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 1⁄20 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.
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 1⁄310^-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
%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.
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. 1⁄2 die younger than 75, 1⁄2 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 1⁄2 (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.
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 I Google it, that turns out to be irel pybfr gb gur ahzore bs Jny-Zneg Fhcrepragref, gur ynetr syntfuvc fgberf gung gur cuenfr “Jny-Zneg” oevatf gb zvaq. Ubjrire, Jny-Zneg nyfb bcrengrf n fznyyre ahzore bs “qvfpbhag fgber”, “arvtuobeubbq znexrg”, naq “rkcerff” ybpngvbaf gung funer gur fnzr oenaqvat. Vs jr vapyhqr “qvfpbhag” naq “arvtuobeubbq” ybpngvbaf, gur gbgny vf nobhg guerr gubhfnaq rvtug uhaqerq. V pna’g svaq gur ahzore bs “rkcerff” fgberf, ohg gur sbezng jnf perngrq va 2011 fb gurer cebonoyl nera’g gbb znal.
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.
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 1⁄20 of all Americans are in high school, which is 15 million.
...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 1⁄310^-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.
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
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.
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. 1⁄2 die younger than 75, 1⁄2 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.
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 1⁄2 (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 I Google it, that turns out to be irel pybfr gb gur ahzore bs Jny-Zneg Fhcrepragref, gur ynetr syntfuvc fgberf gung gur cuenfr “Jny-Zneg” oevatf gb zvaq. Ubjrire, Jny-Zneg nyfb bcrengrf n fznyyre ahzore bs “qvfpbhag fgber”, “arvtuobeubbq znexrg”, naq “rkcerff” ybpngvbaf gung funer gur fnzr oenaqvat. Vs jr vapyhqr “qvfpbhag” naq “arvtuobeubbq” ybpngvbaf, gur gbgny vf nobhg guerr gubhfnaq rvtug uhaqerq. V pna’g svaq gur ahzore bs “rkcerff” fgberf, ohg gur sbezng jnf perngrq va 2011 fb gurer cebonoyl nera’g gbb znal.
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.