10**( log(1/(1-rand()))/log(2) ) - 1

[Update: You may want to shorten the range if you get too many small (0.x) values or values that are too large. Replace 1-rand() with, for example .9-.8*rand().]

10**( -log(1-rand())/log(2) ) - 1 # (same thing) (1-rand())**(log(.1)/log(2)) - 1 # (same thing)

- tye        

I'd show the (interesting) derivation, except the more words I include in a reply to you, the more likely the whole thing will turn ugly.


In reply to Re: Rand() with ?log10? distribution? (math) by tye
in thread Rand() with ?log10? distribution? by BrowserUk

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