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.
Dialectical Bootstrapping
"Dialectical Bootstrapping" is a simple procedure that may improve your estimates. This is how it works:
- Estimate the number in whatever manner you usually would estimate. Write that down.
- Assume your first estimate is off the mark.
- Think about a few reasons why that could be. Which assumptions and considerations could have been wrong?
- What do these new considerations imply? Was the first estimate rather too high or too low?
- Based on this new perspective, make a second, alternative estimate.
Herzog and Hertwig find that average of the two estimates (in a historical-date estimating task) is more accurate than the first estimate, (Edit: or the average of two estimates without the "assume you're wrong" manipulation). To put the finding in a OB/LW-centric manner, this procedure (sometimes, partially) avoids Cached Thoughts.
= 783df68a0f980790206b9ea87794c5b6)
Subscribe to RSS Feed
= f037147d6e6c911a85753b9abdedda8d)