Riddles Round 2

[Legion]

OK Firemeboy is up in the new thread.........

[Firemeboy]

Ok, here is a fun one.

What percentage of all real numbers contain the number three?

[Eric James Stone]

Well, since it involves dividing one infinite (number of real numbers contain 3 as a digit) by another (number of real numbers), and the infinites are of the same order, the answer is technically undefined.

But, in a less rigorously mathematical way of looking at things, you could say that almost all real numbers contain 3 as at least one of their digits.

As an approximation to show this, we can look at percentages of real numbers that do not have 3 at a particular spot.

In the 1's column, 90% of real numbers do not have a 3.
Of that 90%, 90% do not have a 3 in the 10s column.
Of that 90% of 90%, 90% do not have a 3 in the 100s column.
So for each additional column that you examine (whether in front of the decimal point or after) the pool of real numbers without a 3 in them is reduced to 90% of what it was.

It takes only 8 digits to get below 50%.  At 23 digits, you're under 10%.  At 45 digits, you're under 1%.  And as the number of digits goes out to infinity, the percentage of numbers that do not have a 3 in them becomes ever smaller.

[Firemeboy]

So...  what is your answer?  

You are right.  If we don't deal with technicalities, the answer is "All of them".  100% of real number contain the number three...

But you won't find a mathemetician who is willing to back that up.  But technically you can prove that all numbers contain a three, and you can also prove that all numbers contain an 8.  

[Legion]

so then 1 being a real number contains a 3 in.....or are you saything that since there is an infinite amount of numbers then the fact that 1 does not contain a 3 is to small to relize, but then you have to think about all the other numbers that do not contain 3.....which yes is infinite, so there is no possible way that 100% of the numbers contain a 3.  In fact the highest I think you could go would be 10% if that.

[Eric James Stone]

If we don't deal with technicalities, the answer is "All of them".  100% of real number contain the number three...

... But technically you can prove that all numbers contain a three, and you can also prove that all numbers contain an 8.

I'm afraid I don't quite get this.  It is not possible to prove something is true for all (100%) members of a group if it is possible to find even one counter-example.  Even if the percentage of real numbers that do not contain a 3 is infinitely small, the set of real numbers that do not contain a 3 is not null, and therefore it cannot be said that all real numbers contain a 3.

Unless I'm missing something, of course, and there's some sort of wordplay involved that's just going over my head.

[Eric James Stone]

In fact the highest I think you could go would be 10% if that.

No, it's quite easy to go over 10%.  Just consider the numbers from 0 to 99:

0 1 2 4 5 6 7 8 9
10 11 12 14 15 16 17 18 19
20 21 22 24 25 26 27 28 29
40 41 42 44 45 46 47 48 49
50 51 52 54 55 56 57 58 59
60 61 62 64 65 66 67 68 69
70 71 72 74 75 76 77 78 79
80 81 82 84 85 86 87 88 89
90 91 92 94 95 96 97 98 99

3 13 23 30 31 32 33 34 34
36 37 38 39 43 53 63 73 83
93

19 out of 100 numbers have a 3 in them, which is 19%.  And the percentage keeps getting higher as you add more digits.

[Firemeboy]

19 out of 100 numbers have a 3 in them, which is 19%.  And the percentage keeps getting higher as you add more digits.

 

It's a mathematical game.  You are right, the higher you get, the more numbers that contain 3 you find.  Since infinity has no end, you can argue that the percentage of numbers that has three in them must also be rising, and eventually you will reach %100.

But not really.  Percentage deals with a finite number, and infinity, as the name implies, is not finite.  So it's like an unmovable object meeting an unstoppable force.  You can't have both of them in the same universe.  They are incompatible.

I guess I shouldn't have use the word 'prove' so loosely.  You cetainly can't prove it, but you could argue it.

[Eric James Stone]

You cetainly can't prove it, but you could argue it.

Oh, yes.  I could argue just about anything.

[Firemeboy]

If you wanted to show this mathematically, you could say that the percentage of numbers with a 3 in them can be expressed as 1 - (.9)^n where n equals the number of digits.  When you hit about 42 digits, 99 percent of the numbers have a three in them.

So, if you have 1-(.9)^(infinity), then you would have 1, or one hundred percent.

[Legion]

Yes the percents increase from10% to 19% to 27.1% when you go from 10 numbers to 100 number to 1,000 numbers and it increases again when you go to 10,000 (I am not doing the math to figure that out sorry guys).  But every time it increase the amount it increases get smaller, by the time it reaches the amount it is increasing is going down exponentially.  It starts off with big jumps but will soon get to so small of a jump that it is uncalcuable.  There for you can never really make it to 100%.

[Legion]

If you wanted to show this mathematically, you could say that the percentage of numbers with a 3 in them can be expressed as 1 - (.9)^n where n equals the number of digits.  When you hit about 42 digits, 99 percent of the numbers have a three in them.

So, if you have 1-(.9)^(infinity), then you would have 1, or one hundred percent.

This is very similar math thought that is behind the fact that you can prove 1 is equal to 1 or that 1/3 + 2/3 does not equal 1.  It has a lot to do with rounding errors and generalizing

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

you would argue it and lose quite badly. There are several ways to disprove it. The least of which is showing the number 1.

A percentage is an expression of a comparison, a way to show a decimal or a fraction. You can have a decimal following a percentage, so you could have 99.9999 (repeating). You can have 100% if you round, but that's it.

The percentage of 3 in them increases as you increase the number of digits true, but the rate at which it increases DEcreases. Of one-digit number (0-9) 10% have three. Of 2 digit numbers, 19 percent do. Of three digit numbers, it's (i think) 27.1%. so jumping from to 1 to 2 digit numbers, the rate increases by 9%, but only increases by 8.1% when you go from 2 to 3 digit numbers. Thus you can get Ifinitely close to 100%, but you will never actually reach it.

Draw it graphically, it makes a curve, always bending away from 100% as it gets closer.

[Firemeboy]

Actually, .99999999 (ad infinitum) = 1.  With no rounding needed.

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

0.9 repeating may be as close to 1 as to make no difference in any real world calculation, but it is *not* actually 1.