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# Alex gets even with writersblock. Or does he?

35 replies to this topic

### #1 bonanova

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Posted 14 October 2007 - 08:00 AM

After being done out of a pint of O'Doule's by writersblock
last night at Morty's, Alex conjured up a question calculated
to get him even.

After WB had downed his cool one, Alex proposed a double or
nothing puzzle. To the nearest percentage point, he asked,
what percentage of all numbers contain at least one 3?

For example, 13, 31, 33 and 103 all contain the digit 3 at least once.
But 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ... well
you get the idea ... don't contain the digit 3 even once.

Now I know there's a lot of numbers to check, so I'll make it
easy for ya, Alex said. I'll give you five multiple choices.
That gives you a 20% chance even if you guess, and much
better odds if you yer' the genius ya' make yerself out to be.

The percent of all numbers containing at least a single "3" is:

[1] 10%
[2] 13%
[3] 33%
[4] 67%
[5] 100%

What was writersblock's choice, and did he win another pint?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #2 Writersblock

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Posted 14 October 2007 - 09:13 AM

Being slightly tipsy from the nice cool beverage he just won, Writersblock thinks for a second. First he wonders how it can be double or nothing after he already has downed his pint. It can't be nothing if he's already consumed something, right? That confuses him a little and then he thinks on the numbers. At first brush he considers it couldn't possibly be 100% so throw that out. Then he thinks a little more. He smiles a bit and answers:

Spoiler for solution

IF that isn't right, I blame it entirely on the frosty pint I just consumed.
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### #3 bonanova

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Posted 14 October 2007 - 11:50 AM

Alex slides another cold one down the bar to WB.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #4 unreality

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Posted 18 October 2007 - 01:23 AM

Not sure if this is relevant but just say, for example, 10% of infinity, ?/10, is still ? but a differnet level of infinity correct?

like

(?/10) / ?

it wouldnt be ?/? as in 1 as in 100%.... it would be .1 as in 10%

Am I just shootin in the dark here or do I have a slembance of a point? Help me out here lol
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### #5 tunedslick

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Posted 18 October 2007 - 01:31 AM

But, the question was what percentage of ALL numbers consists of a number 3? Not what percentage of sets. I thought it was 33% because:

3 is in 1 to 10, 10% of the time. (3)
3 is in 1 to 100, 19% of the time. (3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93)
3 is in 1 to 1000, 26% of the time because every 300th number?

Am I making sense or completely off? Doesn't there need to be a range?
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### #6 unreality

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Posted 18 October 2007 - 01:34 AM

Yeah, look at your numbers. It increases. The percent will increase as the total number goes up. That's what they're saying. But I semi-disagree, if you look at my previous post.
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### #7 bonanova

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Posted 18 October 2007 - 08:25 AM

But, the question was what percentage of ALL numbers consists of a number 3? Not what percentage of sets. I thought it was 33% because:

3 is in 1 to 10, 10% of the time. (3)
3 is in 1 to 100, 19% of the time. (3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93)
3 is in 1 to 1000, 26% of the time because every 300th number?

Am I making sense or completely off? Doesn't there need to be a range?

The OP gives the range as all numbers.

As the number of digits in the number increases, the likelihood that it contains a three increases.
In fact the fraction of N-digit numbers that contain a 3 is 1 - [.9]**N.
Note that [.9]**N becomes 0 at infinite N.

Spoiler for solution

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #8 brhan

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Posted 18 October 2007 - 02:08 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%. That is very funny ... because there are infinite numbers that contains neither 3 nor 4. (Hope no one will ask for an example).
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### #9 Writersblock

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Posted 19 October 2007 - 01:33 AM

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.
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### #10 unreality

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Posted 19 October 2007 - 05:07 AM

strengths of infinity?

Not sure if this is relevant but just say, for example, 10% of infinity, ?/10, is still ? but a differnet level of infinity correct?

like

(?/10) / ?

it wouldnt be ?/? as in 1 as in 100%.... it would be .1 as in 10%

Am I just shootin in the dark here or do I have a slembance of a point? Help me out here lol

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