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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?

## 35 answers to this question

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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:

To the nearest percentage, 100% of numbers contain at least one digit 3. I figure it this way. In a set of 10, only 3 has a 3. That is 10%. But if you jump to 100, then all the sets of ten have a single three except for the thirty set. This would have 10 3's. So in 100, we have 19 or 19%. If you jump to 1000, then the 300's each have a three, as do the thirty set for each of the other 100 sets, etc. So about 29% of 1000 numbers have a 3. Jump to 100 million and we are about 79 percent or so. Therefore as we approach infinity, I assume this number will jump each time we set the base set to another power of 10. Eventually at really large sets we'll be at some number approaching but never reaching 100%, but the nearest percentage will be 100%.

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

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Alex slides another cold one down the bar to WB.

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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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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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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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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.

Fraction of 1-digit numbers with a 3 is 1-.9 = .1

Fraction of 2-digit numbers with a 3 is 1-.81 = .19

Fraction of 3-digit numbers with a 3 is 1-.729 = .271

Fraction of 4-digit numbers with a 3 is 1-.6561 = .3439

...

Fraction of 10-digit numbers with a 3 is 1-.3486784401 = .6513215599

..

Fraction of 44-digit numbers with a 3 is 1-.00969773... = .99030227

..

Fraction of infinite-digit numbers with a 3 is 1-.0000000000... = 0.9999999999.... -> 1.0

Note there are an order of magnitude [multiple of 10] more numbers in each successive group,

outweighing the combined numbers in all of the previous groups.

It doesn't mean that all numbers with infinite digits contain a 3! Consider the sequence 1, 11, 111, 1111, 11111, ...

It just means that for every infinite-digited number that does not contain a 3 there are an infinite number that do.

Finally, the same percentages hold for numbers containing a 2, 4, or 5, ... as well.

So the fraction of all numbers that contain a 0, 1, 2, 3, 4, 5, 6, 7, 8 and a 9 is ... 1.

Which means, if you believe in talking about percentages of infinite groups,

that ... all numbers contain all ten digits.

Even tho there are an infinite number of numbers that don't.

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

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The answer isn't 10%. Where does that come from

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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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That's sick. I think my mind just asploded...

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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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unless????????

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All I can do is just read and smile.

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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

Oh, you have ventured into the Land of Indeterminate Forms. Consult our guide, Monsieur L'HÃ´pital.

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Oh, you have ventured into the Land of Indeterminate Forms. Consult our guide, Monsieur L'HÃ´pital.

Why is it indeterminate?

Is 10% of infinity not 10% just the same?

What percent of all integers are even? Is that indeterminate?

Alex gave an understandable and meaningful set of choices ... which one is right?

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The chance of geting a number that contains a 3 is better the larger your number is becuse you have more numbers containing three this increase in percentage stops at 50/50 becuse there are an infinite numbers that contain a3 and an infinite number of numbers that do not there for they are even. The problem with infinite is you just keep going till you reach the solution you want.

Edited by johnnyleonclark

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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?

The series this forms is

Term 1 = 10^0

Term 2 = 10^1 + (Term1 * 9)

Term 3 = 10^2 + (Term2 * 9)

Term 4 = 10^3 + (Term3 * 9)

so on and so forth... the percentage would go pretty close to 100% i think..... for any single digit.. not just 3

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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.

Except zero? I think this riddle could be made with any digit, including 0. Once you get to really big numbers it is statisticly very difficult to get numbers without any particular digit. We are dealing with infinity so you need to be able to think in big numbers. After all, lets say a 1000000000 digit number is still small when talking infinity.

To kind of prove the point I made a much smaller scenario. I made a simple quickbasic program (I never thought I was going to use QB again!) in which I would get a point for every 10000 random numbers from 0 to 9 in which no 0s were created. It ran all night without a point. It got plenty of points on the test runs with numbers that were 20 and 30 digits but once you get to higher numbers is statisticly impossible to get a number without a 3 even though like a poster says there are infinite numbers without any particular digit. Now, if it ran all night without any points, imagine the odds with 10000000000000000 digits or much higher. We are dealing with infinity so it get really really close to 100% fairly fast.

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Okay, maybe I am missing something really important because I think the answer is 20%. Or, of the answers given, 13%. From 1 to 10 it is 10%. From 1 to 100 it is 19%. From 1 to 1000, it is 19.9%. From 1 to 10000, it is 19.99%

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