could come up with — and for those interested, I provide the math. For those of you just interested in the cool little science factoids,** just skip ahead to the red parts**.

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**DNA Length **

One of the startling things you learn doing these kind of Fermi problems is how much *length* there is in a *volume*. Yep, that’s exactly what I meant to say – how much length in a volume. MS Word is not liking the grammar of that statement, but bear with me. Imagine a big jar full of pennies. Now imagine taking out all the pennies and laying them side-by-side in a line. No, don’t just imagine the concept – I want you to visualize every single tedious step of the process… done? Good. Now look how *long* the line of pennies came out to be – a lot longer than you probably would have guessed. When you take something linear and pack it into a 3D volume, you get a surprising amount of compression of space. Perhaps this seems a little dull compared to the other topics on timeblimp, so let me illustrate with an example – the amount of DNA contained in your cells.

As it happens, each chromosome of your DNA is a single gigantic molecule – and it’s just one long line, tangled and wound around itself to compactify into your cells’ nuclei. Let’s say you took out all the DNA from each nucleus of every cell in your body, and laid all the DNA molecules end-toend. How long would the DNA stretch? 20 feet? 100 feet? Sure, it might sting for a bit. So let’s not do it to you – let’s unravel the DNA from the active roster of the Pittsburgh Steelers. Tease out the DNA from all 53 guys on the Steelers roster, and carefully line each piece of DNA end-to-end. I’ll stay over here at one end of this disgusting line of DNA – just give me a shout when you’re done lining them all up. How far away do you think you’ll be when you’re done? Are you so far away that you should bring a cell phone? Could we be talking *miles?*

It turns out, we’re talking *light years*! By my math, **the total length of DNA contained in one football team works out to be about half of a light year**. A light year! A distance so far, that it takes a beam of light a whole year to traverse it! And that means the total length of DNA in just one person is *still* an astronomical length. Literally – in just one Steeler, the total DNA length is about 700 times the distance between the Earth and the Sun (an “Astronomical Unit”).

*The math*

Here’s the first startling fact: if you take all the DNA in just *one human cell*, the total length of that DNA is about a meter. From something so microscopic you can’t see it with an unaided eye, you get over three feet of stuff. A rough estimate is that there are about 50 trillion cells in a typical human’s body. (I rounded up to a nice even 100 trillion, since the Steelers are so big. If that seems unfair, add the practice squad to the roster.) Holy crap! And there’s more than just DNA in a typical cell. If you’re feeling adventurous, work out the math for lining up *everything* in a human body (atom by atom) in a line, and see how long that line is.

**Making a Humanball**

Let’s continue our marveling at the strange compactification of space when you go from length to volume. But this time let’s go in the opposite direction, and pack together something that is usually spread out and diffuse. Say, humans. If you were to pack all humans on earth into one giant ball, how big would that ball be? We won’t allow any air gaps – squeeze the ball of humans tightly, so we’re all packed in close. By now, you’ve absorbed the deep lessons inherent in the DNA length example, and you know it’s likely smaller than intuition would first say. But it’s still a pretty big ball of annoyed humans. Turns out, **all the humans in the world can be packed together into a ball about half a mile across. **

*The math*

I got my answer by using sophisticated laser-based optical scanning techniques to measure the volume of every human on earth. HA! To hell with that – I took the easy way out, and divided the average human weight (I assumed 70 kg) by the average density of a human body (1010 kg per m^3). (This is actually pretty close to the density of water, which is 1000 kg per m^3, a handy fact to have if you’re ever forced to do Fermi problems without being able to look up random data like the exact density of human flesh). Now I know the volume of a human body, sidestepping the tricky problem of modeling the volume of protuberances like fingers, and I can simply add up the volumes for all 7 billion people on earth and convert that to a sphere.

**What force of gravity do we feel from Neptune?**

Everything in the universe that has mass will exert a gravitational force on all other mass-riffic stuff. That means the pile of laundry in the corner actually has its own gravity, and is pulling on you right now. In fact, the lump of laundry in the corner of *my* hallway is also pulling on *you*, right now. But as it happens, gravity is a really weak force, so we don’t notice it unless enough mass accumulates to make the force appreciable. So the force from the planet we stand on greatly outweighs the force from anything else in the room.

What about large things that are far away? The force of gravity from the moon isn’t really noticeable to us, though you can see evidence of it in the form of tides. (Yes, the laundry detergent. The moon’s force of gravity is responsible for that particular brand of laundry soap.) What about the other planets? They’re huge, but they’re far away – what is the force of gravity we feel from Neptune, for example? Work out the math, and you find that **the force of gravity we feel from Neptune is about equal to the force of gravity that one of your eyeballs exerts on the other eyeball.**

*The math*

Neptune is about 1 x 10^26 kg in mass, and is about 4.5 x 10^9 km from the sun. It turns out that from Neptune’s point of view, Earth’s orbit is tiny – so I approximate the distance to Neptune to just be its orbital radius, conveniently ignoring all the details about where the Earth is relative to Neptune. From here, it’s some easy math to figure out the force of gravity that Neptune exerts on us. A typical human eyeball, meanwhile, is about 7.5 grams (according to Wikipedia – I didn’t test this out myself). Now, we set their forces of gravity equal to find the distance from an eyeball at which the force equals the pull from Neptune, and we get the distance where the two pulls are equal – R_eye = (Mass_eye / Mass_Neptune)^1/2 * R_Neptune. Put in all the data, and you get that R_{eye} is about 3.9 centimeters – so anything 3.9 centimeters away from your eyeball feels the same tug as from Neptune, approximately. And it just so happens that the inter-eye distance for humans is about 4 – 5 cm. Close enough for this webpage! Of course we’re ignoring the gravity from the rest of the stuff in your body, e.g. the skull in which the eyeball is embedded. So while the math here is technically true, you wouldn’t be able to put it to practical use unless you have a disembodied floating eyeball and a Neptune.

**How big is a Mole? How big is a Mole of Moles?**

Lots of words have more than one definition, but does any word have a greater diversity of meanings than “mole”? The common two definitions are 1) a small rodent creature that burrows in your lawn, and 2) a unit of measure meaning 6.02×10^23 of something. We usually learn about the second definition in chemistry class, because a mole is a convenient way of measuring something microscopic in a macroscopic way. It’s easier to say “4 moles of carbon” than “4 quintillion zillion molecules of carbon.” But, let’s say you had a mole of moles – you literally had 6.02×10^23 of the little shrew-like rodent. How many moles is that?

Let’s go with the Eastern Mole, which is the first to come up on Wikipedia for “mole”. Moles are about 16 cm long (though that includes the tail) and weighs 75 grams. Hmm, I thought they were smaller… so one mole weighs as much as about ten human eyeballs. Good to know. That means **a mole (6.02 x 10^23) of moles would be a giant ball of moles bigger than the moon. **Ick! And not so fun for the moles in the center – you thought we humans had it bad, in the core of a half-mile high humanball? Try being the mole in the middle of that thing. It would look romantic, shining up in the sky in summer evenings, faintly squeaking…

*The math*

We can use the same mass-and-density trick as described earlier in the ball of humans to determine that the volume of 6.02×10^23 moles, packed tightly together with no gaps, is 4.47 x 10^19 m^3. If this volume were in a perfect sphere, we can solve for the radius it must have. Interestingly, the ball will be larger than the moon (4400 km diameter compared to 3700 for the moon), but will weigh less (4.5 x 10^22 kg, compared to 7.35 x 10^22 kg for the moon). Mole flesh, of course, is less dense than moon rock. Want to push this idea further? Try figuring out how many people would be employed by a gogol of Google’s. (But let’s not pack them all into a ball – that’s gross.)

**How big are the bytes on one of those old 5 inch floppy disks?**

Remember floppy disks? Humor me, junior, and pretend that you do. If it rings a bell at all, you probably remember the little 3-inch disks that some older computers still have drives for. But if you’re really old, like me, you’ll remember the 5-inch floppy disks that actually were floppy. They were made of a disc of cellophane-like magnetic storage in a thick plastic envelope, so they felt kind of like a thick piece of junk mail. Very reassuring when your entire PhD thesis is stored on one.

The first iterations of the 5 inch disk could hold 160 kb of data, which sounds laughable now in the era of terabyte thumb drives. So little storage in such a big package? HA! I bet you could *see* the bits! Wait… could you? How much room would each bit get? Given how much surface area the disk contained, and how many bits were stored, could the bits theoretically be visible to the naked eye?

Astoundingly, yes – **on those old 5-inch floppy disks, each bit gets about a quarter of a square millimeter of room, and so theoretically could be seen with the naked eye!**

*The math*

The memory on these old discs gets written onto a circle of diameter 132 mm, with a hole in the middle of about a quarter of that. That gives us a total usable surface area of about 13,000 mm^2. Divide by the number of bytes (160 x 1024), and you get 0.07 mm^2 per byte. That’s just a smidge more area than a square of ¼ mm on a side. And the size limit of an object that can be seen with the naked eye is about a tenth of a millimeter (at typical reading distance). Of course in reality the bits surely weren’t deposited in neat little squares, so who knows if they were actually visible. But we won’t let details like physical reality distract us from a cool result.

**Exponentially Expanding March Madness**

Every year I try to pick some random part of the calendar, like mid-July or Thanksgiving, and ask my friends how their NCAA basketball brackets are doing. Why? Because every March, I have to read my friends incessant whining on facebook and twitter about the status of their brackets. “Oh, my brackets are in trouble!” “My brackets are in flames!” “My brackets are a dystopian wasteland, where rival gangs roam the deserts fighting over precious reserves of gasoline!” Ugh, enough already. Like a proper snobby nerd, I shower my disdain in my smartass postings months too early.

But what if the NCAA tournament did start earlier? Say, we keep up the same tournament pace, but add enough rounds on the beginning to start a couple months earlier? We’d have to add enough hypothetical teams to append more elimination rounds on the front of the tournament – for example, adding another round would require a total of 128 teams, instead of the 64 that the tournament currently starts with.

Here’s where we get into the fun of exponential expansion – each round we add doubles the required number of teams, so this blows up fast. **If we doubled the length of the NCAA tournament to 2 months, we’d need 4096 teams**. The current tournament marches through 6 rounds in 4 weeks, so doubling it to 12 rounds would require a lot of players. In fact, assuming an average of 15 players per college basketball team, **starting the NCAA basketball tournament on Thanksgiving instead of early March would require 2^28 teams, totaling about 4 billion players.** Given there are only 7 billion people on earth, this means we need everyone to pitch in and volunteer. Who wants to join my team? Our school colors are going to be mauve and dark mauve, and our mascot is a graphing calculator.

**Catch Some Air — On the Moon!**

This one is guiltily recycled from one of my book reviews, Insultingly Stupid Movie Physics. In my review, I offered up a couple free ideas to be used in the next sci-fi blockbuster, dramatic yet scientifically-sound concepts that I’m proud to see absolutely no one has used yet. One of them I repeat here, because it gives me a chance to thrown in some Dukes-of-Hazzard-On-The-Moon jokes.

In every (and I mean EVERY) episode of the Dukes of Hazzard, them Duke boys wound up jumping a ramp in the General Lee — there was always some conveniently-placed flatbed truck that for some reason was tilted at just the perfect angle to serve as a perfect car ramp, and the Duke’s would invariably catch some air off it, leaving Enos in their dust. The flatbed was always hauling bales of hay. Where’s all this hay needed in downtown Hazzard? And since when do flatbed trucks tilt that way? And the deputy’s name was ENOS? What the everloving hell?

Aaaaanyway. The Duke boys always caught about 20 feet of air before landing safely and speeding away. (Was the General Lee’s suspension made of adamantium or something? What the hell?) That got me thinking — on what type of planet could you drive off a ramp like this, and actually *never fall down again?* If it’s a light enough planet, with sufficiently weak gravity, could a well-equipped car achieve escape velocity and therefore go into orbit off a ramp? Let’s envision a somewhat-smaller and less-dense planet, and figure out just how less dense it would have to be to make this happen. Here the math is pretty easy — we just need to find a planetary mass for which the gravitational escape velocity is within the reach of a nice car — let’s say 200 mph.

Turns out, * if the moon were made of cotton, a car going off a ramp would achieve escape velocity and go into orbit!* OK, it’s a little unlikely to find a planet made of cotton. But put your belief in suspense for just a skosh, and picture how much better the Dukes of Hazzard scene would be. Bo and Luke gunning the engine, Enos hot on their tail… Luke’s eyes widening, not too sure about this, but Bo’s not backing down… they hit the ramp, overhead shot of the General Lee soaring through the air, a couple of rebel yells, and… they just keep drifting through the air! Bo and Luke are now in orbit around our cotton moon, a couple dozen feet high, indefinitely. Cut to a couple days later — Bo and Luke are drifting above the fluffy lunar landscape. Unshaven, bored, getting a little desperate to get out of orbit. Luke’s getting a little hungry, thinking about at what point he’ll need to resort to cannibalism. Waylon Jennings narrates, “Anybody got a ladder?”

**Hey, that guy ruined my photo! And he ruined yours, too!**

Every once in a while, I presume I must ruin someone else’s family photo, purely by accident — someone’s trying to capture their little angel on film, and I’m in the background scratching my ass or something. Somewhere out in the suburbs of Phoenix, some family’s vacation photos have *me* in the background. I’ll never get a chance to see these photos of me, and the family that cherishes them probably doesn’t even notice me in them. Maybe once in a while, somebody takes a moment to remark on my asymmetric nostrils or baggy cargo shorts or something. Maybe they even share a laugh with a friend over the dumb look on my face in the crowd at the Eiffel Tower. But what if that friend recognized me from *their *family photos? This is an odd happenstance to consider, but could it be possible that I ruin the family photos for two people who know each other, purely by coincidence? If they got together to peruse each other’s Disneyland pics, what are the chances they’ll say “hey, there’s that same stupid doofus in both of our pictures!”

Who the hell cares, you might be asking yourself. Well, nobody, but I only bring it up as a fun exercise in probability. I’m going to seriously consider this question, and use simple assumptions and probabilistic reasoning to estimate the likelihood. And let’s not cheat here — we’re not considering where two people who know each other are at the same event, taking pictures around the same time, and I happen to wander through the field of view for both of them. I’m talking about completely independent accidental portraiture — Bob catches me in the background of his selfie at the Eiffel Tower, years later Jessica tries and fails to crop me out of her picture of the Grand Canyon, and Bob and Jessica happen to know each other. What are the chances of that? Given that I’m spending time blogging about it, the answer turns out to be pretty surprising.

First, the assumptions. Let’s make some simple guesses about people’s photography habits:

- Each person takes 1,000 photos a year. Sounds like a lot, but this is the digital age, after all…
- 5% of these have background people in them. That’s probably about right for me, as 95% of my photos are selfies taken alone at home.
- For a photo with background people, assume there are 2
*recognizable*people in the background on average. This more of a wild guess, but probably is conservative. - Let’s say the average person has 10 years of photos archived that we can go back through.

Furthermore, let’s assume each person taking photos knows 500 other acquaintances — close friends, family, distant friends, coworkers, old school chums, etc. That perhaps seems a little high, but consider this — I’m also going to assume there are 1 BILLION people that we can choose from. I think that’s probably conservative, but let’s go with it — there are 1 billion total possible people to choose from, any of whom has an equal chance to take a photo of me accidentally. What are the chances that two people I’m in the background of will know each other?

Now let’s work out the math! First of all, let’s make one final assumption — I have no idea how many photos I’ll be in the background of, but let’s assume that *if I have N total people in the background of all my photos, then I’m a background person in N OTHER photos*. Given the assumptions above, I have (1000 photos /yr) x (10 years) x (5% have people) x (2 people / photo) = 1000 total people in the background of all my photos. Hi guys! Therefore, I assume I must appear as a background person in 1000 *other* people’s photo collections.

Let’s corral up all 1000 lucky people who have me in the background of their photos. What are the chances that two of them will know each other? If everything’s completely random, the chances of a person knowing one other randomly-selected person is 500 / 1 billion = 5 x 10^{-7}. Pretty tiny. But consider this — there are 1000^{2} = 1,000,000 pairs of people in my set of lucky 1000 backgrounders. The chances that *none* of them know each other are therefore (1 – 5 x 10^{-7}) ^{(1 million)}, which is actually a pretty middle-of-the-road number — it’s equal to 60.7%. For our 1000 people selected at random from 1 billion total people, there’s only a 60% chance that they’ll all be strangers to each other. So, the chances that at least two people will know each other is almost 40%!

So — the answer to this odd question is, **the chances that I’m accidentally in the background of photos taken by two different people who know each other, purely by coincidence, is almost 40%**. If I’ve gotten some of the assumptions wrong (a slight possibility, I admit), then perhaps this percentage is off a bit. But it’s not astronomically unlikely — the surprising result is that this is probably happening all the time.

In fact, a few months after originally writing this, I came across a cute news story *about this actually happening* — a married couple discovered that the husband appeared in the background of a picture of the wife, taken at Disneyland, when *they were children and didn’t know each other.* They discovered this *years later*, and it must have freaked them right out. The husband happened to be looking at the wife’s childhood photos, and recognized *his own father* in the background of one shot, *pushing him in a stroller.* But according to my calculations, it’s not as rare as you might first think. Furthermore, my calculations indicate that *I’m abusing the italics just a bit, and should probably *stop using them.

**I’ll add more of these** as I think of them. Do you have an idea Fer me to consider? (HAHAHAHAAAA AAAAAHAHAHAH!!!) Drop me a line on twitter, and I’ll add to this list! But before I go, did you do the homework I assigned at the beginning of the article? How many other people in the world have precisely the same number of hairs on their head as you? You either solved the hair problem by scouring the earth with a clipboard and tweezers, or made a simple approximation to find the surprisingly high number. There are about 300,000 hairs on the typical head, tops, and 7 billion people on the earth — so on average, each possible hair count is shared among 7 billion / 300,000 = 23,000 people! And that’s likely an underestimate, as it’s not as though all possible hair counts between 1 and 300,000 are equally likely. So chances are you have precisely the same number of hairs on your head as perhaps 50,000 people!

Other places to check out:

- Guesstimation, the book. Somebody convince the author to let me play him in the movie
- Fermi Questions Dot Com