"Darren New" <dne### [at] sanrrcom> wrote:
> And I'm saying no, it isn't technically possible, unless the probability 
> of getting any digit other than zero is literally impossible.  As I said, 
> I understand what you're asserting - each number is equally possible, so 
> all zeros is possible. I'm saying if each number is equally possible and 
> you do an *infinite* number of trials and don't get an *infinite* number 
> of each digit, then they're not all equally possible. The only way to make 
> an infinite number of choices and not get an infinite number of 9's is if 
> 9's cannot appear, because any non-zero number times infinity is infinity.

That assertion would mean that randomly choosing any terminating decimal or 
a decimal that doesn't include all ten digits is impossible.  But it's not, 
so...yeah.

> I don't personally know which is right, but from what I understand the 
> experts to be saying, my assertions are right. (I.e., I think I'm saying 
> what I read the experts to be saying.) That's why I'm asking you if you 
> have any actual expert descriptions that might shed light on why *you* are 
> right, beyond your personal logic.

If my personal logic is insufficiently transparent to need an outside expert 
to be determined valid, it's obviously not very good.

>> Besides, there's a third option:  we're both right.
> I'm not sure I follow how that can be. :-)

Math is strange that way.

-- 
Tim Cook
http://empyrean.freesitespace.net

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