Brian Yennie gave the explanation. When you use random(3) as a sort
key you have a high chance (in fact I think it's 50%) that two of the
items will be assigned the same sort key, and thus their relative
position will be preserved, giving a decidedly non-random sort. If you
sort by random(1000000) or some suitably high number the chances of
getting the same sortkey in your three iterations is miniscule.
-- Peter
Peter M. Brigham
[email protected]
http://home.comcast.net/~pmbrig
On Feb 11, 2010, at 10:20 AM, [email protected] wrote:
Thanks for the responses, but I think there is an issue here.
I figured a sort value was assigned to each item, and certainly a
larger
value of n in random(n) gives, what, more room to move?
I would expect 1000 iterations of random(3) to give an even spread
of 1's,
2's and 3's. It does, of course.
Howwever, randomizing, through 1000 iterations, my three items with
random(3) yields a list heavily weighted in favor of item 1. It
appears far more
often, repeatably, than it ought to. Why item 1? With random(100) the
dispersion is as expected. The value of n has to be about 15 or more
to yield what
looks like a reasonable output, at least with three or four items,
the only
options I tested.
I don't get why.
Craig Newman
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