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Tim Cook wrote:
> While the probability is equal, it doesn't mean all possible
> combinations necessarily show up an infinite number of times in an
> arbitrary sequence.
Why do you say that?
> The aforementioned 0102030405..9596979899 is
> technically a possible random number with each digit 0-9 having an equal
> probability of occurring. An infinite sequence could be considered that
> fails to meet the 'all possible combinations' feature.
I'm not sure I follow wht this example has to do.
>> If you have an infinite number of trials and the letter 'a' never
>> shows up, it means it's impossible for the letter 'a' to show up.
>
> Not really. Randomly picking an infinite amount from the set {a,b}
> *could* result in nothing but bbb..bbb.
Do you have a cite to support this contention?
> That doesn't mean it's
> impossible for the letter 'a' to show up, only that the bbb..bbb
> sequence isn't very likely...except it's exactly the same probability as
> any other sequence of infinite length: 1/infinity.
That's not my understanding of how the math works. Do you have any citation
as evidence for this? Because if the letter 'a' doesn't show up after an
*infinite* number of trials, you clearly don't have any probability for it
to show up at all, and indeed that's what the math pages I've cited already say.
--
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] san rrcom> wrote:
>> While the probability is equal, it doesn't mean all possible combinations
>> necessarily show up an infinite number of times in an arbitrary sequence.
> Why do you say that?
Because 0.50000000... is one possible arbitrary sequence, and nowhere in it
contains the sequence 123.
>>> If you have an infinite number of trials and the letter 'a' never shows
>>> up, it means it's impossible for the letter 'a' to show up.
>> Not really. Randomly picking an infinite amount from the set {a,b}
>> *could* result in nothing but bbb..bbb.
> Do you have a cite to support this contention?
Observe: the number set of decimals from 0 to 1 (inclusive) is infinitely
large, with each element containing an infinite amount of decimal places,
each having an equal probability of being a digit 0-9. This set contains
the number 0.000... QED.
>> That doesn't mean it's impossible for the letter 'a' to show up, only
>> that the bbb..bbb sequence isn't very likely...except it's exactly the
>> same probability as any other sequence of infinite length: 1/infinity.
> That's not my understanding of how the math works. Do you have any
> citation as evidence for this? Because if the letter 'a' doesn't show up
> after an *infinite* number of trials, you clearly don't have any
> probability for it to show up at all, and indeed that's what the math
> pages I've cited already say.
See above. The probability of selecting any particular number is
effectively zero, but that number exists and so *can* be chosen.
--
Tim Cook
http://empyrean.freesitespace.net
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Tim Cook wrote:
> "Darren New" <dne### [at] sanrrcom> wrote:
>>> While the probability is equal, it doesn't mean all possible
>>> combinations necessarily show up an infinite number of times in an
>>> arbitrary sequence.
>> Why do you say that?
>
> Because 0.50000000... is one possible arbitrary sequence, and nowhere in
> it contains the sequence 123.
The probability of a 1 showing up anywhere is zero, so yes, the probability
of 123 showing up anywhere is zero.
> Observe: the number set of decimals from 0 to 1 (inclusive) is
> infinitely large, with each element containing an infinite amount of
> decimal places, each having an equal probability of being a digit 0-9.
Yep.
> This set contains the number 0.000... QED.
And that's why the probability of picking a number like that is zero. :-)
>> That's not my understanding of how the math works. Do you have any
>> citation as evidence for this? Because if the letter 'a' doesn't show
>> up after an *infinite* number of trials, you clearly don't have any
>> probability for it to show up at all, and indeed that's what the math
>> pages I've cited already say.
>
> See above. The probability of selecting any particular number is
> effectively zero, but that number exists and so *can* be chosen.
Sorry, that's not a citation. I.e., I read experts who say you're wrong.
You're trying to convince me using intuitive reasoning that infinity works
in a way the experts say it doesn't work. I see that both arguments are
perfectly reasonable given certain assumptions about how infinity works.
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. :-)
--
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:
> 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. ;)
Besides, there's a third option: we're both right.
--
Tim Cook
http://empyrean.freesitespace.net
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Tim Cook wrote:
> "Darren New" <dne### [at] sanrrcom> 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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