# Probability of infinite non-random subsets in an infinite set of random integers

Discussion in 'Physics & Math' started by Fafnir665, Dec 14, 2009.

1. ### Fafnir665You just got served.Registered Senior Member

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Say you have an infinite set of random integers, what is the probability of having an infinite subset of one integer, say 9?

3. ### PeteIt's not rocket surgeryRegistered Senior Member

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I'm not sure that "random integer" is meaningful. What exactly did you have in mind?

5. ### AlphaNumericFully ionizedRegistered Senior Member

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Do you mean "I pick an integer from 0 to 9 at random (ie each one is equally likely) and I write it down. I then do it again and again and again, infinitely many times and that's my 'set of random integers'. What is the probability the set is made entirely out of 9's?"

7. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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If you have an infinite set of random integers, the subset of 9's should also be infinite, no?

8. ### Walter L. WagnerCosmic Truth SeekerValued Senior Member

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Quite right, Ben. Otherwise, if one of the subsets were finite in members, then the choosing of the integers would be determinable to be non-random.

9. ### QuarkHeadRemedial Math StudentValued Senior Member

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Nah, formal set theory ain't going to help you here. Why? Because, by the definition of a set, no set element, here an integer is allowed to be double-counted. id est the "set" $\{9,9,9,...\}\,\not\subset \mathbb{Z}$.

Let's instead call this "subset" a string, and start easy. Given a starting pool of, say 2 integers, the probability of drawing any predetermined string of length 2 is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{2^2}$. It makes not a jot of difference, as Alpha hinted, whether the elements in this string are all the same, all different, or some and some.

So , the probability of drawing any predetermined string of length $n$ from a set of size $n$ will be $\frac{1}{n^n}$ - so figure the probability if each is infinite.

Rather small, wouldn't you say?

10. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Certainly it makes since if n is finite.

But, suppose I were to write pi to infinite precision, and take each digit as an element in my set. How many times does 9 appear in that set?

Maybe set is the wrong word to use here?

I dunno...I always tend to get myself into trouble when I talk about math

11. ### iceauraValued Senior Member

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My guess:

The probability of "having" an infinite string of 9s somewhere in there is one. The probability of selecting it at random is 0 (its "measure" is 0).

12. ### temurman of no wordsRegistered Senior Member

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It is not known. (You can say sequence instead of set.)

Lu Chao knows 67,890 digits of pi.

13. ### PeteIt's not rocket surgeryRegistered Senior Member

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Why are we discussing sets of random digits, when the original post specified random integers?

14. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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Because any other interpretation of the implied distribution in the wording "random integers" is either poorly defined or felled by Occam's Razor?

Last edited: Dec 16, 2009
15. ### Fafnir665You just got served.Registered Senior Member

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Random digits or random integers doesn't matter. I was thinking of a sequence of integers selected at random, but you can think of the digits of an irrational as the sequence as well.

The problem is all the proofs deal with finite sets, so you could say, yeah by ergoticity the probability that in an infinite set or sequence there is a finite non random homogeneous subset/sequence, but they don't handle infinite sets within infinite sets.

16. ### PeteIt's not rocket surgeryRegistered Senior Member

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I think it matters a great deal, since there are finite digits and infinite integers (which makes selecting random integers problematic).

But, anyway. We have a infinite set of random digits.
You want to know the probability that this set contains an infinite number of 9's, right?
Or, do you want the probability that an infinite sequence of random digits contains an infinite consecutive sequence of 9's?

17. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Ahh ok.... if you like a set of random integers between 0 and 9.

I just chose pi because it's seemingly irrational, so the digits should be a truly random set.

18. ### AlphaNumericFully ionizedRegistered Senior Member

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Irrationality does not imply a randomness in the digits. The first number to be proved transcendental (and thus also irrational) was $\sum_{n}10^{-n!}$. The digits (base 10) are only 1 or 0, there are absolutely no 2,3,4,5,6,7,8,9 digits in the decimal expansion so it's not random in that sense and I can definitely tell you how many digits in the first k digits are equal to 1 (or 0).

Irrationality doesn't imply random in the strictest sense.

19. ### temurman of no wordsRegistered Senior Member

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Just for the fun of it, to continue the trend "criticize BenTheMan", I don't understand why you said "seemingly" because we know for sure that pi is irrational.

20. ### Fafnir665You just got served.Registered Senior Member

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Well, the probability that an infinite sequence would contain an infinite subsequence of 9's is 1. We know that because how a subsequence is defined, you can grab the value from anywhere in the sequence as long as its further on in the sequence.

What I'm looking at is the probability that an infinite sequence would contain an infinite consecutive sequence of 9's, yeah..

If you look back at the erogotic proofs it follows that in finite systems, you wouldn't get finite non-random consecutive sequences if you want the system as a whole to remain random, but with infinite sets it seems like you can allow for infinite non-random consecutives (I've been calling them consecutive homogeneous sub sequences) without upsetting the random nature of the system as a hole.

21. ### D HSome other guyValued Senior Member

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Surely Ben meant that pi is seemingly normal rather than seemingly irrational.

It makes all the difference in the world. The first is a well-defined concept. The latter is not.

22. ### PeteIt's not rocket surgeryRegistered Senior Member

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That would surely be zero.
An infinite consecutive sequence of a given digit would necessarily be the complete tail of the random sequence - there could not be any other digits after.

So your question is the same as this one:
What is the probability that a decimal point followed by an infinite sequence of random digits would produce a rational number?

I'm sure the answer is zero, because there are countable rationals, and uncountable infinite sequences of random digits.

23. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's another way of considering it. An infinite consecutive homogeneous sub-sequence would upset the random nature of the whole sequence, because such a subsequence would necessarily terminate the whole sequence.
The remainder of the infinite random sequence (the "random part", so to speak) must be finite.
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