3 ms·
There was this estimation technique one of the Sixty Symbols professors talked about, where you do all of the calculations, but you only keep one significant di
by stcredzero 10y ago
There was this estimation technique one of the Sixty Symbols professors talked about, where you do all of the calculations, but you only keep one significant digit. It's something you can do quickly in your head while conversing with someone, but tells you pretty reliably if something is feasible.
- jonsen 10y agoWhen the first digit becomes 1 keep two digits. A small effort for much higher precision. (Something to do with Benford's Law? Maybe.)
- TrainedMonkey 10y agoThis makes sense. If the number is 5.5 by throwing away the .5 you lose less than 10% of precision. On the other hand if the number is 1.5 and you throw away the .5 you lose 50% precision.
- Terr_ 10y agoSounds Benford-y, yeah. The biggest relative "error" in [0,99] is going to be at 19, where the estimate (1 * 10^0) at only ~52% of the actual value (1.9 * 10^0). That local minima for accuracy recurs at 199, 1999, etc. From some quick spreadsheet-fu, the mean underestimation for [0,9999] is 88%. Meanwhile, with the "teens" trick, the local minimum is 29 (68.9%) and then at 299, 2999, etc, and the mean underestimation is boosted to... huh, only 91%? Then again, I'm assuming a completely even distribution, which probably won't be true in some contexts.
- stcredzero 10y agoI like it! It's small additional effort for more precision. However, the estimation technique above is optimized for very small effort. You can do the very lightweight version in the middle of a conversation in your head without unduly derailing the conversation. It's a different tool with different trade offs. It's the keychain multitool of estimation methods.
- dpflan 10y agoLike "Fermi Problems" and back-of-the-envelope math? > https://en.m.wikipedia.org/wiki/Fermi_problem https://en.m.wikipedia.org/wiki/Fermi_problem