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Alex gets even with writersblock. Or does he?


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35 replies to this topic

#11 Writersblock

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Posted 19 October 2007 - 05:23 AM

Yeah. I am no mathmatician but there is the concept of infinities tucked within infinities. Your example is one of them, but also, you can have an infinite number of fractions between any two whole numbers as well as having infinite whole numbers.

1< infinite <2
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#12 brhan

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Posted 19 October 2007 - 04:59 PM

It wouldn't be 200% it would still be 100% to the nearest percentage. That is what is cool about this whole concept. To the nearest percentage point, 100% of numbers contain any given digit except zero.

If you don't believe it, follow this.

In a series of 1 -10 there is 1 digit that contains a 3. 1 of 10 = 10%.
In a series of 1 - 100, each set of 10 contains 1 as above, except for the 30's series which contains 10. 10 for the 30's series plus 1 for all the other sets of 10 = 19. 19 of 100 = 19%. You can continue this sequence for every power of 10. At 1000, there are 271 3's. 19 for each series of 100 plus 100 for the 300's series. Extrapolate this far enough and you will see that at 1E66 (1 followed by 66 zeros), 99.9E65 numbers contain at least one digit 3. This means that 99.9 percent of the numbers less than 99.9E65 contain a digit 3. Therefore to the nearest percentage, 100% of numbers contain a digit 3. As you approach infinity this number will get closer and closer to 100%, but never reach it.

And yes, this is true for all digits except zero.



I have got the idea that as a number contains more and more digits, it is more likely that every digit will appear there. Just a couple of comments: 99.9E65 should be 9.99E65 (otherwise it will be bigger than the limit -- 1E66). I don't understand (this is for sure, but the previous one was for a discussion in different direction) why you exclude zero. It should have the same percentile as the other digits ... which you put it 100%.
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#13 Writersblock

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Posted 19 October 2007 - 06:33 PM

You are right, there is a typo in the 9.99E65.

You are also right that zero should be included. I only considered the first and last digits in a sequence as that is all that mattered for the 3. I forgot to figure that there is a "zero" set of tens, one hundreds, etc. when you are counting 101, or 1001 etc. Zero would be exactly the same.
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#14 tunedslick

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Posted 19 October 2007 - 09:56 PM

Okay, I think understand how the answer can be almost 100%, but never 100%. So, how is the answer 10%?
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#15 Writersblock

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Posted 20 October 2007 - 02:19 AM

The answer isn't 10%. Where does that come from :unsure:
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#16 bonanova

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Posted 20 October 2007 - 03:52 PM

To the nearest percentage, 100% of numbers contain at least one digit 3.

Guys guys, I think this math stuff is going crazy. You guys are saying it is 100% to contain a number at least one digit 3.

Well, lets look it this way. What about a number to contain 3 OR 4. According to the argument it is going to be 200%.
No, because a number can contain a 3 AND 4.
That is very funny ... because there are infinite numbers that contains neither 3 nor 4. (Hope no one will ask for an example).


An example is easy, mentioned in a previous post.

The sequence of numbers 1, 11, 111, 1111, 11111, 111111, 1111111, ...
is infinite and contains no 3's or 4's.


Still, the fraction of all numbers that contain a 3, or that contain a 4,
or that contain any specified digit is 1 - [.9]**N where N is the length
of the number. As N -> infinity, [.9]**N -> 0 and the fraction -> 1.

By the same reasoning, the fraction of all numbers that contain all the digits, 0-9, is also 1.

Mind bending, isn't it?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#17 Taidaishar

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Posted 02 November 2007 - 07:02 PM

That's sick. I think my mind just asploded...
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#18 Topher

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Posted 25 February 2008 - 10:49 PM

When dealing with limits as things get indefinitely large (as this problem effectively does) mathematicians often use phrases like except for a vanishingly small number of cases.

If you still don't "get it", imagine that you are looking at all billion digit numbers (including those starting with one or more zero). If you choose one of those numbers 10^1000000000 numbers completely at random, what is the probability that none of the billion digits in it happen to be a 3? Pretty close to zero, right? If you went to a trillion digits, the probability that none of the digits in a randomly selected number would be even smaller.

When dealing with infinite sets of possibilities, contrary to intuition a probability of zero does not mean that the thing is impossible. In fact, zero probability events occur all the time. Imagine a "uniform distribution of all real numbers in the unit interval". Don't let your eyes glaze over, this just means that you get to choose, completely at random, any number between 0 and 1, with every number just as likely as any other. Since there are an infinite number of numbers, the probability of any particular one is zero -- it would take an infinite number of random choices before you could expect any particular number to come up. Yet some number is drawn, even though the probability of drawing that number before hand was zero.
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#19 Lost in space

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Posted 25 February 2008 - 11:00 PM

Spoiler for There are no 3s in binary

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#20 Jkyle1980

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Posted 26 February 2008 - 02:04 AM

All I can do is just read and smile.
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