unforum

[Chris K]

I think it's actually more complicated.

The chance of finding a 3 digit string in a random 1000 digit string is a lot lower than that, I think. Probably about 60%

I'll try and work it out now, wait a sec.

[neon snake]

[thebruce]

[Chris K]

I decided it depends on what the string is.

If the string is symetrical (11011, 5555, 4554) then there are more possible locations for it in the 1000 character long string.

If the string is 1234, and it appears at the beginning, then it cannot appear again until the 5th digit.

This will screw up the probabilty because it no longer follows a binomial distribution.

It might be impossible to work out what the probabilty of a 3 digit string appearing in a 1000 digit string is, if it's not then it's very very hard.
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[neon snake]

[Sasuntsi Davit]

I think what has happened here is that we are mixing theory of infinity and the applications of it.
Infinity is exactly what it says on the tin...it goes on for ever...and ever..and ever..

So with regards to the possibility of a given sequence of numbers appearing in an infinetly long string of numbers being 100%...that is right. If the number goes on for ever and never stops, then you will find everything in it...because it is in theory, and "at the moment" it is no number.

However, once you say that the string could be all 5s, (which is a possibility) you are making an application of the theory, which seems to 'disprove' the 100% thing. you can't find the numbers 0-4 and 6-9 in that sequence. But that was just one event...we will be crossing over into statistics then.

Its again, like the lottery, the chance of winning the "lotto" here in the uk is something like...
1/49 * 1/48 * 1/47 * 1/46 * 1/45 * 1/44 ~= 1*10^-10....which is a very tinty probabilty, and yet people win.

Some people in america have won the lottery twice in 10 years, which is an even lower probabilty...

Anyway, i might have been talking bs, in which case feel free to correct me. Hopefully however, what I said made sense.

[neon snake]

[thebruce]

Re: the chance of a 5 appearing within 2 random digits.

Ok, I knew I'd confuse it with the two chances 19/100 and 10/55

-If the order of the digits count, then there are 100 unique numbers from 00-99.
-If the order of the digits does not count, then there are 55 unique number combinations between 00-99.

-If the order of the digits count, then there are 19 numbers that include at least one 5 (05,15,25,35,45,55,65,75,85,95,50,51,52,53,54,56,57,58,59)
-If the order of the digits does not count, then there are 10 combinations that include at least one 5. (5/0, 5/1, 5/2, 5/3, 5/4, 5/5, 5/6, 5/7, 5/8, 5/9)

So, I can't remember which one to use, but it depends on the wording of the problem. The chance is either 19/100 or 10/55 that 5 will appear once within 2 random digits. One of them is right for this problem .

[rose]

are we there yet?

[neon snake]

Everyone's back at work/college/uni/school!

[thebruce]

hehe good call...


not to beat a dead horse, but I'm risking a thing again with this 0.9~=1 thing. in another discussion in PM, I ended up going through a step by step proof, retaining all aspects of values and concepts in the equation, showing how one on side, some people come to 0.9~ -> 1 and some people come to 0.9~ = 1... and the difference between the two... lemme know what you think. Or correct the proof 'form' if it's not official I'm just using logic and language here... heh

Anyway, this is long, but it's cool... hopefully

------
I'll use numerics for what I consider to be absolute - the points I believe we all agree on, and alphabet for possible points in contention...

Given R represents a difference between any two values - a remainder...
Consider A value we want to equate to B, where R is the difference.
Can we agree that:
1) if A is a geometric infinite series, lim A = B, or A -> B
2) At any given point in the series, to equate the sum of the previous terms to the series' limit, the remainder must be added (as in B-sum(A to n terms) = R), so sum(A to n terms) + R = B
Thus,
3) sum + remainder = value, or A + R = B

Now, given that 0.9~ is an infinite string of 9's, we can thus write an equation that will be the process of generating the decimal number, without having to calculate it:
A) sum(9/10^n as n=1 to infinity) -> 0.9~
Why ->? Because summing an infinite series is a process that will never end, so the process of the summing can never give the exact value that 0.9~ represents. In order to equate the geometric sum to the known value, the remainder needs to be added (see #2)

So in order to make #A "=0.9~" rather than "->0.9~", the R must be added (see points #1 and #2)

So we need to determine R. Let's first determine for the next couple of points that B=0.9~
At any point in the number 0.9~, there are n digits, followed by an infinite number of digits. So how do we write A - the sum of 0.9~ to n digits, plus the remainder - the rest of the infinite digits?
Well, let's work with 3 terms.
4) From #2, B=(A to n terms)+R - 0.9~ = (0.9 + 0.09 + 0.009) + (0.0009~)
since we can't (or want to) calculate using an infinite decimal, we must convert the decimals to workable equations.
5) 0.9~ = sum(9/10^n where n=1 to 3) + sum(9/10^n where n=4 to infinity)

Now let's find an equation for 1 that we can compare to the answer for 0.9~
So, let's work with this... let's say, to 3 terms.
6) From #3, A+R=B - (0.9 + 0.09 + 0.009) + (0.001) = 1
which can be further expressed as:
7) (9/10 + 9/10^2 + 9/10^3) + (1/10^2-9/10^3) = 1
or
8 ) sum(9/10^n as n=1 to 3) + (1/10^(3-1)-9/10^3) = 1
thus
9) sum(9/10^n as n=1 to X) + (1/10^(X-1)-9/10^X) = 1

THIS is the crux of the problem. essentially, with X=infinity, we end up with a value (infinity-1) - one less than infinity, which has the same problem as an infinite number that ends in a number, which means it can't be infinite, so the remainder can't exist. The problem is, they're now trying to equate an infinite process to a static value, which isn't possible (see #A). So the answer for many people is to scrap the impossible value, the concept of 'infinitely small remainder'. They are trying to convert #1, A->B, to A=B, through #3, A+R=B, by ignoring the impossible R, which should remain a concept. which means that they will be able to say that A -> B is the same as A = B. So by removing R, they now have...
sum(9/10^n as n=1 to infinity) -> 1 becoming
B) sum(9/10^n as n=1 to infinity) = 1

Now consider #5, if we are calculating to any length X of A (=0.9~), then
10) 0.9~ = sum(9/10^n where n=1 to X) + sum(9/10^n where n=X+1 to infinity)
if we are calculating an infinite length of A (X=infinity) then:
11) 0.9~ = sum(9/10^n where n=1 to infinity) + sum(9/10^n where n=infinity+1 to infinity)

For the same reason we end up at #B, an impossible remainder, the remainder for #11 is impossible. How can you have a sum from the infinith+1 iteration to the infinith iteration? It's an impossible number. At least in this case, impossibly small. So in the end, they end up removing the remainder, which should remain a concept, to have:
C) 0.9~ = sum(9/10^n where n=1 to infinity)

Thus, they can equate #B and #C for
D) since sum(9/10^n as n=1 to infinity) = 1
and sum(9/10^n where n=1 to infinity) = 0.9~
therefore 0.9~ = 1

Of course, that's precisely what I'm debating. Before removing any impossible values - the remainders, we have #9 and #10:
12)
sum(9/10^n where n=1 to X) + (1/10^(X-1)-9/10^X) = 1
sum(9/10^n where n=1 to X) + sum(9/10^n where n=X+1 to infinity) = 0.9~
Based on #3, sum+remainder=value, or A+R=B, both of these equations work for any value X
* Assuming A=0.9~ and B = 1,
* If I am proving, by #1, that 0.9~ -> 1, or by #3, that 0.9~ + R = 1,
* then the question remains, what's the value of R, without removing 'impossible' numbers.
Since we know from #12 that:
B = 1 = sum(9/10^n as n=1 to X) + (1/10^(X-1)-9/10^X)
and
A = 0.9~ = sum(9/10^n where n=1 to X) + sum(9/10^n where n=X+1 to infinity)
and that R = B - A, we end up with
R = sum(9/10^n as n=1 to X) + (1/10^(X-1)-9/10^X) - sum(9/10^n where n=1 to X) + sum(9/10^n where n=X+1 to infinity)
which reduces to
13) R = (1/10^(X-1)-9/10^X) + sum(9/10^n where n=X+1 to infinity)

So, I can now use #13 as the equation to calculate the value of R, in order to prove that (0.9~) + R = 1

Let's first try calculating R without the equation where X = 3 (3 digits)
From #12 we know that:
1 = sum(9/10^n where n=1 to X) + (1/10^(X-1)-9/10^X)
0.9~ = sum(9/10^n where n=1 to X) + sum(9/10^n where n=X+1 to infinity)
With X = 3 we now have 2 paths:
14)
1 = 9/10 + 9/100 + 9/1000 + (1/100-9/1000)
0.9~ = 9/10 + 9/100 + 9/1000 + sum(9/10^n where n=4 to infinity)
From which we can now extrapolate for R as R=B-A
15) 1-(0.9~) = (9/10+9/100+9/1000 + (1/100-9/1000)) - (9/10+9/100+9/1000 + sum(9/10^n where n=4 to infinity))
which becomes
16) 1-(0.9~) = 0.999 + 0.01-0.009 - 0.999 - sum(9/10^n where n=4 to infinity)
17) 1-(0.9~) = 0.001 - sum(9/10^n where n=4 to infinity)
18 ) R = 1/10^3 - sum(9/10^n where n=4 to infinity)
which is also the answer solving for R from #13

Now, inputting various test values for X, we find that #18 can be expressed as
19) R = (1/10^X - sum(9/10^n where n=X+1 to infinity)
or, R = (0.001 - 0.0009~) where X=3

And wha-la, what do we have? In theory, the impossible calculation for R, in this case 0.001-0.0009~ gives the impossible answer 0.0~1 for any value X.
As X -> infinity, R -> 0.0~1
Thus, If X = infinity, R = 0.0~1
As absolute: A+R=B equates to (0.9~) + (0.0~1) = 1, or, removing the infinitely small and impossible R, by point #2, (0.9~) -> 1

This is the basis for the entire argument - that theoretically, the difference between 0.9~ and 1 is not a number. It's not infinite, because it has to end with a 1, and it's not finite because it contains an infinite amount of 0's. There is no real number between 0.9~ and 1, therefore 0.9~ must = 1.

This is where the definition argument comes in. Speaking in real number terms, yes 0.9~ = 1. Speaking in absolute terms, 0.9~ + R = 1, where R is an impossible value.
(don't be confused by my naming conventions 'real number' terms making 0.9~=1 is theory because it's derived from only handling 'real' numbers as opposed to impossible numbers. For 'absolute' terms, impossible values are retained as variables, just as e and pi represent infinite values)

Ok, so yeah, the entire post was written by me, in notepad, read, edited, without looking up any external references...
I just including everything we talked about here, from my head...
Ok, now I'm going to bed
Oh hey, I can rhyme... but now's not the time.
oh haha


PS: Here's a summary of all the points... #'s are what I assume we all agree on, A-Z are points I'm assuming are in contention, of course, I may be wrong in my labelling either way depending on who reads this.

[SilentAvenger]

I'll let wikipedia do the discussion for me:

http://en.wikipedia.org/wiki/Geometric_series

Look at the infinite geometric series part.

1.111~ = sum(0.1^n)
9 * 1.111~ = 9*sum(0.1^n) = 9.999~
9.999~ = 9 * sum(0.1^n), or a1 = 0.9, x = 0.1

0.9, 0.09, 0.009 etc.

Wikpedia shows that the formula for summing such a series is:
1 / (1-x)

x = 0.1;

1 / (1-0.1) = 1 / 0.9 = 1 / (9 / 10) = 10 / 9;
10/9 * 9 = 10;

9.999~ = 10;

fin

[thebruce]

[SilentAvenger]

thebruce, wikipedia is 100% accurate on formula for infinite series'. This is the way it is.

And, 1/2^infinity is not a number, and therefore cannot be used in calculations.

[Wishi-san]

This thread has degenerated into pointless drivel, it's not a contest of who knows more maths than everyone else. So let's just let this lie.