Fermi problem game thread
I thought it might be fun to make a little game out of asking Fermi problems and trying to work out solutions to others'.
What is a Fermi problem?
A Fermi problem is (to quote wikipedia ):
an estimation problem designed to teach dimensional analysis or approximation, and such a problem is usually a backoftheenvelope calculation. The estimation technique is named after physicist Enrico Fermi as he was known for his ability to make good approximate calculations with little or no actual data. Fermi problems typically involve making justified guesses about quantities and their variance or lower and upper bounds.
Basically, these are questions that would be very difficult to calculate exactly without looking up the answer. The goal is to arrive at a good estimate by making justified assumptions. As such, looking up facts and statistics should be minimized or outright avoided (e.g., if the question is "What is the mass of the water in Lake Michigan?", you shouldn't look up Lake Michigan's volume to aid your answer. This should be estimated from things you know off the top of your head).
The way this thread works
 If you have a Fermi problem, post it as a top level comment.
 If you have a solution to an already posted Fermi problem, post it as a reply to that comment. Be sure to post your reasoning and thought process.
Fermi problems can be fun to come up with, and fun to answer. Examples of Fermi problems might include:

How many piano tuners are there in Chicago? (this one is apparently one that Fermi came up with himself)

How many keystrokes occur worldwide, daily?

What is the mass of the water in Lake Michigan?

How many cars are there in New York City on a given day?

How many blades of grass are there on the National Mall?

How many eggs are consumed in the US each day?
Since many of these have no definitive answer, it is not always possible to score the 'correctness' of an answer. Nonetheless, others can judge when an answer makes reasonable assumptions, so feel free reply to answers with suggestions of which assumptions can be refined or further justified.
How many pixels die each day?
Anything counts as a pixel, be it a CRT or Projector (do these two even die?), LCD, OLED or just a straight up LED in one of those massive outdoor displays.
We discount dead displays, just looking for pixels or sections of pixels that die but leave the display otherwise functional.
I'm gonna discount CRTs and outdoor displays as being statistically insignificant (they wouldn't really change the result one way or another). About 1.9 billion desktop PCs, 2.3 billion androids, 0.7 billion iphones (these numbers are rather suspect, and could be completely wrong). Desktops have an average resolution of about 1920x1080, for 2,073,600 pixels per each. Iphones it's hard to get a good number, so I'm going to say they average around maybe 800x1400, just from eyeballing this; 1,120,000 pixels per device, average. With android, there's a really nice table with all the resolutions; that makes for an average of 1,049,708.096 pixels per device. It is a couple of years out of date, but I doubt numbers have changed that much. Multiplying everything together, that's 7,138,168,620,800,000 living pixels in the world, 7 quadrillion. I'll round up to 10 quadrillion, to account for other devices (CRTs, outdoor displays), improvement in android resolutions since 2017, and to make calculations easier.
As for pixel death, that's harder: I can't find any hard numbers for rate there. Best I can find is this post from 15 years ago, which says 0.0000125% of all pixels are dead; that means 1,250,000,000, or 1.25 billion pixels are dead.
Damn, well nice one.
I'll try finish it off with some completely anecdotal evidence from the devices I've had contact with for the last 5 years. I can think of 6 screens at average of 1.2 million pixels per device,
7.2*10^6
total pixels.These run for lets say an average of 12 hours a day. That's
2.2*10^3
hours of running time.I noticed dead pixels on two devices. That's
2.7^105%
of the pixels that died over 2.2 thousand hours of running time.An average of
1.3*10^8
percent of the pixels under my watch die every hour.Lets scale it up to the entire world
1.3*10^8%
of the worlds pixels is9*10^7
pixels, 90 million die every hour.Finally we get
2.2*10^9
, or simply 2.2 billion pixels die every single day.It's interesting how similar the two figures are to each other, 1.2 billion that are dead vs my 2.2 that die each day, yet so massively different in scope.
Though I took, 1.25*10^5% of 7 trillion and got 89.2 billion dead pixels, my guess if you dropped a couple 0s while using the rounded figure.
At my pixel death rate it would take 40 years to fill the dead pixel quota which is probably a bit high for an expected lifetime of a device. The logic being that after 40 years all devices are renewed, so the rate of pixel deaths need to be able to reach that point.
I'm just shocked these two independently found figures produce a number so reasonable, considering the gymnastics involved getting there.
Hah, I think you dropped a couple zeros! If I do
7138168620800000 * 1.25 * 10^(5)
, then I get your number, but that's not enough; remember, it's 1.25e5 percent, so it has to be7138168620800000 * 1.25 * 10^(5) * 10^(2)
, which comes out to more like 892 million.I'm guessing the discrepency between the two figures comes mostly from the fact that my total death rates are relying on figures from 15 years ago, and displays have improved since.
Ah, yes ofc. I accounted for the % when doing my calculation above this, but forgot to do it here. whoops.
Yeah, also it could be that I just lucky with my screen sample. It's a pretty tiny sample size after all.
I will post one or two to start:
How many keystrokes occur worldwide, daily?
This method is very common in physics, where we call these 'back of the envelope' calculations. Rounding to the nearest power of 10 for all values (coincidentally this is why I switched from astrophysics to material physics) is common in this scenario, because you do not know all data with a greater accuracy than that anyway. So let's try this.
I define a keystroke as taking place on a computer keyboard (so no smartphones etc).
This leads to 10^(3+0+10) = 10^13 or 1000000000000 keystrokes per day.
I worked this one out yesterday in a bit of detail (really just making more explicit and therefore more likely to be wrong guesses for numbers), and got ~10^13. I think your estimate for total people in the world should be 10^9, which would bring us closer in alignment. I think I had a higher estimate for total keystrokes per hour.
Ah yeah, that should definitely be 10^9. Gonna fix that. I rounded down the number of keystrokes per hour because the values quoted in keystrokes per minute or words per minute are usually not sustainable over long periods of time.
Edit: Actually, I'll change it to 10^10 as 7 billion is closer to that then to 10^9.
I read just yesterday that 50% of the world had access to the Internet in 2018, and those are the people who presumably have or can use a device with a keyboard (even on a dumb phone) on a daily basis. For a population of... fuck, we are 7.7 billions already? That comes up to ~4 billions.
Now I'd apply the Pareto principle, and try to estimate how much a random person would be writing each day  10k keypresses come up to 2k words, but that feels too much. 1k/200 words seems somewhat more reasonable, and that would come up to ~3.2e12 keypresses per day for the bottom 80%. Let's double that (80/20), and we get up to ~6.4e12 keypresses per day  rounding up to 7e12.
(That'd mean that the top 20%  the content creators, more or less  only produce 1k words per day. This seems a bit low to me, but it's probably not too off overall.)
What's the volume of all buildings in Manhattan, combined?
Area of Manhattan is 87 km^2. That's 87,000,000m^3
I'm going to say the average building is 10 stories. That's 30m. 90x30=2700 so we have a volume of 2,700,000,000 m^3.
I wish with this stuff there was an actual value to then compare to.
How many bricks are there in London?
How many atoms are there in the human body?
One my college physics professor gave us after he realized we didn't know how to do backoftheenvelope problems: How much would sea level rise if the ice caps melted?
I seem to recall that many of the numbers we used weren't factors of ten, so bonus points if your answer has better than orderofmagnitude precision.
Just because it has more digits that aren't zero doesn't mean it's more precise :) any more than the orders of magnitude calculations are at all a guarantee that it's that order of magnitude (or even 1 either side).
I know the radius of the world is about 6,000km. And I know that the ice in the Arctic and Antarctic caps is measured in kilometres deep. Pretty sure it's not more than 10 so let's say 5. And from what I've seen of maps (that probably used Mercator and are thus useless, but w/e) the ice caps come down to about 5% each, so let's say an angle of ~10 degrees. And we'll assume they're perfectly circular and uniform depth, because meh. Which gives us 2 * ( (1cos(10)) * 2 * Pi * 6000^2 ) * 5 = the cubic kilometres of ice. Or 34,364,053.10042555 cubic kilometres. Now I think liquid water volume is less than the same mass of ice by a factor of .9, so let's make that 30,927,647.790383 km^3 of liquid water. And we'll assume all of that ice was sitting up top rather than floating, since otherwise it wouldn't have any effect at all. Water covers 75% of the planet, so we'll take that and spread it around 75% * 2 * Pi * 6000^2 = 169,646,003.29384 km^2 for a total rise of 182.306963853 metres.
Man, look at all those digits. You could go out and built a seawall right now and get the position accurate to within a nanometre!
Of course a quick google for the answer says it's actually 70m at the high end... Probably because I counted sea ice, which doesn't actually count. Though I also missed all the land ice that isn't in Antarctica, apparently Greenland and extreme northern Europe/Canada has a bunch. And my depth estimate was way too high. But I involved Pi twice! (3x if you count cos), so I have infinite digits to use!
I'll simplify a bit here and there and refine your area estimation. 70% of earth is water, 30% is land. of those 30%, maybe a tenth is antarctica and greenland. We'll estimate 2km of ice sheet, possibly spoiled because you said that was off. That's all we need. 2km on 3% ⁼ 86m on 70%. Without invoking Pi or computing the areas involved.
I didn't anticipate the average thickness of the ice sheet being a stumbling block. Is that not common knowledge?
Also, I think you could refine your estimate of Antarctica's portion of Earth's land surface area (i.e. your onetenth figure) a little bit. Let me know if you want a hint.
I'll have the hint, yeah. Looking at the actual numbers I was fairly spot on.
And wrt thickness: all I wanted to say there is that my guess was tainted by having read the previous answer. I know it's measured in km, that much is common knowledge. I know it can get to the high single digits and I think we can rule out anything beyond 8.8km. but whether 1 or 4 would make for a better average, I didn't know. 2 served me well though.
That's not at all what I was saying. Using a calculator isn't really in the spirit of these problems, and cheating in sigfigs is definitely not what I had in mind.
You're right though, I misspoke. I just meant that you can use common trivia to do better than using powers of 10 to get an estimate.
How many midwives are there in Amsterdam?
I'm going to skip a couple steps from the other commenters.
I know that most developed countries aim for a birth rate of about 2.1% to remain stable, and most are within a few tenths of a percentage point of that. If we assume a population of about 3,000,000 people, like @gpl, then we get 3,000,000 x 0.021 = 63,000 births a year, or 172 births per day. Depending on how many births a midwife can perform in a day or year (probably about 1/day), we would expect between 100 and 300 midwives. This aligns just right with the estimates of @gpl and @wandaseldon of ~10^2!
EDIT: wording
EDIT 2: This is all wrong. I don't know what statistic I was remembering, but this is inaccurate. I shall leave this comment asis for posterity. Thanks to @vektor for pointing out my mistake.
I actually looked it up out of curiosity yesterday. The Netherlands have 500 midwives at a population of 17M. Amsterdam has a twentieth the population, so we can estimate 25 midwives. Structural differences between the rest and Amsterdam should be insignificant(everything is mostly urban), so proportionality is a good guess. We were all a good bit off. I don't think our estimates for birth rates were it, more the productivity of midwives(durch AVG is 2 births a day) and the population of Amsterdam.
Interesting! Now that you mention it, though. I think we all assumed there was always a midwife assisting with a birth. But, given the prevalence of hospital births where doctors and nurses, but not midwives, assist in giving birth, that very well might be inaccurate.
On the other hand, with the new numbers, we can say 1,000,000 people / 80 people a year per birth = 12,500 births per year or 34 births per day. Which, given one million is high, is about right for an estimate of 1 midwife per birth per day (definitely the right order of magnitude, which is all Fermi estimate tends to ask for). Now, they likely don't work that often, but also there are lots of hospital births.
Is the birth rate you mentioned per woman or per person? Because if it's 2.1 births per person, that would induce a life expectancy of around 50 (see my calculation under Wanda's comment). Which is abysmal. If we take it to mean per woman, that'd get you closer to 100. Easier to get to the actual life expectancy from there, particularly if you add in some kids dying before having children. That would halve the number of midwives you need.
You're right! I don't know what statistic I was remembering, but my math is very off. Sorry about that!
Let's see:
I don't know the population of Amsterdam, but it's probably in the millions so that gives us a factor of 10^6. I'll guess 3 million just to give this the semblance of accuracy. So we have 3e6 people.
Half of those are women, and I'll guess that half of those women are in the age range where they could give birth. So that is 7.5e5, or on the order of 10^5. I know that the median age worldwide is around 30, so I think it's fair to guess half of people are in the range 1550. In the Netherlands the population probably skews older than the worldwide average but for a back of the envelope calculation that's splitting hairs.
This is where it gets sketchy. Let's say that ~10% of those women will give birth in a given year. That's ~10^4 births a year, or ~10^2 a day. Each midwife can probably manage only one birth a say, maybe two. Regardless, each midwife oversees 10^0 births a day. So to manage all of the births in Amsterdam in a day, you probably need 10^2 = ~100 midwives.
I came up with the following:
So you would need around 10^(20) = 100 midwives in Amsterdam.
You could eliminate a step there. You don't need to estimate the number of women in the age range you specified, nor do you need the age range. A average woman will give birth twice a lifetime. (Or a human once..) So you'd need to estimate the lifespan. 80 years? Well, one in 80 people give birth this year.
How does the total weight of all insects on Earth compare with the total weight of all humans?
That's a good one. There's 10^10 people on earth, weighing 10^12 kg. The land area of earth is 0.3 (land proportion) * 4 pi * (6000km^2) = 1.4*10^8km² = 1.4 * 10^14 m². Or 1 kg of human per 140m². (Would you look at that, your apartment is statistically only inhabited by your forearm.) That's 7 grams per m^2. I think the Insects should be able to beat that, I'd have given them 100 grams per m², maybe a hint less, but certainly 10 grams. However, I don't think they're beating us by a factor of 100. My experience mostly pertaining to highly cultivated landscapes doesn't help guesstimating though.
There area bout 10^10 people on earth. There are about 10^19 insects. Insects weight about 10^6 kg and humans weight about 10^2 kg. Dividing out, and we get that they're beating us by a factor of about 10.
How many gas stations are there in Wyoming?
Bonus, how many gallons of gas do they sell in an average day?
I'm going to assume that a gas station will have an average of one customer at any given time during the day. That seems like a good baseline approximation for this level of estimate.
I'm also going to say that a given person gets gas one time per week, for a duration of 5 minutes.
That means that, with 12 hours of operation per day, 7 days a week, there's 84 hours to fill. 5 minutes per person means 12 people per hour. I'll say 84*12=1000 (Edit: Wow, this is actually incredibly close) so that means each gas station serves 1000 people. I think Wyoming has something like 600,000 people, so that would make for 600 gas stations in Wyoming.
And then if an average day has 12 people per hour, for 12 hours, that's 150 sales made, and if each one is 12 gallons or so, then that's 2000 gallons per day. Wikipedia indicates tanker trucks have a capacity of 6000 to 12000 gallons or so, so I feel like 2000 per day is a reasonable guess.