[williamkuszmaul]

One estimation trick that I've found effective is the following: (1) determine the smallest number that your sure is larger than the true answer. (2) determine the largest number that you are sure is smaller than the true answer. (3) take the geometric average of the two. (i.e., sqrt(a * b))

The reason this does well, is that oftentimes, (1) overestimates the true answer by roughly the same multiplicative factor as (2) underestimates the true answer by. So the geometric mean cancels the over and under estimates in order to get an estimation that does pretty well.

I find that this works remarkably well for estimating the dimensions of buildings, trees, etc.

[redcalx]

So basically, if there's a distribution of possible outcomes (with estimated duration plotted on the x-axis, and probability on the y-axis), then we put bounds on either side of the distribution and take a 'central' point on the x-axis between them. But instead of a simple average, we can use a log scale on the x-axis, so that we represent estimates of different magnitudes well. Now the simple half way point on the x-axis is on a log scale, and (I think) equivalent to the approach you described.

[infogulch]

The geometric mean being equivalent to the arithmetic mean of values transformed under a log scale is an interesting perspective that I haven't considered, but appears to be well known. https://en.wikipedia.org/wiki/Geometric_mean

[dhimes]

Thank you. I love this.