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Tim Cook wrote:
> "Darren New" <dne### [at] san rrcom> wrote:
>> Now, if you find something that explains *why* it's reasonable to pick
>> an infinite number of random digits and get all zeros, I'll look at
>> it, but right now we're both just asserting the truth of our own
>> positions. :-)
>
> I never said it was *reasonable* to get all zeroes, just that it's
> technically possible. ;)
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.
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.
> Besides, there's a third option: we're both right.
I'm not sure I follow how that can be. :-)
--
Darren New, San Diego CA, USA (PST)
There's no CD like OCD, there's no CD I knoooow!
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"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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Tim Cook wrote:
> 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.
At random? Are you sure it's not? Remember, the irrational numbers are
infinitely more populous in any given segment than the rationals. There are
more irrational numbers between any two rational numbers than there are
rational numbers period.
>> 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.
Well, your personal logic seems just as valid as my personal logic, but they
can't both be right. :-) Therefore, I turn to experts.
--
Darren New, San Diego CA, USA (PST)
There's no CD like OCD, there's no CD I knoooow!
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"Darren New" <dne### [at] sanrrcom> wrote:
> At random? Are you sure it's not? Remember, the irrational numbers are
> infinitely more populous in any given segment than the rationals. There
> are more irrational numbers between any two rational numbers than there
> are rational numbers period.
Except there's the conundrum: the amount of rational numbers in any given
segment is *also* infinite!
Boggles the mind, eh?
--
Tim Cook
http://empyrean.freesitespace.net
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Tim Cook wrote:
> 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.
Actually, probability theory states it **is** zero if some digit is
missing. I was going to write a separate post about it providing *some*
rigor, but am quite busy - hopefully some time this week.
(Essentially, if you take the interval from 0 to 1, write the decimal
expansion, and ask "what is the probability to get *any* number that
doesn't have the digit 4 in it", the answer is 0. The set of all numbers
in [0,1] that don't have the digit 4 anywhere in their expansion is a
set of measure zero, and thus the integral is 0).
--
Wear short sleeves! Support your right to bare arms!
/\ /\ /\ /
/ \/ \ u e e n / \/ a w a z
>>>>>>mue### [at] nawazorg<<<<<<
anl
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Tim Cook wrote:
> "Darren New" <dne### [at] sanrrcom> wrote:
>> At random? Are you sure it's not? Remember, the irrational numbers are
>> infinitely more populous in any given segment than the rationals.
>> There are more irrational numbers between any two rational numbers
>> than there are rational numbers period.
>
> Except there's the conundrum: the amount of rational numbers in any
> given segment is *also* infinite!
'Tis a bigger infinity.<G>
--
Wear short sleeves! Support your right to bare arms!
/\ /\ /\ /
/ \/ \ u e e n / \/ a w a z
>>>>>>mue### [at] nawazorg<<<<<<
anl
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