Fermi Estimates

Just be­fore the Trinity test, En­rico Fermi de­cided he wanted a rough es­ti­mate of the blast’s power be­fore the di­ag­nos­tic data came in. So he dropped some pieces of pa­per from his hand as the blast wave passed him, and used this to es­ti­mate that the blast was equiv­a­lent to 10 kilo­tons of TNT. His guess was re­mark­ably ac­cu­rate for hav­ing so lit­tle data: the true an­swer turned out to be 20 kilo­tons of TNT.

Fermi had a knack for mak­ing roughly-ac­cu­rate es­ti­mates with very lit­tle data, and there­fore such an es­ti­mate is known to­day as a Fermi es­ti­mate.

Why bother with Fermi es­ti­mates, if your es­ti­mates are likely to be off by a fac­tor of 2 or even 10? Often, get­ting an es­ti­mate within a fac­tor of 10 or 20 is enough to make a de­ci­sion. So Fermi es­ti­mates can save you a lot of time, es­pe­cially as you gain more prac­tice at mak­ing them.

Es­ti­ma­tion tips

Th­ese first two sec­tions are adapted from Gues­ti­ma­tion 2.0.

Dare to be im­pre­cise. Round things off enough to do the calcu­la­tions in your head. I call this the spher­i­cal cow prin­ci­ple, af­ter a joke about how physi­cists over­sim­plify things to make calcu­la­tions fea­si­ble:

Milk pro­duc­tion at a dairy farm was low, so the farmer asked a lo­cal uni­ver­sity for help. A mul­ti­dis­ci­plinary team of pro­fes­sors was as­sem­bled, headed by a the­o­ret­i­cal physi­cist. After two weeks of ob­ser­va­tion and anal­y­sis, the physi­cist told the farmer, “I have the solu­tion, but it only works in the case of spher­i­cal cows in a vac­uum.”

By the spher­i­cal cow prin­ci­ple, there are 300 days in a year, peo­ple are six feet (or 2 me­ters) tall, the cir­cum­fer­ence of the Earth is 20,000 mi (or 40,000 km), and cows are spheres of meat and bone 4 feet (or 1 me­ter) in di­ame­ter.

De­com­pose the prob­lem. Some­times you can give an es­ti­mate in one step, within a fac­tor of 10. (How much does a new com­pact car cost? $20,000.) But in most cases, you’ll need to break the prob­lem into sev­eral pieces, es­ti­mate each of them, and then re­com­bine them. I’ll give sev­eral ex­am­ples be­low.

Es­ti­mate by bound­ing. Some­times it is eas­ier to give lower and up­per bounds than to give a point es­ti­mate. How much time per day does the av­er­age 15-year-old watch TV? I don’t spend any time with 15-year-olds, so I haven’t a clue. It could be 30 min­utes, or 3 hours, or 5 hours, but I’m pretty con­fi­dent it’s more than 2 min­utes and less than 7 hours (400 min­utes, by the spher­i­cal cow prin­ci­ple).

Can we con­vert those bounds into an es­ti­mate? You bet. But we don’t do it by tak­ing the av­er­age. That would give us (2 mins + 400 mins)/​2 = 201 mins, which is within a fac­tor of 2 from our up­per bound, but a fac­tor 100 greater than our lower bound. Since our goal is to es­ti­mate the an­swer within a fac­tor of 10, we’ll prob­a­bly be way off.

In­stead, we take the ge­o­met­ric mean — the square root of the product of our up­per and lower bounds. But square roots of­ten re­quire a calcu­la­tor, so in­stead we’ll take the ap­prox­i­mate ge­o­met­ric mean (AGM). To do that, we av­er­age the co­effi­cients and ex­po­nents of our up­per and lower bounds.

So what is the AGM of 2 and 400? Well, 2 is 2×100, and 400 is 4×102. The av­er­age of the co­effi­cients (2 and 4) is 3; the av­er­age of the ex­po­nents (0 and 2) is 1. So, the AGM of 2 and 400 is 3×101, or 30. The pre­cise ge­o­met­ric mean of 2 and 400 turns out to be 28.28. Not bad.

What if the sum of the ex­po­nents is an odd num­ber? Then we round the re­sult­ing ex­po­nent down, and mul­ti­ply the fi­nal an­swer by three. So sup­pose my lower and up­per bounds for how much TV the av­er­age 15-year-old watches had been 20 mins and 400 mins. Now we calcu­late the AGM like this: 20 is 2×101, and 400 is still 4×102. The av­er­age of the co­effi­cients (2 and 4) is 3; the av­er­age of the ex­po­nents (1 and 2) is 1.5. So we round the ex­po­nent down to 1, and we mul­ti­ple the fi­nal re­sult by three: 3(3×101) = 90 mins. The pre­cise ge­o­met­ric mean of 20 and 400 is 89.44. Again, not bad.

San­ity-check your an­swer. You should always san­ity-check your fi­nal es­ti­mate by com­par­ing it to some rea­son­able analogue. You’ll see ex­am­ples of this be­low.

Use Google as needed. You can of­ten quickly find the ex­act quan­tity you’re try­ing to es­ti­mate on Google, or at least some piece of the prob­lem. In those cases, it’s prob­a­bly not worth try­ing to es­ti­mate it with­out Google.

Fermi es­ti­ma­tion failure modes

Fermi es­ti­mates go wrong in one of three ways.

First, we might badly over­es­ti­mate or un­der­es­ti­mate a quan­tity. De­com­pos­ing the prob­lem, es­ti­mat­ing from bounds, and look­ing up par­tic­u­lar pieces on Google should pro­tect against this. Over­es­ti­mates and un­der­es­ti­mates for the differ­ent pieces of a prob­lem should roughly can­cel out, es­pe­cially when there are many pieces.

Se­cond, we might model the prob­lem in­cor­rectly. If you es­ti­mate teenage deaths per year on the as­sump­tion that most teenage deaths are from suicide, your es­ti­mate will prob­a­bly be way off, be­cause most teenage deaths are caused by ac­ci­dents. To avoid this, try to de­com­pose each Fermi prob­lem by us­ing a model you’re fairly con­fi­dent of, even if it means you need to use more pieces or give wider bounds when es­ti­mat­ing each quan­tity.

Fi­nally, we might choose a non­lin­ear prob­lem. Nor­mally, we as­sume that if one ob­ject can get some re­sult, then two ob­jects will get twice the re­sult. Un­for­tu­nately, this doesn’t hold true for non­lin­ear prob­lems. If one mo­tor­cy­cle on a high­way can trans­port a per­son at 60 miles per hour, then 30 mo­tor­cy­cles can trans­port 30 peo­ple at 60 miles per hour. How­ever, 104 mo­tor­cy­cles can­not trans­port 104 peo­ple at 60 miles per hour, be­cause there will be a huge traf­fic jam on the high­way. This prob­lem is difficult to avoid, but with prac­tice you will get bet­ter at rec­og­niz­ing when you’re fac­ing a non­lin­ear prob­lem.

Fermi practice

When get­ting started with Fermi prac­tice, I recom­mend es­ti­mat­ing quan­tities that you can eas­ily look up later, so that you can see how ac­cu­rate your Fermi es­ti­mates tend to be. Don’t look up the an­swer be­fore con­struct­ing your es­ti­mates, though! Alter­na­tively, you might al­low your­self to look up par­tic­u­lar pieces of the prob­lem — e.g. the num­ber of Sikhs in the world, the for­mula for es­cape ve­loc­ity, or the gross world product — but not the fi­nal quan­tity you’re try­ing to es­ti­mate.

Most books about Fermi es­ti­mates are filled with ex­am­ples done by Fermi es­ti­mate ex­perts, and in many cases the es­ti­mates were prob­a­bly ad­justed af­ter the au­thor looked up the true an­swers. This post is differ­ent. My ex­am­ples be­low are es­ti­mates I made be­fore look­ing up the an­swer on­line, so you can get a re­al­is­tic pic­ture of how this works from some­one who isn’t “cheat­ing.” Also, there will be no se­lec­tion effect: I’m go­ing to do four Fermi es­ti­mates for this post, and I’m not go­ing to throw out my es­ti­mates if they are way off. Fi­nally, I’m not all that prac­ticed do­ing “Fer­mis” my­self, so you’ll get to see what it’s like for a rel­a­tive new­bie to go through the pro­cess. In short, I hope to give you a re­al­is­tic pic­ture of what it’s like to do Fermi prac­tice when you’re just get­ting started.

Ex­am­ple 1: How many new pas­sen­ger cars are sold each year in the USA?

The clas­sic Fermi prob­lem is “How many pi­ano tuners are there in Chicago?” This kind of es­ti­mate is use­ful if you want to know the ap­prox­i­mate size of the cus­tomer base for a new product you might de­velop, for ex­am­ple. But I’m not sure any­one knows how many pi­ano tuners there re­ally are in Chicago, so let’s try a differ­ent one we prob­a­bly can look up later: “How many new pas­sen­ger cars are sold each year in the USA?”

As with all Fermi prob­lems, there are many differ­ent mod­els we could build. For ex­am­ple, we could es­ti­mate how many new cars a deal­er­ship sells per month, and then we could es­ti­mate how many deal­er­ships there are in the USA. Or we could try to es­ti­mate the an­nual de­mand for new cars from the coun­try’s pop­u­la­tion. Or, if we hap­pened to have read how many Toy­ota Corol­las were sold last year, we could try to build our es­ti­mate from there.

The sec­ond model looks more ro­bust to me than the first, since I know roughly how many Amer­i­cans there are, but I have no idea how many new-car deal­er­ships there are. Still, let’s try it both ways. (I don’t hap­pen to know how many new Corol­las were sold last year.)

Ap­proach #1: Car dealerships

How many new cars does a deal­er­ship sell per month, on av­er­age? Oofta, I dunno. To sup­port the deal­er­ship’s ex­is­tence, I as­sume it has to be at least 5. But it’s prob­a­bly not more than 50, since most deal­er­ships are in small towns that don’t get much ac­tion. To get my point es­ti­mate, I’ll take the AGM of 5 and 50. 5 is 5×100, and 50 is 5×101. Our ex­po­nents sum to an odd num­ber, so I’ll round the ex­po­nent down to 0 and mul­ti­ple the fi­nal an­swer by 3. So, my es­ti­mate of how many new cars a new-car deal­er­ship sells per month is 3(5×100) = 15.

Now, how many new-car deal­er­ships are there in the USA? This could be tough. I know sev­eral towns of only 10,000 peo­ple that have 3 or more new-car deal­er­ships. I don’t re­call towns much smaller than that hav­ing new-car deal­er­ships, so let’s ex­clude them. How many cities of 10,000 peo­ple or more are there in the USA? I have no idea. So let’s de­com­pose this prob­lem a bit more.

How many coun­ties are there in the USA? I re­mem­ber see­ing a map of coun­ties col­ored by which na­tional an­ces­try was dom­i­nant in that county. (Ger­many was the most com­mon.) Think­ing of that map, there were definitely more than 300 coun­ties on it, and definitely less than 20,000. What’s the AGM of 300 and 20,000? Well, 300 is 3×102, and 20,000 is 2×104. The av­er­age of co­effi­cients 3 and 2 is 2.5, and the av­er­age of ex­po­nents 2 and 4 is 3. So the AGM of 300 and 20,000 is 2.5×103 = 2500.

Now, how many towns of 10,000 peo­ple or more are there per county? I’m pretty sure the av­er­age must be larger than 10 and smaller than 5000. The AGM of 10 and 5000 is 300. (I won’t in­clude this calcu­la­tion in the text any­more; you know how to do it.)

Fi­nally, how many car deal­er­ships are there in cities of 10,000 or more peo­ple, on av­er­age? Most such towns are pretty small, and prob­a­bly have 2-6 car deal­er­ships. The largest cities will have many more: maybe 100-ish. So I’m pretty sure the av­er­age num­ber of car deal­er­ships in cities of 10,000 or more peo­ple must be be­tween 2 and 30. The AGM of 2 and 30 is 7.5.

Now I just mul­ti­ply my es­ti­mates:

[15 new cars sold per month per deal­er­ship] × [12 months per year] × [7.5 new-car deal­er­ships per city of 10,000 or more peo­ple] × [300 cities of 10,000 or more peo­ple per county] × [2500 coun­ties in the USA] = 1,012,500,000.

A san­ity check im­me­di­ately in­val­i­dates this an­swer. There’s no way that 300 mil­lion Amer­i­can cit­i­zens buy a billion new cars per year. I sup­pose they might buy 100 mil­lion new cars per year, which would be within a fac­tor of 10 of my es­ti­mate, but I doubt it.

As I sus­pected, my first ap­proach was prob­le­matic. Let’s try the sec­ond ap­proach, start­ing from the pop­u­la­tion of the USA.

Ap­proach #2: Pop­u­la­tion of the USA

There are about 300 mil­lion Amer­i­cans. How many of them own a car? Maybe 13 of them, since chil­dren don’t own cars, many peo­ple in cities don’t own cars, and many house­holds share a car or two be­tween the adults in the house­hold.

Of the 100 mil­lion peo­ple who own a car, how many of them bought a new car in the past 5 years? Prob­a­bly less than half; most peo­ple buy used cars, right? So maybe 14 of car own­ers bought a new car in the past 5 years, which means 1 in 20 car own­ers bought a new car in the past year.

100 mil­lion /​ 20 = 5 mil­lion new cars sold each year in the USA. That doesn’t seem crazy, though per­haps a bit low. I’ll take this as my es­ti­mate.

Now is your last chance to try this one on your own; in the next para­graph I’ll re­veal the true an­swer.


Now, I Google new cars sold per year in the USA. Wikipe­dia is the first re­sult, and it says “In the year 2009, about 5.5 mil­lion new pas­sen­ger cars were sold in the United States ac­cord­ing to the U.S. Depart­ment of Trans­porta­tion.”


Ex­am­ple 2: How many fatal­ities from pas­sen­ger-jet crashes have there been in the past 20 years?

Again, there are mul­ti­ple mod­els I could build. I could try to es­ti­mate how many pas­sen­ger-jet flights there are per year, and then try to es­ti­mate the fre­quency of crashes and the av­er­age num­ber of fatal­ities per crash. Or I could just try to guess the to­tal num­ber of pas­sen­ger-jet crashes around the world per year and go from there.

As far as I can tell, pas­sen­ger-jet crashes (with fatal­ities) al­most always make it on the TV news and (more rele­vant to me) the front page of Google News. Ex­cit­ing footage and mul­ti­ple deaths will do that. So work­ing just from mem­ory, it feels to me like there are about 5 pas­sen­ger-jet crashes (with fatal­ities) per year, so maybe there were about 100 pas­sen­ger jet crashes with fatal­ities in the past 20 years.

Now, how many fatal­ities per crash? From mem­ory, it seems like there are usu­ally two kinds of crashes: ones where ev­ery­body dies (mean­ing: about 200 peo­ple?), and ones where only about 10 peo­ple die. I think the “ev­ery­body dead” crashes are less com­mon, maybe 14 as com­mon. So the av­er­age crash with fatal­ities should cause (200×1/​4)+(10×3/​4) = 50+7.5 = 60, by the spher­i­cal cow prin­ci­ple.

60 fatal­ities per crash × 100 crashes with fatal­ities over the past 20 years = 6000 pas­sen­ger fatal­ities from pas­sen­ger-jet crashes in the past 20 years.

Last chance to try this one on your own...

A Google search again brings me to Wikipe­dia, which re­veals that an or­ga­ni­za­tion called ACRO records the num­ber of air­line fatal­ities each year. Un­for­tu­nately for my pur­poses, they in­clude fatal­ities from cargo flights. After more Googling, I tracked down Boe­ing’s “Statis­ti­cal Sum­mary of Com­mer­cial Jet Air­plane Ac­ci­dents, 1959-2011,” but that re­port ex­cludes jets lighter than 60,000 pounds, and ex­cludes crashes caused by hi­jack­ing or ter­ror­ism.

It ap­pears it would be a ma­jor re­search pro­ject to figure out the true an­swer to our ques­tion, but let’s at least es­ti­mate it from the ACRO data. Luck­ily, ACRO has statis­tics on which per­centage of ac­ci­dents are from pas­sen­ger and other kinds of flights, which I’ll take as a proxy for which per­centage of fatal­ities are from differ­ent kinds of flights. Ac­cord­ing to that page, 35.41% of ac­ci­dents are from “reg­u­lar sched­ule” flights, 7.75% of ac­ci­dents are from “pri­vate” flights, 5.1% of ac­ci­dents are from “char­ter” flights, and 4.02% of ac­ci­dents are from “ex­ec­u­tive” flights. I think that cap­tures what I had in mind as “pas­sen­ger-jet flights.” So we’ll guess that 52.28% of fatal­ities are from “pas­sen­ger-jet flights.” I won’t round this to 50% be­cause we’re not do­ing a Fermi es­ti­mate right now; we’re try­ing to check a Fermi es­ti­mate.

Ac­cord­ing to ACRO’s archives, there were 794 fatal­ities in 2012, 828 fatal­ities in 2011, and… well, from 1993-2012 there were a to­tal of 28,021 fatal­ities. And 52.28% of that num­ber is 14,649.

So my es­ti­mate of 6000 was off by less than a fac­tor of 3!

Ex­am­ple 3: How much does the New York state gov­ern­ment spends on K-12 ed­u­ca­tion ev­ery year?

How might I es­ti­mate this? First I’ll es­ti­mate the num­ber of K-12 stu­dents in New York, and then I’ll es­ti­mate how much this should cost.

How many peo­ple live in New York? I seem to re­call that NYC’s greater metropoli­tan area is about 20 mil­lion peo­ple. That’s prob­a­bly most of the state’s pop­u­la­tion, so I’ll guess the to­tal is about 30 mil­lion.

How many of those 30 mil­lion peo­ple at­tend K-12 pub­lic schools? I can’t re­mem­ber what the United States’ pop­u­la­tion pyra­mid looks like, but I’ll guess that about 16 of Amer­i­cans (and hope­fully New York­ers) at­tend K-12 at any given time. So that’s 5 mil­lion kids in K-12 in New York. The num­ber at­tend­ing pri­vate schools prob­a­bly isn’t large enough to mat­ter for fac­tor-of-10 es­ti­mates.

How much does a year of K-12 ed­u­ca­tion cost for one child? Well, I’ve heard teach­ers don’t get paid much, so af­ter benefits and taxes and so on I’m guess­ing a teacher costs about $70,000 per year. How big are class sizes these days, 30 kids? By the spher­i­cal cow prin­ci­ple, that’s about $2,000 per child, per year on teach­ers’ salaries. But there are lots of other ex­penses: build­ings, trans­port, ma­te­ri­als, sup­port staff, etc. And maybe some money goes to pri­vate schools or other or­ga­ni­za­tions. Rather than es­ti­mate all those things, I’m just go­ing to guess that about $10,000 is spent per child, per year.

If that’s right, then New York spends $50 billion per year on K-12 ed­u­ca­tion.

Last chance to make your own es­ti­mate!

Be­fore I did the Fermi es­ti­mate, I had Ju­lia Galef check Google to find this statis­tic, but she didn’t give me any hints about the num­ber. Her two sources were Wolfram Alpha and a web chat with New York’s Deputy Sec­re­tary for Ed­u­ca­tion, both of which put the figure at ap­prox­i­mately $53 billion.

Which is definitely within a fac­tor of 10 from $50 billion. :)

Ex­am­ple 4: How many plays of My Bloody Valen­tine’s “Only Shal­low” have been re­ported to last.fm?

Last.fm makes a record of ev­ery au­dio track you play, if you en­able the rele­vant fea­ture or plu­gin for the mu­sic soft­ware on your phone, com­puter, or other de­vice. Then, the ser­vice can show you charts and statis­tics about your listen­ing pat­terns, and make per­son­al­ized mu­sic recom­men­da­tions from them. My own charts are here. (Chuck Wild /​ Liquid Mind dom­i­nates my charts be­cause I used to listen to that artist while sleep­ing.)

My Fermi prob­lem is: How many plays of “Only Shal­low” have been re­ported to last.fm?

My Bloody Valen­tine is a pop­u­lar “in­die” rock band, and “Only Shal­low” is prob­a­bly one of their most pop­u­lar tracks. How can I es­ti­mate how many plays it has got­ten on last.fm?

What do I know that might help?

  • I know last.fm is pop­u­lar, but I don’t have a sense of whether they have 1 mil­lion users, 10 mil­lion users, or 100 mil­lion users.

  • I ac­ci­den­tally saw on Last.fm’s Wikipe­dia page that just over 50 billion track plays have been recorded. We’ll con­sider that to be one piece of data I looked up to help with my es­ti­mate.

  • I seem to re­call read­ing that ma­jor mu­sic ser­vices like iTunes and Spo­tify have about 10 mil­lion tracks. Since last.fm records songs that peo­ple play from their pri­vate col­lec­tions, whether or not they ex­ist in pop­u­lar databases, I’d guess that the to­tal num­ber of differ­ent tracks named in last.fm’s database is an or­der of mag­ni­tude larger, for about 100 mil­lion tracks named in its database.

I would guess that track plays obey a power law, with the most pop­u­lar tracks get­ting vastly more plays than tracks of av­er­age pop­u­lar­ity. I’d also guess that there are maybe 10,000 tracks more pop­u­lar than “Only Shal­low.”

Next, I simu­lated be­ing good at math by hav­ing Qiaochu Yuan show me how to do the calcu­la­tion. I also al­lowed my­self to use a calcu­la­tor. Here’s what we do:

Plays(rank) = C/​(rankP)

P is the ex­po­nent for the power law, and C is the pro­por­tion­al­ity con­stant. We’ll guess that P is 1, a com­mon power law ex­po­nent for em­piri­cal data. And we calcu­late C like so:

C ≈ [to­tal plays]/​ln(to­tal songs) ≈ 2.5 billion

So now, as­sum­ing the song’s rank is 10,000, we have:

Plays(104) = 2.5×109/​(104)

Plays(“Only Shal­low”) = 250,000

That seems high, but let’s roll with it. Last chance to make your own es­ti­mate!


And when I check the an­swer, I see that “Only Shal­low” has about 2 mil­lion plays on last.fm.

My an­swer was off by less than a fac­tor of 10, which for a Fermi es­ti­mate is called vic­tory!

Un­for­tu­nately, last.fm doesn’t pub­lish all-time track rank­ings or other data that might help me to de­ter­mine which parts of my model were cor­rect and in­cor­rect.

Fur­ther examples

I fo­cused on ex­am­ples that are similar in struc­ture to the kinds of quan­tities that en­trepreneurs and CEOs might want to es­ti­mate, but of course there are all kinds of things one can es­ti­mate this way. Here’s a sam­pling of Fermi prob­lems fea­tured in var­i­ous books and web­sites on the sub­ject:

Play Fermi Ques­tions: 2100 Fermi prob­lems and count­ing.

Guessti­ma­tion (2008): If all the hu­mans in the world were crammed to­gether, how much area would we re­quire? What would be the mass of all 108 Mon­gaMillions lot­tery tick­ets? On av­er­age, how many peo­ple are air­borne over the US at any given mo­ment? How many cells are there in the hu­man body? How many peo­ple in the world are pick­ing their nose right now? What are the rel­a­tive costs of fuel for NYC rick­shaws and au­to­mo­biles?

Guessti­ma­tion 2.0 (2011): If we launched a trillion one-dol­lar bills into the at­mo­sphere, what frac­tion of sun­light hit­ting the Earth could we block with those dol­lar bills? If a mil­lion mon­keys typed ran­domly on a mil­lion type­writ­ers for a year, what is the longest string of con­sec­u­tive cor­rect let­ters of *The Cat in the Hat (start­ing from the be­gin­ning) would they likely type? How much en­ergy does it take to crack a nut? If an air­line asked its pas­sen­gers to uri­nate be­fore board­ing the air­plane, how much fuel would the air­line save per flight? What is the ra­dius of the largest rocky sphere from which we can reach es­cape ve­loc­ity by jump­ing?

How Many Licks? (2009): What frac­tion of Earth’s vol­ume would a mole of hot, sticky, choco­late-jelly dough­nuts be? How many miles does a per­son walk in a life­time? How many times can you out­line the con­ti­nen­tal US in shoelaces? How long would it take to read ev­ery book in the library? How long can you shower and still make it more en­vi­ron­men­tally friendly than tak­ing a bath?

Bal­l­park­ing (2012): How many bolts are in the floor of the Bos­ton Gar­den bas­ket­ball court? How many lanes would you need for the out­er­most lane of a run­ning track to be the length of a marathon? How hard would you have to hit a base­ball for it to never land?

Univer­sity of Mary­land Fermi Prob­lems Site: How many sheets of let­ter-sized pa­per are used by all stu­dents at the Univer­sity of Mary­land in one semester? How many blades of grass are in the lawn of a typ­i­cal sub­ur­ban house in the sum­mer? How many golf balls can be fit into a typ­i­cal suit­case?

Stupid Calcu­la­tions: a blog of silly-topic Fermi es­ti­mates.


Fermi es­ti­mates can help you be­come more effi­cient in your day-to-day life, and give you in­creased con­fi­dence in the de­ci­sions you face. If you want to be­come profi­cient in mak­ing Fermi es­ti­mates, I recom­mend prac­tic­ing them 30 min­utes per day for three months. In that time, you should be able to make about (2 Fer­mis per day)×(90 days) = 180 Fermi es­ti­mates.

If you’d like to write down your es­ti­ma­tion at­tempts and then pub­lish them here, please do so as a re­ply to this com­ment. One Fermi es­ti­mate per com­ment, please!

Alter­na­tively, post your Fermi es­ti­mates to the ded­i­cated sub­red­dit.

Up­date 03/​06/​2017: I keep get­ting re­quests from pro­fes­sors to use this in their classes, so: I li­cense any­one to use this ar­ti­cle non­com­mer­cially, so long as its au­thor­ship is noted (me = Luke Muehlhauser).