>  I really can't understand why you are so fixated with that question.
> I never doubted or denied its veracity. My reply clearly implied that it
> is indeed so.

So you would also agree then that 1/2 + 1/4 + 1/8 + ... equals 1 with 
infinitely many terms?  You see where this is going?

>  You don't have to try an infinite number of times to get a value from
> a continuous range. You only have to try once. And the value you get had
> a probability of zero of being chosen. Yet it was chosen.

You are trying to equate the probabilities of 1/infinity with 0/infinity, 
they cannot always be treated as the same (in some situations they can be).

Imagine choosing numbers between 0 and 1.  Getting exactly 0.5 (or any other 
number between 0 and 1) has a probability of 1/infinity, getting exactly 1.5 
has a probability of 0/infinity.  "Normally" they would be the same, but if 
you say something like "what is the probability of getting *any* value 
between 0 and 1" or "what is the probability of getting exactly 0.5 after 
infinitely many tries" then they are not the same.

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scott <sco### [at] scottcom> wrote:
> >  I really can't understand why you are so fixated with that question.
> > I never doubted or denied its veracity. My reply clearly implied that it
> > is indeed so.

> So you would also agree then that 1/2 + 1/4 + 1/8 + ... equals 1 with 
> infinitely many terms?  You see where this is going?

  I agree with it, and no, I don't see your point.

> You are trying to equate the probabilities of 1/infinity with 0/infinity, 
> they cannot always be treated as the same (in some situations they can be).

> Imagine choosing numbers between 0 and 1.  Getting exactly 0.5 (or any other 
> number between 0 and 1) has a probability of 1/infinity, getting exactly 1.5 
> has a probability of 0/infinity.  "Normally" they would be the same, but if 
> you say something like "what is the probability of getting *any* value 
> between 0 and 1" or "what is the probability of getting exactly 0.5 after 
> infinitely many tries" then they are not the same.

  I honestly don't understand.

-- 
                                                          - Warp

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>> So you would also agree then that 1/2 + 1/4 + 1/8 + ... equals 1 with
>> infinitely many terms?  You see where this is going?
>
>  I agree with it, and no, I don't see your point.

That the sequence is also the probability of a head being tossed after N 
tries of throwing a coin.  After 1 try it's 1/2, after 2 tries it's 1/2 + 
1/4, etc.  So after infinitely many tries the probability is *equal* to 1.

Now simply replace "head being tossed" with "this sequence of characters 
being the works".

>  I honestly don't understand.

If you are choosing numbers in the range 0...1, 0.5 has an infinitesimally 
small probability of being chosen, 1.5 has a zero probability of being 
chosen, they are not the same probability.  (In some situations you can call 
the probability of 0.5 being chosen "zero", but you must remember that the 
zero came from 1/infinity and not "real" zero incase you use it in later 
calculations).

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The Internet has already proven that a multitude of monkeys would not 
duplicate the works of Shakespeare, regardless of the amount of time 
allowed.

Regards,
John

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scott <sco### [at] scottcom> wrote:
> >> So you would also agree then that 1/2 + 1/4 + 1/8 + ... equals 1 with
> >> infinitely many terms?  You see where this is going?
> >
> >  I agree with it, and no, I don't see your point.

> That the sequence is also the probability of a head being tossed after N 
> tries of throwing a coin.  After 1 try it's 1/2, after 2 tries it's 1/2 + 
> 1/4, etc.  So after infinitely many tries the probability is *equal* to 1.

> Now simply replace "head being tossed" with "this sequence of characters 
> being the works".

  You talk as if I had said somewhere that in the infinite case the
probability does *not* equal 1. I don't remember saying such a thing.

> >  I honestly don't understand.

> If you are choosing numbers in the range 0...1, 0.5 has an infinitesimally 
> small probability of being chosen, 1.5 has a zero probability of being 
> chosen, they are not the same probability.  (In some situations you can call 
> the probability of 0.5 being chosen "zero", but you must remember that the 
> zero came from 1/infinity and not "real" zero incase you use it in later 
> calculations).

  Now it's you who sounds like saying that 1/2+1/4+... does not equal 1.

-- 
                                                          - Warp

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>> Now simply replace "head being tossed" with "this sequence of characters
>> being the works".
>
>  You talk as if I had said somewhere that in the infinite case the
> probability does *not* equal 1. I don't remember saying such a thing.

You said:

> That's, in fact, the exact same thing as saying "never".

Never is a probability of exactly 0, yet you've just agreed that the 
probability is actually exactly 1.  Make your mind up :-)

Post a reply to this message

scott <sco### [at] scottcom> wrote:
> >> Now simply replace "head being tossed" with "this sequence of characters
> >> being the works".
> >
> >  You talk as if I had said somewhere that in the infinite case the
> > probability does *not* equal 1. I don't remember saying such a thing.

> You said:

> > That's, in fact, the exact same thing as saying "never".

> Never is a probability of exactly 0, yet you've just agreed that the 
> probability is actually exactly 1.  Make your mind up :-)

  Great out of context quote.

  The "never" was not referring to "will it appear?". It referred to
"is it *forced* to appear"?

  No, at no point is it forced to appear. Thus it's *never* forced to
appear. That's not the same thing as "it will never appear".

-- 
                                                          - Warp

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>  The "never" was not referring to "will it appear?". It referred to
> "is it *forced* to appear"?

If the probability is exactly 1, which you agree it is, then of course it is 
forced to appear at some point in the sequence.  If it wasn't then the 
probability would be less than 1.

Or are you going to argue now about the meaning of "forced"? ;-)

Post a reply to this message

scott <sco### [at] scottcom> wrote:
> >  The "never" was not referring to "will it appear?". It referred to
> > "is it *forced* to appear"?

> If the probability is exactly 1, which you agree it is, then of course it is 
> forced to appear at some point in the sequence.  If it wasn't then the 
> probability would be less than 1.

  And if the probability of getting 0.5 is zero, then of course it means
that it will never be chosen.

  Explain to me the exact mechanics which force the works to appear.

-- 
                                                          - Warp

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>> If the probability is exactly 1, which you agree it is, then of course it 
>> is
>> forced to appear at some point in the sequence.  If it wasn't then the
>> probability would be less than 1.
>
>  And if the probability of getting 0.5 is zero, then of course it means
> that it will never be chosen.

Of course.

But the probability of 0.5 turning up is not zero, it is 1/infinity. 
Sometimes they can be used interchangably, but in this situation they 
cannot.  You won't find a mathematician that claims 1/infinity universally 
is equal to zero under all circumstances in normal number systems.

On the other hand, writing an infinite sum like 1/2+1/4+1/8+.. does 
*exactly* equal one, always, never with any doubt, it is universally 
accepted in mathematics.

>  Explain to me the exact mechanics which force the works to appear.

Calculate the probability, you will find it is equal to one.  Exactly one. 
That, by definition, means it is guaranteed, or "forced" if you like, to 
happen.  QED.

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