To muddy the waters further:

http://en.wikipedia.org/wiki/Almost_surely

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Mueen Nawaz <m.n### [at] ieeeorg> wrote:
> http://en.wikipedia.org/wiki/Almost_surely

  Seems like my claims about the monkeys and Shakespeare were not, after all,
completely ridiculous.

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"Mueen Nawaz" <m.n### [at] ieeeorg> wrote in:
> http://en.wikipedia.org/wiki/Almost_surely

So, a probability of 1 doesn't mean 'always', and a probability of 0 doesn't 
mean 'never' when there's infinity involved.  Was that so hard?

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Warp wrote:
> Mueen Nawaz <m.n### [at] ieeeorg> wrote:
>> http://en.wikipedia.org/wiki/Almost_surely
> 
>   Seems like my claims about the monkeys and Shakespeare were not, after all,
> completely ridiculous.

I never thought they were completely ridiculous. Just unconvincing. :-) 
It's still up in the air for me as to who is right. :-)

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Mueen Nawaz wrote:
>         Let's say that instead of a 2 sided coin, I have a three state
> random generator, which generates each state with equal probability.
> 
>         Now I had already shown that the likelihood of getting a
> _particular_ infinite sequence was the same as getting a _particular_
> value between 0 and 1. (That is, the likelihood is zero - I'm arguing
> purely mathematically - let's put the "real" world aside).
> 
>         Intuitively, that's because a point is infinitely smaller than
> the interval [0,1].
> 
>         In fact, if I were to ask the probability of getting a rational
> number, the answer is still 0. While the set of rationals is infinite,
> it's still infinitely smaller than the unit interval - the latter is of
> a higher infinity.

I think this is relevant:
http://en.wikipedia.org/wiki/Almost_surely

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Nicolas Alvarez wrote:
> I think this is relevant:
> http://en.wikipedia.org/wiki/Almost_surely

And I should learn to read the rest of the thread before posting a reply.

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John VanSickle wrote:
> Mueen Nawaz wrote:
> 
>>         It turns out this set has "measure" 0. And any integral of a
>> function (the uniform distribution in this case) over a set of measure
>> zero is 0.
> 
> I understand that the Dirac delta function,
> 
> http://en.wikipedia.org/wiki/Dirac_delta_function
> 
> is an exception to this.

	Well, it's not exactly a function.

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                       >>>>>>mue### [at] nawazorg<<<<<<
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Mueen Nawaz wrote:
> 	Put another way, I showed that the even if I allow some sequences with
> a single 1 to sneak in, the probability is still 0.

Oh yeah, I certainly wasn't debating the accuracy of the result.  It 
appears that you already considered this detail anyway, so my 
nit-picking was entirely unnecessary.

I did like the Cantor set analogy by the way.  I didn't forsee it when I 
was reading your post and was delighted when you pointed it out.

> 
>> I think represent the same number on the real line, but one would be in
>> the set and one wouldn't.  Nevertheless I do think that the Cantor set
> 
> 	One would be in the set of sequences we're interested in, and the other
> wouldn't. In terms of the Cantor set, they're both in it as they are the
> same number.

Yeah, that's what I meant.

>>     lim_{n->inf} a^n = 0 if 0 <= a < 1
> 
> 	(Incidentally, I'm not sure I understand your limit. What is 'a' and
> why is it fixed?

For base m a would be (m-1)/m -- the probability that a single given 
digit of the random sequence doesn't contain a 1 in base m.

>> 1 = sum_{0:inf} p
>>
>> Since no real number has this property, p cannot exist.  Thus there is
>> no probability (zero or otherwise) that a random sequence is in S.  Said
>> in more standard terminology: S has no measure.
> 
> 	I didn't read the whole thing - xors make my head hurt. But I got the
> general idea, having seen similar arguments before. When I was learning
> integration theory using measures, I noticed that integrals were always
> defined over _measurable_ sets.

In that case I'll spare you the trouble of being tempted to read it 
later.  I just adapted the construction of a Vitali set from the real 
line to infinite binary sequences by replacing addition with xor.  A few 
small things changed as a result, but nothing major.

I'm enveious you had an integration theory class that got into measure 
theory.  I don't recall such a course being available back when I was 
taking math classes, but it would have been pretty fun had there been.

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Invisible wrote:
> (For example, if you have a group of size X, it has subgroups of every 
> size that is a factor of X - including 1 and X itself.

I assume by "group" you actually mean to say "cyclic group"?  If not 
that might explain why it's proving hard to understand.

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>> (For example, if you have a group of size X, it has subgroups of every 
>> size that is a factor of X - including 1 and X itself.
> 
> I assume by "group" you actually mean to say "cyclic group"?  If not 
> that might explain why it's proving hard to understand.

I was under the impression the result applies to all groups - which is 
why the groups of prime order are always cyclic.

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