On Sat, May 1, 2010 at 8:26 PM, Rex Allen <rexallen...@gmail.com> wrote:

> On Sat, May 1, 2010 at 7:37 PM, Brent Meeker <meeke...@dslextreme.com>
> wrote:
> >
> > Sure we can, because part of the meaning of "random", the very thing that
> > lost us the information, includes each square having the same measure for
> > being one of the numbers.  If, for example, we said let all the "1"s come
> > first - in which case we can't hit any "not-1"s, that would be
> inconsistent
> > with saying we didn't have any information.
>
> We have two things here.  Random.  And infinite.
>
> Three things actually.  My random aim.  An infinite row of squares.
> And each square's randomly assigned number lying between 1 and 6.
>
> If, due to the nature of infinity, there are the same number of 1's
> and not-1's, then I'd expect the probability of hitting a 1 to be
> 50-50.
>
> But, there are also the same number of 1's and even numbers.
>
> And the same number of evens and odds.
>
> And the same number of 1's and 2's.
>
> And the same number of 2's and not-2's.
>
> AND...I have the *random* aim of the dart that I'm throwing at the
> row.  So it's not a question of saying which number is likely to be
> next in a sequence.  Rather, the question is which number am I likely
> to hit on this infinite row of squares.
>
> SO, I think we have zero information that we can use to base our
> probability calculation on.  Because of the counting issues introduced
> by the infinity combined with the lack of pattern.  There is no usable
> information.
>


Mathematicians do apparently have a well-defined notion of the "frequency"
of different possible finite sequences (including one-digit sequences) in an
infinite digit sequence. For example, see the article at
http://www.lbl.gov/Science-Articles/Archive/pi-random.html which talks about
attempts by mathematicians to prove that the digit sequence of pi has a
property called "normality", which means that any n-digit sequence should
appear with the same frequency as every other n-digit sequence (so in base
2, it would imply that the 2-digit sequences 00, 01, 10 and 11 all appear
equally frequently in the infinite sequence):


'Describing the normality property, Bailey explains that "in the familiar
base 10 decimal number system, any single digit of a normal number occurs
one tenth of the time, any two-digit combination occurs one one-hundredth of
the time, and so on. It's like throwing a fair, ten-sided die forever and
counting how often each side or combination of sides appears."'

'Pi certainly seems to behave this way. In the first six billion decimal
places of pi, each of the digits from 0 through 9 shows up about six hundred
million times. Yet such results, conceivably accidental, do not prove
normality even in base 10, much less normality in other number bases.'

'In fact, not a single naturally occurring math constant has been proved
normal in even one number base, to the chagrin of mathematicians. While many
constants are believed to be normal -- including pi, the square root of 2,
and the natural logarithm of 2, often written "log(2)" -- there are no
proofs.'


So while it hasn't been proved, it sounds like it's at least a well-defined
notion (and the article discusses some approaches to proving it which show
some promise). Perhaps it means that if you look at the frequencies of
different n-digit sequences in the first N digits of a number, the
frequencies all approach equality in the limit as N goes to infinity. It
would presumably be possible to find infinite sequences that *aren't*
"normal" in this sense, like .011011011011...

(Meanwhile, note that the naive idea of just picking a digit randomly from
the entire infinite sequence, with all digits equally likely, doesn't
actually make sense because you can't have a uniform probability
distribution on an infinite series of numbers. It would lead to paradoxes
along the lines of the two-envelope paradox discussed at
http://consc.net/papers/envelope.html except in this variant you'd be given
one of two envelopes which you find to contain N dollars, where N was chosen
at random from the infinite series of natural numbers 1,2,3,... using a
uniform probability distribution so each natural number was equally likely.
Then if you have a choice to exchange it for another sealed envelope chosen
in the same way, you should always bet that the second envelope contains
more money with probability 1 since there are an infinite number of possible
Ns larger than the one you got and only a finite number of Ns smaller. The
paradox is that this argument would seem to work even before you have opened
the first envelope and seen the specific value of N inside, so you're saying
that there's a probability 1 that one of two identical featureless sealed
envelopes has more money in it than the other!)

If this notion of considering the frequency of different finite sequences in
an infinite sequence is a well-defined one, perhaps something similar could
also be applied to an infinite spacetime and the frequency of Boltzmann
brains vs. ordinary observers, although the mathematical definition would
presumably be more tricky. You could consider finite-sized chunks of
spacetime, or finite-sized spin networks or something in quantum gravity,
and then look at the relative frequency of all the ones of a given "size"
large enough to contain macroscopic observers. Suppose you knew the
frequency F1 of "chunks" that appeared to be part of the early history of a
baby universe, with entropy proceeding from lower on one end to higher on
the other end, vs. the frequency F2 of "chunks" that seem to be part of a de
Sitter space that had high entropy on both ends. Then if you could also
estimate the average number N1 of ordinary observers that would be found in
a chunk of the first type, and the average number N2 of Boltzmann brains
that would be found spontaneously arising in a chunk of the second type,
then if F1*N1 was much greater than F2*N2 you'd have a justification for
saying that a typical observer is much more likely to be an ordinary one
than a Boltzmann brain.

Jesse

-- 
You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To post to this group, send email to everything-l...@googlegroups.com.
To unsubscribe from this group, send email to 
everything-list+unsubscr...@googlegroups.com.
For more options, visit this group at 
http://groups.google.com/group/everything-list?hl=en.

Reply via email to