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 Forum index » Archive » Archive: Perplex City » PXC: Project Syzygy Pre-Game
[LOCKED] [PUZZLE?] E Numbers
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LordKinbote
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Joined: 25 Sep 2002
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 		EDIT: I'm trying to figure out your explanation of "infinite chance". I am not familiar with such a topic in probability, so I'm checking to see if your math is correct. So if you see this post, hold off on yelling at how I'm stupid until I've done a little more research on the subject and deleted this edit. Thanks.

SilentAvenger wrote:

For my hand's sake, I = infinity;
Simple:
Assuming the random stream of numbers E (Razz not the real e, just for fun) is of infinite length, and there is a string S of finite length N, the chance of S appearing in the beginning of E is 1/10^N. In E there are I substrings of length N, therefore, in each one, there is a I/10^N chance of the substring being S. Now, because 10^N is a finite number, it pales in the face of I, and runs away (well, not really, but I think of it this way. In math terms, as lim x->I, any number which does not ->I becomes so small in comparison that it does not have to be considered. Its different if you have more than one numbers ->I), so we are left with I - Infinity.

Hehe, if you call that simple.


Your proof is on the right track, but there is not an I/10^N chance of a chosen substring of length N being S, there is a 1/10^N chance. Your way ends up with an infinite probability, and since the probability of an event must take a value between 0 and 1, there must be an error.

So let's create a discrete random variable X that counts the number of substrings in E that are S. We're looking for the case in which there exists at least one substring, which means we want the probability that X is greater or equal to 1, or P{X>=1}. We know this equals 1 - P{X=0}, since the probability of one event is equal to 1 minus the complement of that event.

So what is P{X=0}, or the probability that there are no substrings in E that are S? This means that the first substring of length N is not S, as is the second, as is the third, etc. The probability that a given substring is not S is (1-(1/10)^N). We can consider looking at each substring an independent event, so to get the probability, we need to multiply the probability of each event together, so we get the limit as x->inf of (1-1/10)^N)^x. As we know, a fraction multiplied an infinite number of times is going to go to 0.

So if P{X=0}=0, then P{X>=1}=1-0=1. So it is a sure thing that S is a substring inside E.

It's pretty much the same proof, but it just corrects that one error of infinite probability.

PostPosted: Wed Dec 22, 2004 10:16 pm
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Shish
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 		> Your proof is on the right track, but there is not an I/10^N chance of a
> chosen substring of length N being S, there is a 1/10^N chance.

1/10^n per attempt, multiplied by an infinite number of attempts


> Your way ends up with an infinite probability, and since the probability of
> an event must take a value between 0 and 1, there must be an error.

It's not so much that the probability is infinite, so much as given infinite attempts it will happen infinite times - thus ending as I/I, which you may take to be 1, unless there are special rules about multiplying and dividing infinity that should be applied...

It can't be a direct probablility anyway, else the same math would show that flipping a 3 coins would have a 1.5 chance of a head appearing.
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PostPosted: Wed Dec 22, 2004 11:18 pm
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LordKinbote
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Shish wrote:


1/10^n per attempt, multiplied by an infinite number of attempts


I know that, you know that, but that's not what was said.


Quote:
It's not so much that the probability is infinite, so much as given infinite attempts it will happen infinite times - thus ending as I/I, which you may take to be 1, unless there are special rules about multiplying and dividing infinity that should be applied...

It can't be a direct probablility anyway, else the same math would show that flipping a 3 coins would have a 1.5 chance of a head appearing.


Well, it doesn't come out as I/I, it comes out as I/10^n as you said above.

If you take the limit of something and get I/I, though, you should backtrack and use L'Hospital's rule (take the derivative of the numerator and denominator and try again).

If you saw my edit at the top, I'm trying to figure out this "infinite probability" stuff. It's something I never studied, so I'm looking into it. I don't want to talk more about it until I'm sure I know what I'm talking about.

PostPosted: Wed Dec 22, 2004 11:31 pm
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Shish
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LordKinbote wrote:
If you saw my edit at the top, I'm trying to figure out this "infinite probability" stuff. It's something I never studied, so I'm looking into it. I don't want to talk more about it until I'm sure I know what I'm talking about.


If you don't want to discuss something... don't post it on an internet discussion forum -_-

I replied anyway because I find discussion and debate to be much more educational than typical study - I daresay that my various activities of today've used more brain power, and proved more educational, than the whole of the last term at school...
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PostPosted: Thu Dec 23, 2004 12:04 am
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LordKinbote
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Shish wrote:
LordKinbote wrote:
If you saw my edit at the top, I'm trying to figure out this "infinite probability" stuff. It's something I never studied, so I'm looking into it. I don't want to talk more about it until I'm sure I know what I'm talking about.


If you don't want to discuss something... don't post it on an internet discussion forum -_-

I replied anyway because I find discussion and debate to be much more educational than typical study - I daresay that my various activities of today've used more brain power, and proved more educational, than the whole of the last term at school...


I posted it before I realized there was something I didn't understand. I think it's reasonable to want to put my contribution to the discussion on hold without deleting the fairly lengthy proof I wrote.

I agree, discussion and debate can be very educational, but NOT when there's a chance that the person who seems to know most about the subject is wrong. If that's the case, the only education we'd be getting is miseducation. So I just wanted to pursue other avenues (like, for example, an e-mail to an old probability professor) before I continued. I'd rather build my temple of knowledge on sturdy ground. Wink

PostPosted: Thu Dec 23, 2004 12:13 am
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thebruce
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LordKinbote wrote:
As we know, a fraction multiplied an infinite number of times is going to go to 0.

So if P{X=0}=0,

but it's not 0, it approaches 0... 1/10^N where N=inf, is not 0, it's 1/inf...

Quote:
then P{X>=1}=1-0=1. So it is a sure thing that S is a substring inside E.

It's pretty much the same proof, but it just corrects that one error of infinite probability.


infinite probability just means it's impossible to know what the probability will be. I guess whether you believe that's possible depends on whether you believe there has to be a definitive solution for everything in the universe Smile


just had a HHGTTG flashback about the infinite improbability drive... Shocked
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PostPosted: Thu Dec 23, 2004 6:39 pm
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InspJJ
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 		Glad we have you guys to argue over this stuff - my maths teacher invited me to leave after my 'O' levels...
My knowledge of probability theory stops at 'If tomorrow is Xmas I will be at work by 3pm' - cries......
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PostPosted: Fri Dec 24, 2004 5:23 am
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SilentAvenger
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 		LordKinbote, what I meant by Infinite probablilty, is that the substring S will appear in a string of random letters an infinite number of time, and not only once.

Now, what remains to be proven to END this discussion, is wether e is really an infinite stream of random numbers, and not a structured one.

We have not displayed any proof in this thread that e is indeed truly random.

PostPosted: Fri Dec 24, 2004 8:22 am
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Olorin
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SilentAvenger wrote:
LordKinbote, what I meant by Infinite probablilty, is that the substring S will appear in a string of random letters an infinite number of time, and not only once.

Now, what remains to be proven to END this discussion, is wether e is really an infinite stream of random numbers, and not a structured one.

We have not displayed any proof in this thread that e is indeed truly random.


and indeed, I don't think it is.

e = 1 + 1/1! + 1/2! + 1/3! +...

is a very precise structure !

Same for pi...

F.O.R.

PostPosted: Fri Dec 24, 2004 11:13 am
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tanner
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 		e is NOT random --- it is is unpredictable without calculating --- same as pi
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PostPosted: Fri Dec 24, 2004 11:18 am
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neon snake
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 		Embarassed bugger.

Let me rephrase something I said earlier...

'The chances of any finite number (including E numbers for food) appearing in the infinite string 'e' is pretty high, all things considered.

Of course, if 'e' was truly random (which we of course all know that it isn't), as well as being infinite, then the chance of any finite string appearing is 100% (not an equal chance of it appearing or not appearing, as that would be 50%).'

Ahem. That's better.
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PostPosted: Fri Dec 24, 2004 12:13 pm
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tanner
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 		Smile -- yep
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PostPosted: Fri Dec 24, 2004 12:39 pm
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thebruce
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neon snake wrote:
Embarassed bugger.

Let me rephrase something I said earlier...

'The chances of any finite number (including E numbers for food) appearing in the infinite string 'e' is pretty high, all things considered.

Of course, if 'e' was truly random (which we of course all know that it isn't), as well as being infinite, then the chance of any finite string appearing is 100% (not an equal chance of it appearing or not appearing, as that would be 50%).'

Ahem. That's better.


actually, not specifically regarding e, it was never matter of a sequence having an equal chance to appear or not appear in an infinite random number, but that the chance of it appearing neared 100% but would never reach 100%, therefore there is no way to know if the chance is 100% or if it will never appear, because the only way to know is to find it, which may theoretically take an infinite amount of time. So it's not 1 or 0 chance, it's 1 and 0, essentially, the result cannot be calculated because it's based on an infinite framework.

Now with e, given it's a number that can be calculated, I'm sure out there somewhere a formula can be created that when given a sequence of numbers, will be able to find exactly where in e it exists... e is not an infinite random number... infinite non-repeating? I don't think we will ever know (can we ever know until we find a repeating set?)

interesting... take the number 0.123123123123123123123223etc... if we're only able to 'calculate' the number to 12 digits, it would appear that the number would be repeating, and we may consider it a repeating number. But who knows how far down the line whether or not a digit may differ? How can we know for a fact that any repeating decimal won't in fact have an alternate number way down the line if we never (or just can't) calculate that far? So how can we know that any number that doesn't end is actually a repeating number? In that sense, any repeating number and any (apparently) non-repeating number may or may not be an infinite number - but we can never truly know if it's infinite unless we never come to the end, which is not a concept we can ever grasp.. .theoretically, a number may be as long as it takes thousands of years to calculate... is it then infinite? nope, cuz it ends. But long before that we have essentially come to the conclusion that it's an infinite number.

so, we are currently under the impression that e is a non-repeating infinite number. But it may very well have an end at some point. Because of that uncertainty, we can't draw any conclusions about the chances of a sequence appearing if the conclusion is based on a certainty either way (finite or infinite)... we can only create a calculation for a sequence if the calculation takes into account the uncertainty of the number's finite-ness. ie, if the same result can be reached whether or not it's a finite number.

I hate infinity... it fries too many brains Confused
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PostPosted: Fri Dec 24, 2004 12:43 pm
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tanner
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Quote:
a sequence having an equal chance to appear or not appear in an infinite random number, but that the chance of it appearing neared 100% but would never reach 100%


the %age value approaches 100% at infinity
as its an infinite sequence then it does reach 100%
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PostPosted: Fri Dec 24, 2004 12:52 pm
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neon snake
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Quote:
So it's not 1 or 0 chance, it's 1 and 0, essentially, the result cannot be calculated because it's based on an infinite framework.


Mathematically speaking:
The chance of a finite sequence being found in an infinite random sequence absolutely is 1, with no possibility of it being 0. The proof has been presented several times above.
This is a mathematical principle. It's not something we're unsure of, and are trying to prove - it's something we are 100% sure of, and are trying to present existing proofs in a way that make sense to people who havn't seen them before, or who have not sat through the same amount of maths lectures that we have.

The rephrase I presented above was due to my original off the cuff comment (which started the debate), which was made without a good enough knowledge of the nature of 'e'. I believed it to be random; it has been presented that it is not (ie. the next digit is calculable, which is obvious now that the formula has been shown!).


Infinite numbers are subject to mathematics in the same way as ay other number - they have laws, and can be used in calculations etc etc.


Now then. Next topic for discussion.

The sequence between 1 and infinity is smaller than that between 0 to 1.

I havn't specified any rules, because that spoil it.

There is a way of counting where this is true, and I've been deliberately vague about it.


Discuss.
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PostPosted: Fri Dec 24, 2004 2:45 pm
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