3 ms·
This method breaks rather quickly because, when you multiply two order-of-magnitude estimates, you get a two-orders-of magnitude estimate. (And so forth.) In th
by zeteo 14y ago
This method breaks rather quickly because, when you multiply two order-of-magnitude estimates, you get a two-orders-of magnitude estimate. (And so forth.) In the piano tuners example, there are some five independent quantities being estimated. Of course, the answer "I think there are 500 piano tuners in NYC" doesn't sound as impressive when you add "within five orders of magnitude"...
- pmiller2 14y agoThis is strictly true, but doesn't account for the fact that errors tend to cancel one another. For instance, if 2 of the 5 independent estimates are high, one is right on, and 2 are low, you can end up with a number very close to the actual number. And, by "very close," I mean usually within 1-2 orders of magnitude, depending on the quality of the component estimates.
- deleted 14y ago[deleted]
- deleted 14y ago[deleted]
- erydo 14y agoErrors cancel out when they're errors of magnitude, which is what I believe he was referring to, and is often the most important property of Fermi estimates. If one estimate is off by 2x and another is off by ½x, those cancel out because the exponents are additive. Edit: Parent deleted comment…oh well.
- jimhefferon 14y agoAnd yet it typically yields the right answer (in the hands of an expert). I wonder why?
- deleted 14y ago[deleted]