Mueen Nawaz wrote:
> 	Not sure I'm getting it. Just to clarify, I'm using the word normal in
> the way you originally meant it (normal numbers). I'm using normal
> sequence in the same sense.
> 
> 	I was merely pointing out that you originally said:
> 
> "Given that truly random sequences are normal, and in a normal sequence
> every block of a particular length occurs with equal probability"
> 
> 	I was giving you an example of a truly random sequence that was _not_
> normal.

Yes. I was phrasing it sloppily there, since we hadn't talked about other 
distributions at that point in the conversation. To clarify, a "normal 
sequence" in the way I'm using it means all possible subsequences have the 
same probability distribution in the asymptote. Hence, you can't take a 
gausian distribution of symbols and expect to get a normal sequence from them.

> 
>> To have a "normal number", you need linear distribution, basically.
>> I.e., all substrings of a given length appear with equal asymptotic
>> probability as the number of symbols you look at gets large. *Given*
>> that, every sequence of any given length will appear if you let N be
>> infinity.
> 
> 	Agree - other than with the use of the word "will".<G> What I mean is
> that what you say is how I understand what probability theory says.

OK. I don't think I know enough math to convince someone else of it. I'm 
just taking it on authority. It would seem to be the sort of thing amenable 
to proof, so when a math text says it's so, I'll take their word for it. :-) 
Especially when they point out a bunch of other hypotheses where they say 
"We think so but haven't proven it."

>> In other words, in order to avoid Shakespeare appearing, you would have
>> to have a non-random distribution *because* the sequence is infinite and
> 
> 	I think I get the general idea, but to nitpick, you can avoid it even
> with a random distribution. If I have a distribution where the
> probability of typing the letter 'e' is forbidden, it's still random
> (not "truly" random).

No, it's not random. If you go an infinite number of symbols and "e" doesn't 
appear, then "e" doesn't have the same frequency in the asymptote as 
something that *does* appear an infinite number of times.

What you do in that case is remap the output to encode fewer letters of 
english per bit, and interpret it that way. That's the equivalent of reading 
the number in some other base (binary/hex/etc).  I'm dealing with English as 
base-26 or base-127 or whatever character set you want to use.

> 	I suppose you may object to my referring to it as random, but it is
> consistent with probability theory: A uniform distribution from 0 to 1
> is a valid random distribution - even though you've excluded all numbers
> greater than 1.

Sure. But we're starting with the assumption that all the letters of the 
alphabet can be typed. If you give the monkeys a typewriter with no "e" key, 
of course they're not going to type out shakespeare.

And you can't say "they might never type the word 'the'", because if they 
typed "th" and you knew they never typed "the", then you again wouldn't have 
a random sequence, because you could predict with 1/25 assurance what the 
next letter would be instead of 1/26.

> It's not equal to the RHS because the RHS is
> meaningless. It's undefined.

It's not meaningless. It's merely not a number. That doesn't mean you can't 
work with it. Just like you can work with math where you have functions that 
are uncalculable, corresponding to functions that get stuck in infinite 
loops in a programming language.

In any case, therefore, the statement still holds, because 0 is defined and 
0*inf is not.

But I think I've exhausted my ability to convince you that Shakespeare 
necessarily appears in the output, if the output is infinite and making the 
other normal assumption that each possible character is typed with equal 
probability. :-)

-- 
   Darren New, San Diego CA, USA (PST)
   There's no CD like OCD, there's no CD I knoooow!

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