If you're trying to find a deviation within a context, then you'd have to define the "norm" and the "context". In this case, the norm (i.e. mean) would be your estimate number, but what is the context (i.e. range)?
What if the actual nr of hours was 7 instead of 60? (7 is off by the same number of hours from "estimated" as 60, but in the other direction). Then your formula and his formula would give you the same result.
His formula is defining the range in which the mean is the amount estimated, the low end is 0 and the high-end is 67 (i.e. 33.5 * 2). With his formula you get the same accuracy for actual 7 and actual 60.


>A follow-up....and because I respect you as the "just the facts, Sgt Joe Friday, no nonsense", then maybe you can make sense of this.
>
>I found how someone is calculating this....again, for a bad sprint of 60 hours on an original guess of 33.5 hours, I figured the accuracy was 55.83%. But I've got a project manager saying it's about 21%. Here is how the PM is calcing it...this seems more like a factor than an accuracy percentage.
>
>1 - abs ( amount estimated - remaining work left - time spent)
> -----------------------------------------------------------------------------
> amount estimated
>
>
>so that....
>

>
>1 -   abs (  33.5  - 0 - 60)
>       ----------------------------
>              33.5
>
>1 -  ( 26.5)
>      ----------  
>        33.5
>
>1 -  .791
>
>
>accuracy:   21%
>
>