# 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.” Boo-yah! ### 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. ### Conclusion 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). • Cor­rec­tion: (2 Fer­mis per day)×(90 days) = 200 Fermi es­ti­mates. • It’s prob­a­bly worth figur­ing out what went wrong in Ap­proach 1 to Ex­am­ple 1, which I think is this part: [300 cities of 10,000 or more peo­ple per county] × [2500 coun­ties in the USA] Note that this gives 750,000 cities of 10,000 or more peo­ple in the US, for a to­tal of at least 7.5 billion peo­ple in the US. So it’s already clearly wrong here. I’d say 300 cities of 10,000 peo­ple or more per county is way too high; I’d put it at more like 1 (Edit: note that this gives at least 250 mil­lion peo­ple in the US and that’s about right). This brings down the fi­nal es­ti­mate from this ap­proach by a fac­tor of 300, or down to 3 mil­lion, which is much closer. (Ver­ifi­ca­tion: I just picked a ran­dom US state and a ran­dom county in it from Wikipe­dia and got Bar­tow County, Ge­or­gia, which has a pop­u­la­tion of 100,000. That means it has at most 10 cities with 10,000 or more peo­ple, and go­ing through the list of cities it ac­tu­ally looks like it only has one such city.) This gives about 2,500 cities in the US to­tal with pop­u­la­tion 10,000 or more. I can’t ver­ify this num­ber, but ac­cord­ing to Wikipe­dia there are about 300 cities in the US with pop­u­la­tion 100,000 or more. As­sum­ing the pop­u­la­tions of cities are power-law dis­tributed with ex­po­nent 1, this means that the nth-ranked city has pop­u­la­tion about 30,000,000/​n, so this gives about 3,000 cities in the US with pop­u­la­tion 10,000 or more. And in fact we didn’t even need to use Wikipe­dia! Just as­sum­ing that the pop­u­la­tion of cities is power-law dis­tributed with ex­po­nent 1, we see that the dis­tri­bu­tion is de­ter­mined by the pop­u­la­tion of the most pop­u­lous city. Let’s take this to be 20 mil­lion peo­ple (the num­ber you used for New York City). Then the nth-ranked city in the US has pop­u­la­tion about 20,000,000/​n, so there are about 2,000 cities with pop­u­la­tion 10,000 or more. Edit: Found the ac­tual num­ber. Ac­cord­ing to the U.S. Cen­sus Bureau, as of 2008, the ac­tual num­ber is about 2,900 cities. In­ci­den­tally, this shows an­other use of Fermi es­ti­mates: if you get one that’s ob­vi­ously wrong, you’ve dis­cov­ered an op­por­tu­nity to fix some as­pect of your model of the world. • I feel com­pel­led to re­peat this old physics clas­sic: How Fermi could es­ti­mate things! Like the well-known Olympic ten rings, And the one-hun­dred states, And weeks with ten dates, And birds that all fly with one… wings.  :-) • That is beau­tiful. • I’ve run mee­tups on this topic twice now. Every time I do, it’s difficult to con­vince peo­ple it’s a use­ful skill. More words about when es­ti­ma­tion is use­ful would be nice. In most ex­er­cises that you can find on Fermi calcu­la­tions, you can also ac­tu­ally find the right an­swer, writ­ten down some­where on­line. And, well, be­ing able to quickly find in­for­ma­tion is prob­a­bly a more use­ful skill to prac­tice than es­ti­ma­tion; be­cause it works for non-quan­tified in­for­ma­tion too. I un­der­stand why this is; you want to be able to show that these es­ti­mates aren’t very far off, and for that you need to be able to find the ac­tual num­bers some­how. But that means that your ex­am­ples don’t ac­tu­ally mo­ti­vate the effort of prac­tic­ing, they only demon­strate how. I sus­pect the fol­low­ing kinds of situ­a­tions are fruit­ful for es­ti­ma­tion: • De­cid­ing in un­fa­mil­iar situ­a­tions, be­cause you don’t know how things will turn out for you. If you’re in a re­ally novel situ­a­tion, you can’t even find out how the same de­ci­sion has worked for other peo­ple be­fore, and so you have to guess at ex­pected value us­ing the best in­for­ma­tion that you can find. • Value of in­for­ma­tion calcu­la­tions, like here and here, where you can­not pos­si­bly know the ex­pected value of things, be­cause you’re try­ing to de­cide if you should pay for in­for­ma­tion about their value. • De­cid­ing when you’re not on­line, be­cause this makes ac­cess­ing in­for­ma­tion more ex­pen­sive than com­pu­ta­tion. • De­ci­sions where you have un­usual in­for­ma­tion for a par­tic­u­lar situ­a­tion—the in­ter­net might have ex­cel­lent base-rate in­for­ma­tion about your gen­eral situ­a­tion, but it’s un­likely to give you the pre­cise odds so that you can in­cor­po­rate the ex­tra in­for­ma­tion that you have in this spe­cific situ­a­tion. • Look­ing smart. It’s nice to look smart some­times. Others? Does any­one have ex­am­ples of when Fermi calcu­la­tions helped them make a de­ci­sion? • Fermi’s seem es­sen­tial for busi­ness to me. Others agree; they’re taught in stan­dard MBA pro­grams. For ex­am­ple: • Can our busi­ness (or our non-profit) af­ford to hire an ex­tra per­son right now? E.g., if they re­quire the same train­ing time be­fore use­ful­ness that oth­ers re­quired, will they bring in more rev­enue in time to make up for the loss of run­way? • If it turns out that product X is a suc­cess, how much money might it make—is it enough to jus­tify in­ves­ti­gat­ing the mar­ket? • Is it cheaper (given the cost of time) to use dis­pos­able dishes or to wash the dishes? • Is it bet­ter to pro­cess pay­ments via pay­pal or checks, given the fees in­volved in pay­pal vs. the de­lays, has­sles, and as­so­ci­ated risks of non-pay­ment in­volved in checks? And on and on. I use them sev­eral times a day for CFAR and they seem es­sen­tial there. They’re use­ful also for one’s own prac­ti­cal life: com­mute time vs. rent trade­offs; vi­su­al­iz­ing “do I want to have a kid? how would the time and dol­lar cost ac­tu­ally im­pact me?”, re­al­iz­ing that macademia nuts are ac­tu­ally a cheap food and not an ex­pen­sive food (once I think “per calorie” and not “per ap­par­ent size of the con­tainer”), and so on and so on. • Oh, right! I ac­tu­ally did the co­mute time vs. rent com­pu­ta­tion when I moved four months ago! And wound up with a sur­pris­ing enough num­ber that I thought about it very closely, and de­cided that num­ber was about right, and changed how I was look­ing for apart­ments. How did I for­get that? Thanks! • re­al­iz­ing that macademia nuts are ac­tu­ally a cheap food and not an ex­pen­sive food (once I think “per calorie” and not “per ap­par­ent size of the con­tainer”), But calories aren’t the only thing you care about—the abil­ity to sa­ti­ate you also mat­ters. (Seed oil is even cheaper per calorie.) • The main use I put Fermi es­ti­mates to is fact-check­ing: when I see a statis­tic quoted, I would like to know if it is rea­son­able (es­pe­cially if I sus­pect that it has been mis­quoted some­how). • Qiaochu adds: If you get [an es­ti­mate] that’s ob­vi­ously wrong, you’ve dis­cov­ered an op­por­tu­nity to fix some as­pect of your model of the world. • I also think Fermi calcu­la­tions are just fun. It makes me feel to­tally awe­some to be able to con­jure ap­prox­i­mate an­swers to ques­tions out of thin air. • There’s a free book on this sort of thing, un­der a Creative Com­mons li­cense, called Street-Fight­ing Math­e­mat­ics: The Art of Ed­u­cated Guess­ing and Op­por­tunis­tic Prob­lem Solv­ing. Among the fun things in it: Chap­ter 1: Us­ing di­men­sional anal­y­sis to quickly pull cor­rect-ish equa­tions out of thin air! Chap­ter 2: Fo­cus­ing on easy cases. It’s amaz­ing how many prob­lems be­come sim­pler when you set some vari­ables equal to 1, 0, or ∞. Chap­ter 3: An awful lot of things look like rec­t­an­gles if you squint at them hard enough. Rec­t­an­gles are nice. Chap­ter 4: Draw­ing pic­tures can help. Hu­mans are good at look­ing at shapes. Chap­ter 5: Ap­prox­i­mate ar­ith­metic in which all num­bers are ei­ther 1, a power of 10, or “a few”—roughly 3, which is close to the ge­o­met­ric mean of 1 and 10. A few times a few is ten, for small val­ues of “is”. Mul­ti­ply and di­vide large num­bers on your fingers! … And there’s some more stuff, too, and some more chap­ters, but that’ll do for an ap­prox­i­mate sum­mary. • XKCD’s What If? has some ex­am­ples of Fermi calcu­la­tions, for in­stance at the start of work­ing out the effects of “a mole of moles” (similar to a mole of choc donuts, which is what re­minded me). • Thanks, Luke, this was helpful! There is a sub-tech­nique that could have helped you get a bet­ter an­swer for the first ap­proach to ex­am­ple 1: perform a san­ity check not only on the fi­nal value, but on any in­ter­me­di­ate value you can think of. In this ex­am­ple, when you es­ti­mated that there are 2500 coun­ties, and that the av­er­age county has 300 towns with pop­u­la­tion greater than 10,000, that im­plies a lower bound for the to­tal pop­u­la­tion of the US: as­sum­ing that all towns have ex­actly 10,000 peo­ple, that gets you a US pop­u­la­tion of 2,500x300x10,000=7,500,000,000! That’s 7.5 billion peo­ple. Of course, in real life, some peo­ple live in smaller towns, and some towns have more then 10,000 peo­ple, which makes the true im­plied es­ti­mate even larger. At this point you know that ei­ther your es­ti­mate for num­ber of coun­ties, or your es­ti­mate for num­ber of towns with pop­u­la­tion above 10,000 per county, or both, must de­crease to get an im­plied pop­u­la­tion of about 300 mil­lion. This would have brought your over­all es­ti­mate down to within a fac­tor of 10. • I had the plea­sure the other day of try­ing my hand on a slightly un­usual use of Fermi es­ti­mates: try­ing to guess whether some­thing un­likely has ever hap­pened. In par­tic­u­lar, the ques­tion was “Has any­one ever been kil­led by a fal­ling pi­ano as in the car­toon trope?” Others nearby at the time ob­jected, “but you don’t know any­thing about this!” which I found amus­ing be­cause of course I know quite a lot about pi­anos, things fal­ling, how peo­ple can be kil­led by things fal­ling, etc. so how could I pos­si­bly not know any­thing about pi­anos fal­ling and kil­ling peo­ple? Un­for­tu­nately, our es­ti­mate gave it at around 1-10 deaths by pi­ano-fal­ling so we weren’t able to make a strong con­clu­sion ei­ther way over whether this hap­pened. I would be in­ter­ested to hear if any­one got a sig­nifi­cantly differ­ent re­sult. (We only con­sid­ered fal­ling grands or baby grands to count as up­right pi­anos, key­boards, etc. just aren’t hu­morous enough for the car­toon trope.) • I’ll try. Let’s see, grands and baby grands date back to some­thing like the 1700s; I’m sure I’ve heard of Mozart or Beethoven us­ing pi­anos, so that gives me a time-win­dow of 300 years for fal­ling pi­anos to kill peo­ple in Europe or Amer­ica. What were their to­tal pop­u­la­tion? Well, Europe+Amer­ica right now is, I think, some­thing like 700m peo­ple; I’d guess back in the 1700s, it was more like… 50m feels like a de­cent guess. How many peo­ple in to­tal? A de­cent ap­prox­i­ma­tion to ex­po­nen­tial pop­u­la­tion growth is to sim­ply use the av­er­age of 700m and 50m, which is 325, times 300 years, 112500m per­son-years, and a lifes­pan of 70 years, so 1607m per­sons over those 300 years. How many peo­ple have pi­anos? Visit­ing fam­i­lies, I rarely see pi­anos; maybe 1 in 10 had a pi­ano at any point. If fam­i­lies av­er­age a size of 4 and 1 in 10 fam­i­lies has a pi­ano, then we con­vert our to­tal pop­u­la­tion num­ber to, (1607m /​ 4) /​ 10, 40m pi­anos over that en­tire pe­riod. But wait, this is for fal­ling pi­anos, not all pi­anos; pre­sum­ably a fal­ling pi­ano must be at least on a sec­ond story. If it sim­ply crushes a mover’s foot while on the porch, that’s not very comedic at all. We want gen­uine ver­ti­cal­ity, real free fall. So our pi­ano must be on a sec­ond or higher story. Why would any­one put a pi­ano, baby or grand, that high? Un­less they had to, that is—be­cause they live in a city where they can’t af­ford a ground-level apart­ment or house. So we’ll ask in­stead for ur­ban fam­i­lies with pi­anos, on a sec­ond or higher story. The cur­rent ur­ban per­centage of the pop­u­la­tion is hit­ting ma­jor­ity (50%) in some coun­tries, but in the 1700s it would’ve been close to 0%. Aver­age again: 50+0/​2=25%, so we cut 40m by 75% to 30m. Every build­ing has a ground floor, but not ev­ery build­ing has more than 1 floor, so some ur­ban fam­i­lies will be able to live on the ground floor and put their pi­ano there and not fear a hu­morously mu­si­cal death from above. I’d guess (and here I have no good figures to jus­tify it) that the av­er­age ur­ban build­ing over time has closer to an av­er­age of 2 floors than more or less, since struc­tural steel is so re­cent, so we’ll cut 30m to 15m. So, there were 15m fam­i­lies in ur­ban ar­eas on non-ground-floors with pi­anos. And how would pi­anos get to non-ground-floors...? By lift­ing, of course, on cranes and things. (Yes, even in the 1700s. One as­pect of Am­s­ter­dam that struck me when I was vis­it­ing in 2005 was that each of the nar­row build­ing fronts had big hooks at their peaks; I was told this was for hoist­ing things up. Like pi­anos, I shouldn’t won­der.) Each pi­ano has to be lifted up, and, sad to say, taken down at some point. Even pi­anos don’t live for­ever. So that’s 30m hoist­ings and low­er­ings, each of which could be hilar­i­ously fatal, an av­er­age of 0.1m a year. How do we go from 30m crane op­er­a­tions to how many times a pi­ano falls and then also kills some­one? A pi­ano is se­ri­ously heavy, so one would ex­pect the failure rate to be non­triv­ial, but at the same time, the crews ought to know this and be ex­pe­rienced at mov­ing heavy stuff; off­hand, I’ve never heard of fal­ling pi­anos. At this point I cheated and look at the OSHA work­place fatal­ities data: 4609 for 2011. At a guess, half the USA pop­u­la­tion is gain­fully em­ployed, so 4700 out of 150m died. Let’s as­sume that ‘pi­ano mov­ing’ is not nearly as risky as it sounds and merely has the av­er­age Amer­i­can risk of dy­ing on the job. We have 100000 pi­ano hoist­ings a year, per pre­vi­ous. If a team of 3 can do lifts or hoist­ing of pi­anos a day, then we need 136 teams or 410 peo­ple. How many of these 410 will die each year, times 300? (410 * (4700/​150000000))*300 = 3.9 So ir­ri­tat­ingly, I’m not that sure that I can show that any­one has died by fal­ling pi­ano, even though I re­ally ex­pect that peo­ple have. Time to check in Google. Search­ing for kil­led by fal­ling pi­ano, I see: But no ac­tual cases of pi­anos fal­ling a story onto some­one. So, the calcu­la­tion may be right − 0 is within an or­der of mag­ni­tude of 3.9, af­ter all. • 0 is within an or­der of mag­ni­tude of 3.9, af­ter all. No it’s not! Ac­tu­ally it’s in­finitely many or­ders of mag­ni­tude away! • Nit­pick alert: I be­lieve pi­anos used to be a lot more com­mon. There was a time when they were a ma­jor source of at-home mu­sic. On the other hand, the pop­u­la­tion was much smaller then, so maybe the effects can­cel out. • I won­der. Pi­anos are still re­ally ex­pen­sive. They’re very bulky, need skil­led main­te­nance and tun­ing, use spe­cial high-ten­sion wires, and so on. Even if tech­nolog­i­cal progress, out­sourc­ing man­u­fac­ture to China etc haven’t re­duced the real price of pi­anos, the world is also much wealthier now and more able to af­ford buy­ing pi­anos. Another is­sue is the growth of the pi­ano as the stan­dard Pres­ti­gious In­stru­ment for the col­lege arms races (vastly more of the pop­u­la­tion goes to col­lege now than in 1900) or sig­nal­ing high cul­ture or moder­nity (in the case of East Asia); how many pi­anos do you sup­pose there are scat­tered now across the USA com­pared to 1800? Or in Ja­pan and China and South Korea com­pared to 1900? And on the other side, peo­ple used to make mu­sic at home, yes—but for that there are many cheaper, more portable, more durable al­ter­na­tives, such as cut-down ver­sions of pi­anos. • Pi­anos are still re­ally ex­pen­sive. Con­cert grands, yes, but who has room for one of those? Try sel­l­ing an old up­right pi­ano when clear­ing a de­ceased rel­a­tive’s es­tate. In the UK, you’re more likely to have to pay some­one to take it away, and it will just go to a scrapheap. Of course, that’s pre­sent day, and one rea­son no-one wants an old pi­ano is that you can get a bet­ter elec­tronic one new for a few hun­dred pounds. But back in Vic­to­rian times, as Nancy says el­sethread, a pi­ano was a stan­dard fea­ture of a Vic­to­rian par­lor, and that went fur­ther down the so­cial scale that you are imag­in­ing, and lasted at least through the first half of the twen­tieth cen­tury. Even bet­ter-off work­ing peo­ple might have one, though not the fac­tory drudges liv­ing in slums. It may have been differ­ent in the US though. • Con­cert grands, yes, but who has room for one of those? Try sel­l­ing an old up­right pi­ano when clear­ing a de­ceased rel­a­tive’s es­tate. Cer­tainly: http://​​www.ny­times.com/​​2012/​​07/​​30/​​arts/​​mu­sic/​​for-more-pi­anos-last-note-is-thud-in-the-dump.html?_r=2&ref=arts But like di­a­monds (I have been told that you can­not re­sell a di­a­mond for any­where near what you paid for it), and per­haps for similar rea­sons, I don’t think that mat­ters to the pro­duc­tion and sale of new ones. That ar­ti­cle sup­ports some of my claims about the glut of mod­ern pi­anos and falls in price, and hence the claim that there may be un­usu­ally many pi­anos around now than in ear­lier cen­turies: With thou­sands of mov­ing parts, pi­anos are ex­pen­sive to re­pair, re­quiring long hours of la­bor by skil­led tech­ni­ci­ans whose num­bers are diminish­ing. Ex­cel­lent digi­tal pi­anos and portable key­boards can cost as lit­tle as sev­eral hun­dred dol­lars. Low-end im­ported pi­anos have im­proved re­mark­ably in qual­ity and can be had for un­der$3,000. “In­stead of spend­ing hun­dreds or thou­sands to re­pair an old pi­ano, you can buy a new one made in China that’s just as good, or you can buy a digi­tal one that doesn’t need tun­ing and has all kinds of bells and whis­tles,” said Larry Fine, the ed­i­tor and pub­lisher of Acous­tic & Digi­tal Pi­ano Buyer, the in­dus­try bible.

At least, if we’re com­par­ing against the 1700s/​1800s, since the ar­ti­cle then goes on to give sales figures:

So from 1900 to 1930, the golden age of pi­ano mak­ing, Amer­i­can fac­to­ries churned out mil­lions of them. Nearly 365,000 were sold at the peak, in 1910, ac­cord­ing to the Na­tional Pi­ano Man­u­fac­tur­ers As­so­ci­a­tion. (In 2011, 41,000 were sold, along with 120,000 digi­tal pi­anos and 1.1 mil­lion key­boards, ac­cord­ing to Mu­sic Trades mag­a­z­ine.)

(Queen Vic­to­ria died in 1901, so if this golden age 1900-1930 also pop­u­lated par­lors, it would be more ac­cu­rate to call it an ‘Ed­war­dian par­lor’.)

• We got ~\$75 for one we picked up out of some­body garbage in a garage sale, and given the high in­ter­est we had in it, prob­a­bly could have got­ten twice that. (Had an ex­change stu­dent liv­ing with us who loved play­ing the pi­ano, and when we saw it, we had to get it—it ac­tu­ally played pretty well, too, only three of the chords needed re­place­ment. It was an ex­pe­rience load­ing that thing into a pickup truck with­out any equip­ment. Used a trash length of gar­den hose as rope and a -lot- of brute strength.)

• I was bas­ing my no­tion on hav­ing heard that a pi­ano was a stan­dard fea­ture of a Vic­to­rian par­lor. The origi­nal state­ment of the prob­lem just speci­fies a pi­ano, though I grant that the car­toon ver­sion re­quires a grand or baby grand. An up­right pi­ano just wouldn’t be as funny.

Th­ese days, there isn’t any mu­si­cal in­stru­ment which is a stan­dard fea­ture in the same way. In­stead, be­ing able to play recorded mu­sic is the stan­dard.

Thanks for the link about the lack of new mu­si­cal in­stru­ments. I’ve been think­ing for a while that sta­bil­ity of the clas­si­cal or­ches­tra meant there was some­thing wrong, but it hadn’t oc­curred to me that we’ve got the same sta­bil­ity in pop mu­sic.

• That ar­ti­cle treats all forms of syn­the­sis as one in­stru­ment. This is IMO not an ac­cu­rate model. The ex­plo­sion of elec­tronic pop in the ’80s was be­cause the tech­nol­ogy was on the up­ward slope of the lo­gis­tic curve, and new stuff was be­com­ing available on a reg­u­lar ba­sis for artists to glee­fully seize upon. But even now, there’s stuff you can do in 2013 that was largely out of reach, if not un­known, in 2000.

• But even now, there’s stuff you can do in 2013 that was largely out of reach, if not un­known, in 2000.

Have any handy ex­am­ples? I find that a bit sur­pris­ing (al­though it’s a dead cert that you know more about pop mu­sic than I do, so you’re prob­a­bly right).

• I’m talk­ing mostly about new things you can do mu­si­cally due to tech­nol­ogy. The par­tic­u­lar ex­am­ple I was think­ing of was au­to­tune, but that was ac­tu­ally in­vented in the late 1990s (whoops).

But digi­tal sig­nal pro­cess­ing in gen­eral has benefited hugely in Moore’s Law, and the ease af­forded by be­ing able to ap­ply tens or hun­dreds of filters in real time. The phase change mo­ment was when a mu­si­cian could do this in faster than 1x time on a home PC. The past decade has been mostly on the top of the S-curve, though.

Nev­er­the­less, treat­ing all syn­the­sis as one thing is sim­ply an in­cor­rect model.

• Funny co­in­ci­dence. About a week ago I was tel­ling some­one that peo­ple some­times give au­to­tune as an ex­am­ple of a qual­i­ta­tively new mu­si­cal/​au­ral de­vice, even though Godley & Creme ba­si­cally did it 30+ years ago. (Which doesn’t con­tra­dict what you’re say­ing; just be­cause it was pos­si­ble to mimic au­to­tune in 1979 doesn’t mean it was triv­ial, ac­cessible, or doable in real time. Although au­to­tune isn’t new, be­ing able to au­to­tune on an in­dus­trial scale pre­sum­ably is, ’cause of Moore’s law.)

• Agreed, al­though I don’t know how im­prac­ti­cal or un­known it was in 2000 — I re­mem­ber play­ing with GranuLab on my home PC around 2001.

• I was bas­ing my no­tion on hav­ing heard that a pi­ano was a stan­dard fea­ture of a Vic­to­rian par­lor.

Sure, but think how small a frac­tion of the pop­u­la­tion that was. Most of Vic­to­rian England was, well, poor; coal min­ers or fac­tory work­ers work­ing 16 hour days, that sort of thing. Not wealthy bour­geoisie with par­lors host­ing the sort of high so­ciety ladies who were raised learn­ing how to play pi­ano, sketch, and faint in the arms of suit­ors.

An up­right pi­ano just wouldn’t be as funny.

Un­less it’s set in a sa­loon! But given the low pop­u­la­tion den­sity of the Old West, this is a rel­a­tively small er­ror.

• Aver­age again: 50+0/​2=25%, so we cut 40m by 75% to 30m.

To 10m, surely?

• The work­place fatal­ities re­ally gone down re­cently, with all the safe jobs of sit­ting in front of the com­puter. You should look for work­place fatal­ities in con­struc­tion, prefer­ably his­tor­i­cal (be­fore safety guidelines). Ac­count­ing for that would raise the es­ti­mate.

A much big­ger is­sue is that one has to ac­tu­ally stand un­der the pi­ano as it is be­ing lifted/​low­ered. The rate of such hap­pen­ing can be much (or­ders of mag­ni­tude) be­low that of fatal work­place ac­ci­dents in gen­eral, and ac­count­ing for this would lower the es­ti­mate.

• You should look for work­place fatal­ities in con­struc­tion, prefer­ably his­tor­i­cal (be­fore safety guidelines).

I don’t know where I would find them, and I’d guess that any re­li­able figures would be very re­cent: OSHA wasn’t even founded un­til the 1970s, by which point there’s already been huge shifts to­wards safer jobs.

A much big­ger is­sue is that one has to ac­tu­ally stand un­der the pi­ano as it is be­ing lifted/​low­ered. The rate of such hap­pen­ing can be much (or­ders of mag­ni­tude) be­low that of fatal work­place ac­ci­dents in gen­eral, and ac­count­ing for this would lower the es­ti­mate.

That was the point of go­ing for life­time risks, to avoid hav­ing to di­rectly es­ti­mate per-lift­ing fatal­ity rates—I thought about it for a while, but I couldn’t see any re­motely rea­son­able way to es­ti­mate how many pi­anos would fall and how of­ten peo­ple would be near enough to be hit by it (which I could then es­ti­mate against num­ber of pi­anos ever lifted to pull out a fatal­ity rate, so in­stead I re­versed the pro­ce­dure and went with an over­all fatal­ity rate across all jobs).

• You also need to ac­count for the fact that some pro­por­tion of pi­ano-hois­ter work-re­lated fatal­ities will be to other fac­tors like heat­stroke or heart at­tack or wrap­ping the rope around their arm.

• To a very good first ap­prox­i­ma­tion, the dis­tri­bu­tion of fal­ling pi­ano deaths is Pois­son. So if the ex­pected num­ber of deaths is in the range [0.39, 39], then the prob­a­bil­ity that no one has died of a fal­ling pi­ano is in the range [1e-17, 0.677] which would lead us to be­lieve that with a prob­a­bil­ity of at least 13 such a death has oc­curred. (If 3.9 were the true av­er­age, then there’s only a 2% chance of no such deaths.)

• I dis­agree that the lower bound is 0; the right range is [-39,39]. Be­cause af­ter all, a fal­ling pi­ano can kill nega­tive peo­ple: if a pi­ano had fallen on Adolf Hitler in 1929, then it would have kil­led −5,999,999 peo­ple!

• Sorry. The prob­a­bil­ity is in the range [1e-17, 1e17].

• That is a large prob­a­bil­ity.

• It’s for when you need to be a thou­sand mil­lion billion per­cent sure of some­thing.

• A de­cent ap­prox­i­ma­tion to ex­po­nen­tial pop­u­la­tion growth is to sim­ply use the av­er­age of 700m and 50m

That ap­prox­i­ma­tion looks like this

It’ll over­es­ti­mate by a lot if you do it over longer time pe­ri­ods. e.g. it over­es­ti­mates this av­er­age by about 50% (your es­ti­mate ac­tu­ally gives 375, not 325), but if you went from 1m to 700m it would over­es­ti­mate by a fac­tor of about 3.

A pretty-easy way to es­ti­mate to­tal pop­u­la­tion un­der ex­po­nen­tial growth is just cur­rent pop­u­la­tion 1/​e life­time. From your num­bers, the pop­u­la­tion mul­ti­plies by e^2.5 in 300 years, so 120 years to mul­ti­ply by e. That’s two life­times, so the to­tal num­ber of lives is 700m2. For a smidgen more work you can get the “real” an­swer by do­ing 700m 2 − 50m 2.

• Ce­cil Adams tack­led this one. Although he could find no doc­u­mented cases of peo­ple be­ing kil­led by a fal­ling pi­ano (or a fal­ling safe), he did find one case of a guy be­ing kil­led by a RISING pi­ano while hav­ing sex with his girlfriend on it. What would you have es­ti­mated for the prob­a­bil­ity of that?

From the web­page:

The ex­cep­tion was the case of strip-club bouncer Jimmy Fer­rozzo. In 1983 Jimmy and his dancer girlfriend were hav­ing sex on top of a pi­ano that was rigged so it could be raised or low­ered for perfor­mances. Ap­par­ently in the heat of pas­sion the cou­ple ac­ci­den­tally hit the up switch, where­upon the pi­ano rose and crushed Jimmy to death against the ceiling. The girlfriend was pinned un­der­neath him for hours but sur­vived. I ac­knowl­edge this isn’t a sce­nario you want de­picted in de­tail on the Satur­day morn­ing car­toons; my point is that death due to ver­ti­cal pi­ano move­ment has a ba­sis in fact.

• You did much bet­ter in Ex­am­ple #2 than you thought; the con­clu­sion should read

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

which looks like a Fermi vic­tory (albeit an ar­ith­metic fail).

• Lol! Fixed.

• Thanks for writ­ing this! This is definitely an im­por­tant skill and it doesn’t seem like there was such a post on LW already.

Some mild the­o­ret­i­cal jus­tifi­ca­tion: one rea­son to ex­pect this pro­ce­dure to be re­li­able, es­pe­cially if you break up an es­ti­mate into many pieces and mul­ti­ply them, is that you ex­pect the er­rors in your pieces to be more or less in­de­pen­dent. That means they’ll of­ten more or less can­cel out once you mul­ti­ply them (e.g. one piece might be 4 times too large but an­other might be 5 times too small). More pre­cisely, you can com­pute the var­i­ance of the log­a­r­ithm of the fi­nal es­ti­mate and, as the num­ber of pieces gets large, it will shrink com­pared to the ex­pected value of the log­a­r­ithm (and even more pre­cisely, you can use some­thing like Hoeffd­ing’s in­equal­ity).

Another mild jus­tifi­ca­tion is the no­tion of en­tan­gled truths. A lot of truths are en­tan­gled with the truth that there are about 300 mil­lion Amer­i­cans and so on, so as long as you know a few rele­vant true facts about the world your es­ti­mates can’t be too far off (un­less the model you put those facts into is bad).

• More pre­cisely, you can com­pute the var­i­ance of the log­a­r­ithm of the fi­nal es­ti­mate and, as the num­ber of pieces gets large, it will shrink com­pared to the ex­pected value of the log­a­r­ithm (and even more pre­cisely, you can use some­thing like Hoeffd­ing’s in­equal­ity).

If suc­cess of a fermi es­ti­mate is defined to be “within a fac­tor of 10 of the cor­rect an­swer”, then that’s a con­stant bound on the al­lowed er­ror of the log­a­r­ithm. No “com­pared to the ex­pected value of the log­a­r­ithm” in­volved. Be­sides, I wouldn’t ex­pect the value of the log­a­r­ithm to grow with num­ber of pieces ei­ther: the log of an in­di­vi­d­ual piece can be nega­tive, and the true an­swer doesn’t get big­ger just be­cause you split the prob­lem into more pieces.

So, as­sum­ing in­de­pen­dent er­rors and us­ing ei­ther Hoeffd­ing’s in­equal­ity or the cen­tral limit the­o­rem to es­ti­mate the er­ror of the re­sult, says that you’re bet­ter off us­ing as few in­puts as pos­si­ble. The rea­son fermi es­ti­mates even in­volve more than 1 step, is that you can make the per-step er­ror smaller by choos­ing pieces that you’re some­what con­fi­dent of.

• Oops, you’re ab­solutely right. Thanks for the cor­rec­tion!

• There are 3141 coun­ties in the US. This is easy to re­mem­ber be­cause it’s just the first four digits of pi (which you already have mem­o­rised, right?).

• This re­minds me of the sur­pris­ingly ac­cu­rate ap­prox­i­ma­tion of pi x 10^7 sec­onds in a year.

• This chart has been ex­tremely helpful to me in school and is full of weird ap­prox­i­ma­tion like the two above.

• I will note that I went through the men­tal ex­er­cise of cars in a much sim­pler (and I would say bet­ter) way: I took the num­ber of cars in the US (300 mil­lion was my guess for this, which is ac­tu­ally fairly close to the ac­tual figure of 254 mil­lion claimed by the same ar­ti­cle that you refer­enced) and guessed about how long cars typ­i­cally ended up last­ing be­fore they went away (my es­ti­mate range was 10-30 years on av­er­age). To have 300 mil­lion cars, that would sug­gest that we would have to pur­chase new cars at a suffi­ciently high rate to main­tain that num­ber of ve­hi­cles given that lifes­pan. So that gave me a range of 10-30 mil­lion cars pur­chased per year.

The num­ber of 5 mil­lion cars per year ab­solutely floored me, be­cause that ac­tu­ally would fail my san­ity check—to get 300 mil­lion cars, that would mean that cars would have to last an av­er­age of 60 years be­fore be­ing re­placed (and in ac­tu­al­ity would in­di­cate a re­place­ment rate of 250M/​5M = 50 years, ish).

The ac­tual cause of this is that car sales have PLUMMETED in re­cent times. In 1990, the me­dian age of a ve­hi­cle was 6.5 years; in 2007, it was 9.4 years, and in 2011, it was 10.8 years—mean­ing that in be­tween 2007 and 2011, the me­dian car had in­creased in age by 1.4 years in a mere 4 years.

I will note that this sort of calcu­la­tion was taught to me all the way back in el­e­men­tary school as a sort of “math­emagic”—us­ing math to get good re­sults with very lit­tle knowl­edge.

But it strikes me that you are per­haps try­ing too hard in some of your calcu­la­tions. Of­ten­times it pays to be lazy in such things, be­cause you can eas­ily over­com­pen­sate.

• Tip: frame your es­ti­mates in terms of in­ter­vals with con­fi­dence lev­els, i.e. “90% prob­a­bil­ity that the an­swer is within and ”. Try to work out both a 90% and a 50% in­ter­val.

I’ve found in­ter­val es­ti­mates to be much more use­ful than point es­ti­mates, and they com­bine very well with Fermi tech­niques if you keep track of how much round­ing you’ve in­tro­duced over­all.

In ad­di­tion, you can com­pute a Brier score when/​if you find out the cor­rect an­swer, which gives you a tar­get for im­prove­ment.

• Dou­glas W. Hub­bard has a book ti­tled How to Mea­sure Any­thing where he states that half a day of ex­er­cis­ing con­fi­dence in­ter­val cal­ibra­tion makes most peo­ple nearly perfectly cal­ibrated. As you noted and as is said here, that method fits nicely with Fermi es­ti­mates.

This com­bi­na­tion seems to have a great ra­tio be­tween train­ing time and use­ful­ness.

• 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.

Would it bankrupt the global econ­omy to or­bit all the world’s Sikhs?

• 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.

I’m not sure about this claim about day-to-day life. Maybe there are some lines of work where this skill could be use­ful, but in gen­eral it’s quite rare in day-to-day life where you have to come up with quick es­ti­mates on the spot to make a sound de­sci­sion. Many things can be looked up on the in­ter­net for a rather marginal time-cost nowa­days. Often enough prob­a­bly even less time, than it would ac­tu­ally take some­one to calcu­late the guessti­mate.

If a de­sci­sion or statis­tic is im­por­tant you, should take the time to ac­tu­ally just look it up, or if the in­for­ma­tion you are try­ing to guess is im­pos­si­ble to find on­line, you can at least look up some statis­tics that you can and should use to make your guess bet­ter. As you read above, just get­ting one es­ti­mate in a long line of rea­son­ing wrong (es­pe­cially where big num­bers are con­cerned) can throw off your guess by a fac­tor of 100 or 1000 and make it use­less or even harm­ful.

If your guess is im­por­tant to an ar­gu­ment you’re con­struct­ing on-the-fly, I think you could also take the time to just look it up. (If it’s not an in­ter­view or some con­ver­sa­tion which dic­tates, that us­ing a smart­phone would be un­ac­cept­able).

And if a de­sci­sion or ar­gu­ment is not im­por­tant enough to in­vest some time in a quick on­line search, then why bother in the first place? Sure, it’s a cool skill to show off and it re­quires some ra­tio­nal­ity, but that doesn’t mean it’s truly use­ful. On the other hand maybe I’m just par­tic­u­larly uni­mag­i­na­tive to­day and can’t think of ways, how Fermi es­ti­mates could pos­si­bly im­prove my day-to-day life by a mar­gin that would war­rant the effort to get bet­ter at it.

• Write down your own Fermi es­ti­ma­tion at­tempts here. One Fermi es­ti­mate per com­ment, please!

• One fa­mous Fermi es­ti­mate is the Drake equa­tion.

• A run­ning list of my own: http://​​www.gw­ern.net/​​Notes#fermi-calcu­la­tions (And there’s a num­ber of them float­ing around pre­dic­tion­book.com; Fermi-style loose rea­son­ing is great for con­strain­ing pre­dic­tions.)

• Here’s one I did with Mar­cello awhile ago: about how many high schools are there in the US?

My at­tempt: there are 50 states. Each state has maybe 20 school dis­tricts. Each dis­trict has maybe 10 high schools. So 50 20 10 = 10,000 high schools.

Mar­cello’s at­tempt (IIRC): there are 300 mil­lion Amer­i­cans. Of these, maybe 50 mil­lion are in high school. There are maybe 1,000 stu­dents in a high school. So 50,000,000 /​ 1,000 = 50,000 high schools.

Ac­tual an­swer:

...

...

...

Num­bers vary, I think de­pend­ing on what is be­ing counted as a high school, but it looks like the ac­tual num­ber is be­tween 18,000 and 24,000. As it turns out, the first ap­proach un­der­es­ti­mated the to­tal num­ber of school dis­tricts in the US (it’s more like 14,000) but over­es­ti­mated the num­ber of high schools per dis­trict. The sec­ond ap­proach over­es­ti­mated the num­ber of high school stu­dents (it’s more like 14 mil­lion) but also over­es­ti­mated the av­er­age num­ber of stu­dents per high school. And the ge­o­met­ric mean of the two ap­proaches is 22,000, which is quite close!

• I tried the sec­ond ap­proach with bet­ter suc­cess: it helps to break up the “how many Amer­i­cans are in high school” calcu­la­tion. If the av­er­age Amer­i­can lives for 80 years, and goes to high school for 4, then 120 of all Amer­i­cans are in high school, which is 15 mil­lion.

• there are 300 mil­lion Amer­i­cans. Of these, maybe 50 mil­lion are in high school

...you guessed 1 out of 6 Amer­i­cans is in high­school?

With an av­er­age lifes­pan of 70+ years and a high­school du­ra­tion of 3 years (edit: oh, it’s 4 years in the US?), shouldn’t it be some­where be­tween 1 in 20 and 1 in 25?

• This con­ver­sa­tion hap­pened some­thing like a month ago, and it was Mar­cello us­ing this ap­proach, not me, so my mem­ory of what Mar­cello did is fuzzy, but IIRC he used a big num­ber.

The dis­tri­bu­tion of pop­u­la­tion shouldn’t be ex­actly uniform with re­spect to age, al­though it’s prob­a­bly more uniform now than it used to be.

• Just tried one to­day: how safe are planes?

Last time I was at an air­port, the screen had five flights, three-hour pe­riod. It was peak time, so mul­ti­plied only by fif­teen, so 25 flights from Chen­nai air­port per day.

~200 coun­tries in the world, so guessed 500 ad­justed air­ports (effec­tive no. of air­ports of size of Chen­nai air­port), giv­ing 12500 flights a day and 3*10^6 flights a year.

One crash a year from my news mem­o­ries, gives pos­si­bil­ity of plane crash as 1310^-6 ~ 310^-7.

Prob­a­bil­ity of dy­ing in a plane crash is 3*10^-7 (source). At hun­dred dead pas­sen­gers a flight, fatal crashes are ~ 10^-5. Off by two or­ders of mag­ni­tude.

• Prob­a­bil­ity of dy­ing in a plane crash is 3*10^-7 (source). At hun­dred dead pas­sen­gers a flight, fatal crashes are ~ 10^-5. Off by two or­ders of mag­ni­tude.

If there are 310^6 flights in a year, and one ran­domly se­lected plane crashes per year on av­er­age, with all aboard be­ing kil­led, then the chances of dy­ing in an air­plane crash are $\\left \( \\frac\{1\}\{3\} \\right \$%20\10%5E{-6}), surely?

Yes, there’s a hun­dred dead pas­sen­gers on the flight that went down, but there’s also a hun­dred liv­ing pas­sen­gers on ev­ery flight that didn’t go down. The hun­dreds can­cel out.

• The hun­dreds can­cel out.

Wow, that was stupid of me. Of course they do! And thanks.

• Anna Sala­mon’s Sin­gu­lar­ity Sum­mit talk from a few years ago ex­plains one Fermi es­ti­mate re­gard­ing the value of gath­er­ing more in­for­ma­tion about AI im­pacts: How Much it Mat­ters to Know What Mat­ters: A Back of the En­velope Calcu­la­tion.

• How old is the SECOND old­est per­son in the world com­pared to the old­est? Same for the united states?

I bogged down long be­fore I got the an­swer. Below is the gib­ber­ish I gen­er­ated to­wards bog­ging down.

So OK, I don’t even know off­hand how old is the old­est, but I would bet it is in the 114 years old (yo) to 120 yo range.

Then figure in some hand-wavey way that peo­ple die at ages nor­mally dis­tributed with a mean of 75 yo. We can es­ti­mate how many sigma (stan­dard de­vi­a­tions) away from that is the old­est per­son.

Figure there are 6 billion peo­ple now, but I know this num­ber has grown a lot in my life­time, it was less than 4 billion when I was born 55.95 years ago. So say the 75 yo’s come from a pop­u­la­tion con­sis­tent with 3 billion peo­ple. 12 die younger than 75, 12 die older, so the old­est per­son in the world is 1 in 1.5 billion on the dis­tri­bu­tion.

OK what do I know about nor­mal dis­tri­bu­tions? Nor­mal dis­tri­bu­tion goes as exp ( -((mean-x)/​(2sigma))^2 ). So at what x is exp( -(x/​2sigma)^2 ) = 1e-9? (x /​ 2sigma) ^ 2 = -ln ( 1e-9). How to es­ti­mate nat­u­ral log of a billionth? e = 2.7 is close enough for gov­ern­ment work to the sqrt(10). So ln(z) = 2log_10(z). Then -ln(1e-9) = −2log_10(1e-9) = 29 = 18. So (x/​2sigma)^2 = 18, sqrt(18) = 4 so

So I got 1 in a billion is 4 sigma. I didn’t trust that so I looked that up, Maybe I should have trusted it, in fact 1 in a billion is (slightly more than ) 6 sigma.

mean of 75 yo, x=115 yo, x-mean = 40 years. 6 sigma is 40 years. 1 sigma=6 years.

So do I have ANYTHING yet? I am look­ing for dx where exp(-((x+dx)/​(2sigma))^2) - exp( -(x/​2sigma)^2)

• So, this isn’t quite ap­pro­pri­ate for Fermi calcu­la­tions, be­cause the math in­volved is a bit in­tense to do in your head. But here’s how you’d ac­tu­ally do it:

Age-re­lated mor­tal­ity fol­lows a Gom­pertz curve, which has much, much shorter tails than a nor­mal dis­tri­bu­tion.

I’d start with or­der statis­tics. If you have a pop­u­la­tion of 5 billion peo­ple, then the ex­pected per­centile of the top per­son is 1-(1/​10e9), and the ex­pected per­centile of the sec­ond best per­son is 1-(3/​10e9). (Why is it a 3, in­stead of a 2? Be­cause each of these ex­pec­ta­tions is in the mid­dle of a range that’s 1/​5e9, or 2/​10e9, wide.)

So, the ex­pected age* of death for the old­est per­son is 114.46, us­ing the num­bers from that post (and com­mit­ting the sin of re­port­ing sev­eral more sig­nifi­cant figures), and the ex­pected age of death for the sec­ond old­est per­son is 113.97. That sug­gests a gap of about six months be­tween the old­est and sec­ond old­est.

* I should be clear that this is the age cor­re­spond­ing to the ex­pected per­centile, not the ex­pected age, which is a more in­volved calcu­la­tion. They should be pretty close, es­pe­cially given our huge pop­u­la­tion size.

But ter­mi­nal age and cur­rent age are differ­ent- it could ac­tu­ally be that the per­son with the higher ter­mi­nal age is cur­rently younger! So we would need to look at per­mu­ta­tions and a bunch of other stuff. Let’s ig­nore this and as­sume they’ll die on the same day.

So what does it look like in re­al­ity?

The longest lived well-recorded hu­man was 122, but note that she died less than 20 years ago. The to­tal pop­u­la­tion whose births were well-recorded is sig­nifi­cantly smaller than the cur­rent pop­u­la­tion, and the num­bers are even more pes­simistic than the 3 billion figure you get at; in­stead of look­ing at peo­ple al­ive in the 1870s, we need to look at the num­ber born in the 1870s. Our model es­ti­mates she’s a 1 in 2*10^22 oc­cur­rence, which sug­gests our model isn’t tuned cor­rectly. (If we re­place the 10 with a 10.84, a rel­a­tively small change, her age is now the ex­pec­ta­tion for the old­est ter­mi­nal age in 5 billion- but, again, she’s not out of a sam­ple of 5 billion.)

The real gaps are here; about a year, an­other year, then months. (A de­crease in gap size is to be ex­pected, but it’s clear that our model is a bit off, which isn’t sur­pris­ing, given that all of the co­effi­cients were re­ported at 1 sig­nifi­cant figure.)

• Upvoted (among other things) for a way of de­ter­min­ing the dis­tri­bu­tion of or­der statis­tics from an ar­bi­trary dis­tri­bu­tion know­ing those of a uniform dis­tri­bu­tion which sounds ob­vi­ous in ret­ro­spect but to be hon­est I would never have come up with on my own.

• I am look­ing for dx where exp(-((x+dx)/​(2sigma))^2) - exp( -(x/​2sigma)^2)

As­sum­ing dx << x, this is ap­prox­i­mated by a differ­en­tial, (-xdx/​sigma^2) * exp( -(x/​2sigma)^2, or the rel­a­tive drop of dx/​sigma^2. You want it to be 12 (lost one per­son out of two), your x = 4 sigma, so dx=1/​8 sigma, which is un­der a year. Of course, it’s rather op­ti­mistic to ap­ply the nor­mal dis­tri­bu­tion to this prob­lem, to be­gin with.

• I es­ti­mated how much the pop­u­la­tion of Helsinki (cap­i­tal of Fin­land) grew in 2012. I knew from the news that the growth rate is con­sid­ered to be steep.

I knew there are cur­rently about 500 000 habitants in Helsinki. I set the up­per bound to 3 % growth rate or 15 000 res­i­dents for now. With that rate the city would grow twen­tyfold in 100 years which is too much. But the rate might be steeper now. For lower bound i chose 1000 new res­i­dents. I felt that any­thing less couldnt re­ally pro­duce any news. AGM is 3750.

My sec­ond method was to go through the num­ber of new apart­ments. Here I just checked that in re­cent years about 3000 apart­ments have been built yearly. Guess­ing that the house­hold size could be 2 per­sons I got 6000 new res­i­dents.

It turned out that the pop­u­la­tion grew by 8300 res­i­dents which is high­est in 17 years. Other­wise it has re­cently been around 6000. So both meth­ods worked well. Both have the benefit that one doesnt need to care whether the growth comes from births/​deaths or peo­ple flow. They also didn’t re­quire con­sid­er­ing how many peo­ple move out and how many come in.

Ob­vi­ously i was much more con­fi­dent on the sec­ond method. Which makes me think that ap­ply­ing con­fi­dence in­ter­vals to fermi es­ti­mates would be use­ful.

• For the “Only Shal­low” one, I couldn’t think of a good way to break it down, and so be­gan by ap­prox­i­mat­ing the to­tal num­ber of listens at 2 mil­lion. My fi­nal es­ti­mate was off by a fac­tor of one.

• Matt Ma­honey’s es­ti­mate of the cost of AI is a sort-of Fermi es­ti­mate.

• Out of the price of a new car, how much goes to buy­ing raw ma­te­ri­als? How much to cap­i­tal own­ers? How much to la­bor?

• How many Wall-Marts in the USA.

• That sounds like the kind of thing you could just Google.

But I’ll bite. Wal-Marts have the ad­van­tage of be­ing pretty evenly dis­tributed ge­o­graph­i­cally; there’s rarely more than one within easy driv­ing dis­tance. I re­call there be­ing about 15,000 towns in the US, but they aren’t uniformly dis­tributed; they tend to cluster, and even among those that aren’t clus­tered a good num­ber are go­ing to be too small to sup­port a Wal-Mart. So let’s as­sume there’s one Wal-Mart per five towns on av­er­age, tak­ing into ac­count clus­ter­ing effects and towns too small or iso­lated to sup­port one. That gives us a figure of 3,000 Wal-Marts.

When V Google it, that gheaf out to or very close to the num­ber of Wal-Mart Su­per­centers, the large flag­ship stores that the phrase “Wal-Mart” brings to mind. How­ever, Wal-Mart also op­er­ates a smaller num­ber of “dis­count store”, “neigh­bor­hood mar­ket”, and “ex­press” lo­ca­tions that share the same brand­ing. If we in­clude “dis­count” and “neigh­bor­hood” lo­ca­tions, the to­tal is about three thou­sand eight hun­dred. I can’t find the num­ber of “ex­press” stores, but the for­mat was cre­ated in 2011 so there prob­a­bly aren’t too many.

• Differ­ent method. As­sume all 300 mil­lion us cit­i­zens are served by a Wal Mart. Any pop­u­la­tion that doesn’t live near a Wal-Mart has to be small enough to ig­nore. Each Wal-mart prob­a­bly has be­tween 10,000 and 1 mil­lion po­ten­tial cus­tomers. Both fringes seem un­likely, so we can be within a fac­tor of 10 by guess­ing 100000 peo­ple per Wal-Mart. This also leads to 3000 Wal-Marts in the US.

• The guys at last.fm are usu­ally very will­ing to help out with in­ter­est­ing re­search (or at least were when I worked there a cou­ple of years ago), so if you par­tic­u­larly care about that in­for­ma­tion it’s worth try­ing to con­tact them.

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• You can try es­ti­mat­ing, or keep list­ing the rea­sons why it’s hard.

• One of my fa­vorite num­bers to re­mem­ber to aid in es­ti­ma­tions is this: 1 year = pi * 10^7 sec­onds. Its re­ally pretty ac­cu­rate.

Of course for Fermi es­ti­ma­tion just re­mem­ber 1 Gs (gi­gasec­ond) = 30 years.

• I spend prob­a­bly a pretty un­usual amount of time es­ti­mat­ing things for fun, and have come to use more or less this ex­act pro­cess on my own over time from do­ing it.

One thing I’ve ob­served, but haven’t truly tested, is my ge­o­met­ric means seem to be much more effec­tive when I’m will­ing to put a more tight guess on them. I started off bound­ing them with what I thought the an­swer con­ceiv­ably could be, which seemed ob­jec­tive and of­ten felt eas­ier to es­ti­mate. The prob­lem was that of­ten ei­ther the lower or up­per bound was too ar­bi­trary rel­a­tive to it’s weight on my fi­nal es­ti­mate. Say, av­er­age times an av­er­age 15 year old sends an mms photo in a week. My up­per bound may be 100ish but my lower bound could be 2 al­most as eas­ily as it could be 5 which ranges my fi­nal es­ti­mate quite a bit, be­tween 14 and 22.

• I just wanted to say, af­ter read­ing the Fermi es­ti­mate of cars in the US, I liter­ally clapped—out loud. Well done. And I highly ap­pre­ci­ate the hon­est poor first at­tempt—so that I don’t feel like such an idiot next time I com­pletely fail.

• Po­ten­tially use­ful: http://​​in­sta­calc.com/​​

• I’m happy to see that the Great­est Band of All Time is the only rock band I can re­call ever men­tioned in a top-level LessWrong post. I thought ra­tio­nal­ists just sort of listened only to Great Works like Bach or Mozart, but I guess I was wrong. Clearly luke­prog used his skills as a ra­tio­nal­ist to ra­tio­nally de­duce the band with the great­est tal­ent, cre­ativity, and artis­tic im­pact of the last thirty years and then de­cided to put a refer­ence to them in this post :)

• If you check out Me­dia posts, you’ll see that LWers like a range of mu­sic. It wouldn’t sur­prise me too much if they tend to like con­tem­po­rary clas­si­cal bet­ter than clas­si­cal clas­si­cal.

• I like a spe­cific sub­set of clas­si­cal clas­si­cal, but I sus­pect not at all typ­i­cal.

• I thought ra­tio­nal­ists just sort of listened only to Great Works like Bach or Mozart

Why?

• Be­cause of the images of differ­ent mu­si­cal gen­res in our cul­ture. There is an as­so­ci­a­tion of clas­si­cal mu­sic and be­ing aca­demic or up­per class. In pop­u­lar me­dia, lik­ing clas­si­cal mu­sic is a cheap sig­nal for these char­ac­ter types. This nat­u­rally trig­gers con­fir­ma­tion bi­ases, as we view the ra­tio­nal­ist listen­ing to Bach as typ­i­cal, and the ra­tio­nal­ist listen­ing to The Rol­ling Stones as atyp­i­cal. Peo­ple also use mu­si­cal prefer­ence to sig­nal what type of per­son they are. If some­one wants to be seen as a ra­tio­nal­ist, they of­ten men­tion their love of Bach and don’t men­tion gen­res with a differ­ent image, ex­cept to dis­par­age them.

• I think you’re con­flat­ing “ra­tio­nal­ist” and “in­tel­lec­tual.” I agree that there is a stereo­type that in­tel­lec­tu­als only listen to Great Works like Bach or Mozart, but I’m cu­ri­ous where the OP picked up that this stereo­type also ought to ap­ply to LW-style ra­tio­nal­ists. I mean, Eliezer takes pains in the Se­quences to make anime refer­ences speci­fi­cally to avoid this kind of thing.

• I mean, Eliezer takes pains in the Se­quences to make anime refer­ences speci­fi­cally to avoid this kind of thing.

Well, he also likes tends to like anime, and anime has a ten­dency to deal with some fu­ture-ish is­sues.

• When­ever some­one com­pli­ments “Eliezer Yud­kowsky”, they are re­ally com­pli­ment­ing “Eliezer Yud­kowsky’s writ­ing” or “Eliezer Yud­kowsky’s best writ­ing that stands out most in my mind”. Peo­ple who met me in per­son were of­ten shocked at how much my in-per­son im­pres­sion de­parted from the pic­ture they had in their minds. I think this mostly had to do with imag­in­ing me as be­ing the sort of ac­tor who would be cho­sen to play me in the movie ver­sion of my life—they imag­ined way too much dig­nity. That forms a large part of the rea­son why I oc­ca­sion­ally toss in the de­liber­ate anime refer­ence, which does seem to have fixed the di­ver­gence a bit.

• I’m just point­ing out the way such a bias comes into be­ing. I know I don’t listen to clas­si­cal, and al­though I’d ex­pect a slightly higher pro­por­tion here than in the gen­eral pop­u­la­tion, I wouldn’t guess it wold be a ma­jor­ity or sig­nifi­cant plu­ral­ity.

If I had to guess, I’d guess on varied mu­si­cal tastes, prob­a­bly trend­ing to­wards more niche gen­res than broad spec­trum pop than the gen­eral pop­u­la­tion.

• Well, Eliezer men­tions Bach a bunch in the se­quences as an ex­am­ple of a great work of art. I used stereo­types to ex­trap­o­late. :p

(the state­ment was tongue-in-cheek, if that didn’t come across. I am gen­uinely a lit­tle sur­prised to see MBV men­tioned here though.)

• I re­cently ran across an ar­ti­cle de­scribing how to find a rough es­ti­mate of the stan­dard de­vi­a­tion of a pop­u­la­tion, given a num­ber of sam­ples, which seems that it would be suit­able for Fermi es­ti­mates of prob­a­bil­ity dis­tri­bu­tions.

First of all, you need a large enough pop­u­la­tion that the cen­tral limit the­o­rem ap­plies, and the dis­tri­bu­tion can there­fore be as­sumed to be nor­mal. In a nor­mal dis­tri­bu­tion, 99.73% of the sam­ples will be within three stan­dard de­vi­a­tions of the mean (ei­ther above or be­low; a to­tal range of six stan­dard de­vi­a­tions). There­fore, one can roughly es­ti­mate the stan­dard de­vi­a­tion by tak­ing the largest value, sub­tract­ing the small­est value, and di­vid­ing the re­sult by 6.

Source

This is use­ful, be­cause in a nor­mal dis­tri­bu­tion, around 7 in 10 of the sam­ples will be within one stan­dard de­vi­a­tion of the mean, and around 19 in ev­ery 20 will be within two stan­dard de­vi­a­tions of the mean.

• If you want a range con­tain­ing 70% of the sam­ples, why wouldn’t you just take the range be­tween the 15th and 85th per­centile val­ues?

• That would also work, and prob­a­bly be more ac­cu­rate, but I sus­pect that it would take longer to find the 15th and 85th per­centile val­ues than it would to find the ends of the range.

• How long can the In­ter­na­tional Space Sta­tion stay up with­out a boost? I can think of a cou­ple of ways to es­ti­mate that.

• I recom­mend try­ing to take the har­monic mean of a phys­i­cal and an eco­nomic es­ti­mate when ap­pro­pri­ate.

• I recom­mend do­ing ev­ery­thing when ap­pro­pri­ate.

Is there a par­tic­u­lar rea­son why the har­monic mean would be a par­tic­u­larly suit­able tool for com­bin­ing phys­i­cal and eco­nomic es­ti­mates? I’ve spent only a few sec­onds try­ing to think of one, failed, and had trou­ble mo­ti­vat­ing my­self to look harder be­cause on the face of it it seems like for most prob­lems for which you might want to do this you’re about equally likely to be find­ing any given quan­tity as its re­cip­ro­cal, which sug­gests that a gen­eral prefer­ence for the har­monic mean is un­likely to be a good strat­egy—what am I miss­ing?

• Can I have a rep­re­sen­ta­tive ex­am­ple of a prob­lem where this is ap­pro­pri­ate?

• So, what you’re say­ing is that the larger num­ber is less likely to be ac­cu­rate the fur­ther it is from the smaller num­ber? Why is that?

• Out of the price of a new car, how much goes to buy­ing raw ma­te­ri­als? How much to cap­i­tal own­ers? How much to la­bor?