RE: SQL RAND function question

Adam,

Thanks for the explanation, that's what I was thinking, but you spelled it out nicely.

Like you, I'm a little confused about getting half the 1's I expected. Getting half the 0's would
have made sense.

Still I'm confused, IIRC every example I've seen about using _any_ built in random number function has
said basically, they return [0,1) or [0,1] and to get a number between a given range multiply by the
range. (IBM's SQL manual included!)

Obviously, that doesn't hold true for the reasons you've explained.

So what is the "right" answer to using such a built in random number function?

Charles Wilt
Software Engineer
CINTAS Corporation - IT 92B
513.701.1307
wiltc@xxxxxxxxxx

-----Original Message-----
From: midrange-l-bounces@xxxxxxxxxxxx [mailto:midrange-l-
bounces@xxxxxxxxxxxx] On Behalf Of Adam Glauser
Sent: Tuesday, February 12, 2008 1:59 PM
To: midrange-l@xxxxxxxxxxxx
Subject: Re: SQL RAND function question

Wilt, Charles wrote:

I'm thinking perhaps the way I'm going from the RAND() results to a

integer from 0-11 is the problem.

Anybody understand what I'm doing wrong?

myDate = date('2008-03-01') + cast(round((rand(12) * 11),0) as int)

months

I don't think round() is what you want, although I'm confused a bit by
the distribution of results that you stated. Anyway, here is why I
think that you got about half as many 11s as you expected:

rand(12) * 11 gives a double float number r, in the range [0,11].
round(r) gives an integer in the set Z' (0,1,2,...,11).

What values result in each number in the set Z'?
if r is in [0,0.5), r' = 0
if r is in [0.5,1.5), r' = 1
if r is in [1.5,2.5), r' = 2
...
if r is in [9.5 ,10.5), r' = 10
if r is in [10.5,11] , r' = 11

So, there are only half as many values that will lead to a result of
either 0 or 11 as there are that lead to any of the other numbers.

Of course, the problem with this conclusion is that it doesn't explain
why you got half as many 1s as you were expecting.
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