Fermi Estimate

Build a chain of rough guesses and say how wrong each one could be. The tool shows what everyone expects the answer to be worth, and what it is actually worth.

Worked example — Fermi Estimate

Estimate	100
Usually within	×2.8
Typical miss	×2
Worst case, if every guess goes the same way	×46

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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The worst case is 729. What happens is 2.85. 🖖

Set six factors to ×3 each and read the worst case and the typical miss side by side. The worst case, every guess wrong in the same direction, is 3⁶ = 729. The typical miss is 2.85, and 86% of the time you land within a factor of ten. That gap is the whole method. Fermi's own chain is tighter still, because he knew Chicago's population better than he knew how many households own a piano: it lands on 100 tuners with a band of 2.8, and 98% of the time within a factor of ten. Six numbers, none of them looked up, and an answer good enough to argue with.

Multiplying is adding, once you take logs 🖖

Errors in a product do not pile up, because multiplying numbers means adding their logarithms. Independent things that add combine in quadrature, so the spread of n factors grows like √n, not n. Six factors are 2.4 times as uncertain as one, not six times. Which means adding a seventh factor improves the estimate per factor. Change the chain length in the presets and watch the band grow far slower than the worst case beside it.

Where it fails, and why Drake is the famous example 🖖

Press When it stops working. Five factors at ×3 and one at ×10,000, which is roughly what the Drake equation asks you to accept about how long a civilisation stays audible. The band goes to ×246, the chance of landing within a factor of ten drops to 32%, and the share column puts 93% of the doubt on that one row. Cancellation needs bounded errors. You genuinely do know Chicago's population to within 50%, so being high on one factor and low on another buys you something. Nobody knows the last Drake factor to within four orders of magnitude, so there is nothing to cancel against, and √n cannot help. The Drake equation page states the other half of this: a product collapses on its weakest link.

Problems solved in full

Litres of water a city of 500,000 drinks in a year 7 steps
How many litres of water does a city of 500,000 drink in a year? Guess every number, then work out what the guessing cost you.
1. Start with the chain. Drinking water only, not showers: about 2 litres a person a day, 365 days, half a million people.
  
  $2 \times 365 \times 500, 000 = 365, 000, 000$
2. Three hundred and sixty-five million litres. How wrong is each factor allowed to be? The population is a census figure, so call it ×1.2. The 2 litres is a real guess — some people drink 3, some drink 1.3 — so ×1.5. Days in a year is exact.
  
  $σ_{i} = \frac{ln k _{i}}{3} : \frac{ln 1.5}{3} = 0.234, \frac{ln 1.2}{3} = 0.105$
3. Each factor uncertain by ×k contributes ln(k)/√3 to the spread of the logarithm, and independent contributions add in quadrature.
  
  $σ = 0.23 4^{2} + 0.10 5^{2} = 0.257$
4. One standard deviation is a factor of 1.3. So the honest answer is somewhere around 365 million litres, and you would be surprised to be out by more than about 30%.
  
  $e^{0.257} = 1.29$
5. Compare that with the naive fear. If both guesses went wrong the same way you would be out by 1.2 × 1.5 = 1.8. The actual spread is 1.3, and the difference between those two numbers is the only reason estimating is worth doing.
  
  $1.2 \times 1.5 = 1.8 vs 1.29$
6. The √n argument needs the errors to be independent. If you estimated the population from an out-of-date figure and then used the same figure to work out household count, you have made one mistake twice, and it does not cancel with itself.
  
  $σ_{tot} = \sum σ_{i}^{2} ⟸ Cov (i, j) = 0$
7. That is the failure mode a Fermi estimate cannot see from the inside, and it is why the method wants factors drawn from different places: a census, a physiological rate, a calendar. Three sources that can be wrong independently. One source used three times looks identical on the page and gives you a band that is a fiction.
  
  $n σ ⟶ n σ (ρ = 1)$
Answer

About 365 million litres, and you are unlikely to be out by more than 30%. The worst case was 1.8, the real band is 1.3, and the gap between them is the method. The number the tool cannot check for you is whether your guesses were really independent.
Heartbeats in a lifetime, and whether a mouse really gets fewer 7 steps
Work out how many times a human heart beats in a lifetime and how wide the answer’s band is. Then decide whether a mouse genuinely gets fewer beats than a person, or only appears to.
1. Five rows, multiplied in any order you like. 70 beats a minute for 78 years comes to about 2.9 × 10⁹.
  
  $70 \times 60 \times 24 \times 365 \times 78 = 2, 869, 776, 000 \approx 2.9 \times 1 0^{9}$
2. Only two of the five rows can be wrong. Minutes in an hour, hours in a day and days in a year are definitions, and the share column gives each of them 0% of the doubt while the heart rate and the lifespan take 50% each. Five rows with the uncertainty of two.
  
  $σ_{i} = \frac{ln 1.2}{3} = 0.105 (\times 2), σ = 0.105 2 = 0.149$
3. One standard deviation comes to a factor of 1.16, which the panel rounds to ×1.2. Both guesses going wrong the same way at once gives ×1.4, and because each was declared as a bound rather than a tendency, that is a wall and not a tail.
  
  $e^{0.149} = 1.16, 1.2 \times 1.2 = 1.44$
4. So the answer lies between 2.0 and 4.1 billion and nowhere else.
  
  $[\frac{2.87 \times 1 0 ^{9}}{1.44}, 2.87 \times 1 0^{9} \times 1.44] = [1.99 \times 1 0^{9}, 4.13 \times 1 0^{9}]$
5. Now the same five rows for two other animals: the three definitions unchanged, two new guesses. A mouse at 600 beats a minute living 2 years, an elephant at 30 beats a minute living 60.
  
  $600 \cdot 60 \cdot 24 \cdot 365 \cdot 2 = 6.3 \times 1 0^{8}, 30 \cdot 60 \cdot 24 \cdot 365 \cdot 60 = 9.5 \times 1 0^{8}$
6. The elephant’s two guesses are looser than the human ones, so allow them ×1.5 each. Its range then reaches 2.1 × 10⁹, and the human range begins at 2.0 × 10⁹. The two touch, and the comparison settles nothing.
  
  $k = 1.5 : [\frac{9.5 \times 1 0 ^{8}}{2.25}, 9.5 \times 1 0^{8} \times 2.25] = [4.2 \times 1 0^{8}, 2.1 \times 1 0^{9}]$
7. Tighten them. A heart rate and a lifespan are both measurable, so ×1.2 is honest for an animal anyone has studied, and the elephant then tops out at 1.4 × 10⁹ against a human floor of 2.0 × 10⁹. Now the ranges separate.
  
  $k = 1.2 : [\frac{9.5 \times 1 0 ^{8}}{1.44}, 9.5 \times 1 0^{8} \times 1.44] = [6.6 \times 1 0^{8}, 1.4 \times 1 0^{9}]$
Answer

About 2.9 × 10⁹ beats, with an honest range of 2.0 to 4.1 × 10⁹. Three of the five rows are definitions and carry none of the doubt, which is why five factors behave here like two. A human does get several times more beats than a mouse or an elephant, but the estimate only says so once the animal’s own guesses are tightened to ×1.2. At ×1.5 the ranges overlap and the gap between the two point estimates means nothing, which is the question to ask of any two estimates before believing the difference between them.

$σ = \sum_{i} σ_{i}^{2}, σ_{i} = 0 ⟺ k_{i} = 1$

Learning path

Orders of magnitude

References (2)

Why independent errors combine in quadrature rather than accumulating Taylor, J. R. (1997). An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements (2nd ed.). University Science Books — Chapter 3, on propagation for products and quotients.
The piano-tuner problem itself Weinstein, L. and Adam, J. A. (2008). Guesstimation: Solving the World's Problems on the Back of a Cocktail Napkin. Princeton University Press.

By the 42Math team · Updated 2026-08-25

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Piano tuners in Chicago - Fermi's own chain, and not one number in it was looked up. It lands on 100 tuners, and the honest band is a factor of 2.8 — a real answer, from six guesses.
Heartbeats in a lifetime - Every factor here is known to within 20% or better, so the spread nearly vanishes: the answer is 2.9 billion beats and you are inside a factor of 1.2 of it.
When it stops working - Five factors at ×3 and one at ×10,000. The band blows out to ×246 and the chance of landing within a factor of ten falls to 32%. This is the Drake equation's problem, and no amount of averaging fixes it.