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Alex gets even with writersblock. Or does he?

Question

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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WB gave the method of obtaining the correct answer it in the first post.

The formula is:

f(n) = (f(n-1)*9 + 1)/10 where f(n) is the percentage of numbers out of 10^n that contain any non-zero digit.

The formula is simply a weighted average: 1/10 of 10^n will be have the digit in 100% of its numbers and the other 9/10 will have it in (f(n-1)*100)% of its numbers. (Visualize it the 10 blocks).

You can create a spreadsheet to quickly find the trend.

f(100) = 0.999973439...That means that almost 100% of all the numbers less than a googol contain at least one 3.

Note: If the question is "What percentage of ALL numbers contain at least a 3?", the answer isn't "close to but never 100%", its exactly 100%.

This is because the "ALL" makes the formula limn->∞f(n), which is 1.

While this might not make sense (because you can list out numbers that don't contain a 3), it still is the correct answer. The reason is you're trying to take a percentage of a percentage of infinity, which isn't a number.

Edited by vinays84

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For every number that you show me with a three, I will show you at least eight others that don't. If you look at it as the probability that one of the digits is not three, you must stop looking as soon as one of the digits is three, otherwise you are counting them more than once.

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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).

No, the trick here is that the percentage is approx 100 percent and not exact 100 percent.. For eg: 99 percent put of 10000 leaves around 100 numbers without 3...

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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?

it has to be 100% (almost), since we have infinite numbers and we hence have infinite nos. beginning with 3 which increases the %s one after another, till 300 % is about 0.19, due to 300-400, % becomes 0.39 and then 3000-3999, makes it as high as 46% ans so on........

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ur r right however the answer would be i gues 99.99999 something, so nearest point of % is 100 only

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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is there a number that does not contain any digit?

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Here is my solution.He picked 4, becouse he said: Hey!Not all numbers have a 3 in them.

But he lost.Why?Becouse numbers are infinite, how can you get a precentage of something you can't count?

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Here is my solution.He picked 4, becouse he said: Hey!Not all numbers have a 3 in them.

But he lost.Why?Becouse numbers are infinite, how can you get a precentage of something you can't count?

Very simple. What is the percentage of even numbers in the set of all natural numbers? The answer is 50%. Even though both sets are infinite, since one set is a subset of another set it can be expressed as a percentage of the total set. In some cases it can be calculated precisely, but in some (like this problem) it can only be approximated.

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13%

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Looking the other way round: What is the probability that a number doesn't include '3'?

90% ^ number of digits

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