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PostPosted: Sat Jul 31, 2010 00:50 
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The way I see it, is that anything less than 1 cubed equals less than 1, anything over 1 cubed equals greater than 1. There are equal numbers between 0 and 1 to 1 and 2 so the percentage of numbers cubed below 1 (including 1) which will equal 1 or less is 50% and the percentage of numbers between 1 and 2 which will equal a cubic volume between 1 and 2 is also 50%.

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PostPosted: Sat Jul 31, 2010 08:39 
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8 cubic centimeters is a cube with 2 cm per side.

A cube of 1 cubic cm is a cube with 1 cm per side.

The answer is 50%

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PostPosted: Sat Jul 31, 2010 09:27 
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graball wrote:
The way I see it, is that anything less than 1 cubed equals less than 1, anything over 1 cubed equals greater than 1. There are equal numbers between 0 and 1 to 1 and 2 so the percentage of numbers cubed below 1 (including 1) which will equal 1 or less is 50% and the percentage of numbers between 1 and 2 which will equal a cubic volume between 1 and 2 is also 50%.


But the number of numbers between 0 and 0.1 is equal to the number of numbers between 1 and 2 and is also equal to the number of numbers between 1 and five thousand million: its is \aleph_1 - the cardinality of the set of all countable ordinal numbers. As an infinity is not a number doing arithmetic with it, as you do, can lead to all kinds of contradictions, such as the famous proof that 1 = 0.

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PostPosted: Sat Jul 31, 2010 09:36 
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Big Tone wrote:
8 cubic centimeters is a cube with 2 cm per side.
A cube of 1 cubic cm is a cube with 1 cm per side.
The answer is 50%


8 cubic centimeters is a cube with face area 4 cm^2.
A cube of 1 cubic cm is a cube with face area 1 cm^2
The answer is 25%
:lol:

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PostPosted: Sat Jul 31, 2010 09:57 
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Read the question again. What is the PROBABILITY of its VOLUME being 1cc or less?

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PostPosted: Sat Jul 31, 2010 10:03 
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graball wrote:
Read the question again. What is the PROBABILITY of its VOLUME being 1cc or less?
As long as dcb doesn't get it now, I'm happy :lol:

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PostPosted: Sat Jul 31, 2010 10:07 
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dcbwhaley wrote:
The answer is 25%

I thought Pete already said the answer is 50%. Grabby got it right. We're looking for a paradox now aren't we? :?

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Last edited by Big Tone on Sat Jul 31, 2010 10:08, edited 1 time in total.

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PostPosted: Sat Jul 31, 2010 10:29 
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Big Tone wrote:
I thought Pete already said the answer is 50%. Grabby got it right. We're looking for a paradox now aren't we? :?


I don't think that Pete has given an answer yet. And there are no mathematical paradoxes: just bad mathematicians :)

The flaw with your reasoning is that the problem is stated in terms of volume not edge length. And there are seven times as many possible volumes for cubes with edges between one and two than for cubes with edges less than one.

I have just done a simulation on this problem and, with a sample size of 10^7, get answers between 12.48% and 12.51%

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PostPosted: Sat Jul 31, 2010 10:50 
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dcbwhaley wrote:
Big Tone wrote:
I thought Pete already said the answer is 50%. Grabby got it right. We're looking for a paradox now aren't we? :?
I don't think that Pete has given an answer yet.


Pete317 wrote:
Grabby is right, it is 50% (or 0,5 for the pedantic). The unit's don't matter.
In answer to Steve, it doesn't matter whether the "size" is in volume or length.
But can someone explain this apparent paradox?


Keep up dcb ;) :D

dcbwhaley wrote:
And there are no mathematical paradoxes: just bad mathematicians :)
:whome:

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PostPosted: Sat Jul 31, 2010 11:58 
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Pete317 wrote:
Grabby is right, it is 50% (or 0,5 for the pedantic). The unit's don't matter.
In answer to Steve, it doesn't matter whether the "size" is in volume or length.
But can someone explain this apparent paradox?


Of course the units matter. Volume is length cubed. Area is length squared.

If the question is about edge length the answer is 1/2 50%
If the question is about face area the answer is 1/2^2 25%
If the question is about volume the answer is 1/2^3 12.5%.

What was the question again?

If you generate 1000 random numbers in the range 0 to 8, representing a 1000 random cubes you wil find ~125 of them are less than one.

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PostPosted: Sat Jul 31, 2010 12:32 
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Ok, it seems that I came close to setting off WWIII here :twisted: so here goes:

If the sides of the cubes are in integer units of 1cm, the only two possible cubes are 1 and 8 cu cm. You cannot have, for example, a 5 cu cm cube,
If the sides are in mm, you can have 20 possible cubes, with 20 possible volumes.
The same holds true for whatever units you choose, right down to units of one atom, or molecule, or whatever the cubes are made of. You cannot have a cube consisting of 4 atoms, for example. (you could go to even smaller units theoretically, but not physically)
So, the number of possible cubes of less than or equal to 1 cu cm is equal to the number of possible cubes equal to or less than 8 cu cm but greater than 1 cu cm.

I know I'm not very good at explaining things, but I hope this is clear enough :)

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PostPosted: Sat Jul 31, 2010 12:42 
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However, the probability side of the problem is lacking. You seem to be assuming that all distributions are uniform.

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PostPosted: Sat Jul 31, 2010 12:52 
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When I said random I assumed a uniform distribution.
The thing is, you can't have a uniform distribution of volumes. If you have a cube with sides of a trillion trillion light years, and you remove just one atom from it, you no longer have a cube.

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PostPosted: Sat Jul 31, 2010 13:41 
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Pete317 wrote:
When I said random I assumed a uniform distribution.
The thing is, you can't have a uniform distribution of volumes. If you have a cube with sides of a trillion trillion light years, and you remove just one atom from it, you no longer have a cube.


Random does not mean uniform, so add the words "uniform distribution" next time. And you can have a uniform distribution of volumes.

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PostPosted: Sat Jul 31, 2010 13:45 
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Here is a new question, and here I am assuming a uniform distribution.

Take a finite line element (or thin rod, say), and cut it in two random places, assuming independence, and find the probability that you can make a triangle out of the pieces.

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PostPosted: Sat Jul 31, 2010 14:09 
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Pete317 wrote:
If the sides of the cubes are in integer units of 1cm, the only two possible cubes are 1 and 8 cu cm. You cannot have, for example, a 5 cu cm cube,
If the sides are in mm, you can have 20 possible cubes, with 20 possible volumes.
The same holds true for whatever units you choose, right down to units of one atom, or molecule, or whatever the cubes are made of. You cannot have a cube consisting of 4 atoms, for example. (you could go to even smaller units theoretically, but not physically)
So, the number of possible cubes of less than or equal to 1 cu cm is equal to the number of possible cubes equal to or less than 8 cu cm but greater than 1 cu cm.

Finite elements are a subset of continuous variables. I assumed a continuous variable (the length) as nothing was specified, simply because it is the way of the world (until you reach the level of quantum mechanics). So I would say my conclusion was correct and for the right reason - but I would say that :P

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PostPosted: Sat Jul 31, 2010 14:13 
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trakgalvis wrote:
Here is a new question, and here I am assuming a uniform distribution.

Take a finite line element (or thin rod, say), and cut it in two random places, assuming independence, and find the probability that you can make a triangle out of the pieces.

Can this be reduced to "what is the probability that the sum of two numbers is equal or greater than a single number (all from a uniform distribution)"?

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PostPosted: Sat Jul 31, 2010 14:15 
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Steve wrote:
trakgalvis wrote:
Here is a new question, and here I am assuming a uniform distribution.

Take a finite line element (or thin rod, say), and cut it in two random places, assuming independence, and find the probability that you can make a triangle out of the pieces.

Can this be reduced to "what is the probability that the sum of two numbers is equal or greater than a single number (all from a uniform distribution)"?


I will tell you, when you provide an answer! :) Remember the two cuts are in the same line element!

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PostPosted: Sat Jul 31, 2010 14:20 
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Steve wrote:
Finite elements are a subset of continuous variables.


That's one way of looking at it.
Another way is that no matter how many times you subdivide integral units, you always end up with yet smaller integral units.
Continuous variables imply infinity - which only exists in the abstract.

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PostPosted: Sat Jul 31, 2010 14:32 
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trakgalvis wrote:
Take a finite line element (or thin rod, say), and cut it in two random places, assuming independence, and find the probability that you can make a triangle out of the pieces.


Given that any side of the triangle cannot be greater than the sum of the other two sides, I'd say that the probability is 0.5

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