More precisely, you can compute the variance of the logarithm of the final estimate and, as the number of pieces gets large, it will shrink compared to the expected value of the logarithm (and even more precisely, you can use something like Hoeffding’s inequality).
If success of a fermi estimate is defined to be “within a factor of 10 of the correct answer”, then that’s a constant bound on the allowed error of the logarithm. No “compared to the expected value of the logarithm” involved. Besides, I wouldn’t expect the value of the logarithm to grow with number of pieces either: the log of an individual piece can be negative, and the true answer doesn’t get bigger just because you split the problem into more pieces.
So, assuming independent errors and using either Hoeffding’s inequality or the central limit theorem to estimate the error of the result, says that you’re better off using as few inputs as possible. The reason fermi estimates even involve more than 1 step, is that you can make the per-step error smaller by choosing pieces that you’re somewhat confident of.
If success of a fermi estimate is defined to be “within a factor of 10 of the correct answer”, then that’s a constant bound on the allowed error of the logarithm. No “compared to the expected value of the logarithm” involved. Besides, I wouldn’t expect the value of the logarithm to grow with number of pieces either: the log of an individual piece can be negative, and the true answer doesn’t get bigger just because you split the problem into more pieces.
So, assuming independent errors and using either Hoeffding’s inequality or the central limit theorem to estimate the error of the result, says that you’re better off using as few inputs as possible. The reason fermi estimates even involve more than 1 step, is that you can make the per-step error smaller by choosing pieces that you’re somewhat confident of.
Oops, you’re absolutely right. Thanks for the correction!