# re: "Finite" For "Greater Than Zero"page 10

•  197
^n is a positive, non-
zero, number.

The coin being tossed infinitely often is a mathematical coin, needless to say. Let's take the complete works of Shakespeare ... result, then with probability one the string S will appear on the record. And not just once but infinitely often.

Indeed, and there also sequences of throws that do not have a single occurrence of 'to be or not to be' encoded in them. Equally many as sequences that do have those famous words, in fact.
Jan
No, "0.333..." is an exact representation. The "..." is part ... three recurring", which is even more transparently an infinite series.)

The sequence 0.3, 0.33, ..., 0.333... has a limit of 1/3 as the number of digits grows without bound. That's not the same as saying it *is* 1/3.

What is being said is that 0.333... is a notation for 1/3.

Since it is a definition it can't be wrong,
Jan
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I find it hard to get a handle on an infinite probability.

You and me both, man. However, in the archives are very, very, very long discussions about whether such a notion ... from an infinite set). It would probably be a kindness to the group not to go down that path again.

I don't remember everything I learned in my probability and statistics classes forty years ago, but I do seem to remember that in order to put probability theory on a rigorous basis, and in particular to understand why the probability of picking one particular integer from the countably infinite set of all integers is zero, you needed to understand Borel sets.
I'm now content to believe that in the limit as N approaches infinity the probability of picking a particular member of a set of size N is obviously zero.
( lim 1/N = 0 as N > infinity.)
But, as always with limits, I realize that N never really gets to infinity; it just gets to a value that is large enough to make the probability different from zero less than any epsilon you care to mention. So the probability never really gets to be identically zero.
But if N is, say, ***, then the probability is 10**-100, which is close enough to zero for government work.
I'm now content to believe that in the limit as N approaches infinity the probability of picking a particular member ... from zero less than any epsilon you care to mention. So the probability never really gets to be identically zero.

That's the definition of a limit. For every epsilon > 0 there exists a delta > 0 such that L - f(x) Since N can never reach infinity, how could 1/N ever reach 0.
But if N is, say, ***, then the probability is 10**-100, which is close enough to zero for government work.

Yep, you can make it arbitrarily close as long as your arbitrary number is greater then zero.

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Of course it does. The altered model has the same "finite dimensional distributions" as your model, hence the tosses of the die are still independent.

No disagreement on that, and indeed you cannot distinguish the two on basis of a finite number of experiments. But ... drawn is not a good model for throwing a die infinitely many times. You can disagree with that of course,

In both models individual infinite sequences have the same chance of being drawn, namely 0. So no disagreement from me on that. More is true: Any given event concerning te behavior of the entire sequence will have the same chance in either model.

J.
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What is being said is that 0.333... is a notation for 1/3.

Actually, what I'd say is that it's a notation for

Sigma(i=1, infinity) ***^(-i)
which happens to be equal to 0.3 / (1 - 0.1) = 0.3 / 0.9 = 1/3.

It's a notation for the limit, which, as he says, is 1/3.

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BTW, is the notation with dots over the repeating decimals still used? As in . 0.3 = 0.333... . . 0.12345 = 0.123452345...

I've never seen dots used. I learned it (30 years ago) with a line over the repeating digits:

0.3 = 0.333...

0.12345 = 0.123452345...

but I don't know if that's still taught.

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I'm now content to believe that in the limit as ... zero. ( lim 1/N = 0 as N > infinity.)

That's the definition of a limit. For every epsilon > 0 there exists a delta > 0 such that L ... x or in your example a is infinity). Since N can never reach infinity, how could 1/N ever reach 0.

So if we take lim 1/N = 0, what you're saying is that for any epsilon (say, 0.5), there's some delta such that
0 - 1/x 0)exists(M)forall(x>M) L-f(x) 0)exists(M)forall(x>M) L-f(x) > n

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