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#1
01-07-2006, 04:49 PM
 Capt. Ridley's Shooting Party Guest Join Date: Jul 2003
Probability question.

Whilst driving back to university today, I thought about the following:

Suppose that we have a perfect sphere, of unit radius, as well as an infinitely thin knife. We take the sphere and randomly cut it with the knife until we have two volumes. How could we work out the probability of the smallest volume being a specific volume prior to our cutting?
#2
01-07-2006, 04:56 PM
 Xema Guest Join Date: Mar 2002
You'll probably need to specify what "random" means in this context.
#3
01-07-2006, 04:58 PM
 Big Mony Mony Guest Join Date: Jan 2006
The probability of either volume having a specific volume is 0. This is because there is an infinite number of possible cuts giving infinitely many end volumes. You are essentially picking a truly random real number between 0 and 1, and your chance of getting a specific number is 0 (because there are infinitely many).

I'm thinking here of one cut that goes all the way through the sphere, like a plane intersecting it where the plane is randomly chosen from all possible planes intersecting the sphere. Perhaps you had something else in mind.
#4
01-07-2006, 05:04 PM
 ultrafilter Guest Join Date: May 2001
Quote:
 Originally Posted by Xema You'll probably need to specify what "random" means in this context.
Doesn't matter. As Big Mony Mony points out, you're choosing a single point from a set with Lebesgue measure, and it therefore has probability 0.

Now, if the question were changed to ask the probability of the volume being less than a certain amount, we'd need to know the distribution of cuts.
#5
01-07-2006, 05:15 PM
 Capt. Ridley's Shooting Party Guest Join Date: Jul 2003
What's a Lebesgue measure and what is its relevance to the problem?
#6
01-07-2006, 07:49 PM
 David Simmons Charter Member Join Date: Nov 2001 Posts: 12,684
As has been said, the probability of slicing the sphere at any particular place that is selected in advance is zero. The best you can do is give a probability that the volume of the piece will be between two numbers.

For example if you have a sphere of 10 in3 the probability that the piece will have a volume of 0 ≤ volume < 1 in3 is 1/10.
#7
01-07-2006, 10:15 PM
 Saint Cad Guest Join Date: Jul 2005
What would be really cool is to use the Banach-Tarski Paradox and make the sphere as big as the Earth by cutting and reassembly.

To the OP, the probability is 0. Think of it this way, I'm thinking of a number between 0 and 1. What is it?

SPOILER:
(0.947372190452305612 x sqrt(2)) / (e + 2pi)

Did you get it? Do you think that you'd have any chance of getting it?

I know what you're saying. Doesn't a probability of 0 mean the event is utterly impossible? Yes - if you are dealing with a finite number of possible events. If you are dealing with an infinite number of events, then you need to use Lesbegue measure.
#8
01-07-2006, 10:26 PM
 wolf_meister Charter Member Join Date: May 2003 Location: Where the owls say "Whom" Posts: 5,356
SPOILER:

(0.9473721904523056119 x sqrt(2)) / (e + 2pi)

Geez my guess was close but not close enough. Damn
#9
01-07-2006, 10:41 PM
 Saint Cad Guest Join Date: Jul 2005
Quote:
 Originally Posted by David Simmons As has been said, the probability of slicing the sphere at any particular place that is selected in advance is zero. The best you can do is give a probability that the volume of the piece will be between two numbers. For example if you have a sphere of 10 in3 the probability that the piece will have a volume of 0 ≤ volume < 1 in3 is 1/10.

I don't think so. Lay the sphere on a number line perpendicular to the cut so that the number line is the diameter and goes from 0 to 2. (I know these numbers are different from yours but I'm interested in the general principle. the final result will be the same by scaling). We want to find the point u on the number line such that the integral from 0 to u of pi (1 - x[sup]2)dx [if i remember my integral calculus correctly] equals 4pi/3 x 1/10 = 2pi/15.

So (u - (1/3)u[sup]3) x pi = 2pi/15 or 15u - 5u[sup]3 = 2
A numerical analysis indicates that 0.134<u<0.135
by symmetry this gives us a second possibility that cut is at the other end near 2(remember we are looking at the smaller volume).

After some handwaving, we conclude that the probability that the cut will give us 1/10 or less of a volume is slightly less than 13.5%
#10
01-07-2006, 11:23 PM
 David Simmons Charter Member Join Date: Nov 2001 Posts: 12,684
Quote:
 Originally Posted by SaintCad I don't think so. Lay the sphere on a number line perpendicular to the cut so that the number line is the diameter and goes from 0 to 2. (I know these numbers are different from yours but I'm interested in the general principle. the final result will be the same by scaling). We want to find the point u on the number line such that the integral from 0 to u of pi (1 - x[sup]2)dx [if i remember my integral calculus correctly] equals 4pi/3 x 1/10 = 2pi/15. So (u - (1/3)u[sup]3) x pi = 2pi/15 or 15u - 5u[sup]3 = 2 A numerical analysis indicates that 0.134
Different problem. I postulated that the cut has a volume of 1 in3. You have set up a problem that has cuts with different volumes.
#11
01-07-2006, 11:33 PM
 David Simmons Charter Member Join Date: Nov 2001 Posts: 12,684
Make that volume 1 in3

On further thought I'm not sure of my answer. It does seem to me that if you randomly slice out a piece of a sphere of 10 in3 volume the chance that it will be between 0 and 1 is 1/10. But I've been wrong before.
#12
01-07-2006, 11:49 PM
 Saint Cad Guest Join Date: Jul 2005
Quote:
 Originally Posted by David Simmons Different problem. I postulated that the cut has a volume of 1 in3. You have set up a problem that has cuts with different volumes.

No, same problem. What I showed was that to make a cut so that the smaller section has a volume 1/10 that of the original, the cut must take off (approximately) 13.5% of the radius perpendicular to the cut. I don't know why you assume that this solution is not scalable to spheres of any size. I just solved the general case as opposed to your specific case.
#13
01-08-2006, 12:00 AM
 David Simmons Charter Member Join Date: Nov 2001 Posts: 12,684
Quote:
 Originally Posted by SaintCad No, same problem. What I showed was that to make a cut so that the smaller section has a volume 1/10 that of the original, the cut must take off (approximately) 13.5% of the radius perpendicular to the cut. I don't know why you assume that this solution is not scalable to spheres of any size. I just solved the general case as opposed to your specific case.
I see that you are saying that when you randomly cut a sphere you are really randomly picking a point on the diameter. And a random point on the diameter has an equal chance of being in any equal increment of that diameter and different cuts will result in different volumes.

So it looks like my example was in error. Sorry.

I guess what should be said is that you have zero chance of cutting any particular volume and the best you can do is compute the chance of cutting some volume that lies within a range of volumes.
#14
01-08-2006, 12:20 AM
 Saint Cad Guest Join Date: Jul 2005
I did make a small error - not conceptual but mathematical.

The formula X^2 + Y^2 = R^2 gives a circle centered at the origin not at (1,0)

the formula thus would be i.e. (u - (1/3)u^3)pi + (2/3)pi = 2pi/15
the (4/3)pi takes into account that the lower bound is -1 and not 0

so NOW solving for u in: u - (1/3)u^3 = -8/15 gives a u of -0.608 which translates into a probability of 39.2%

Now to test this, I decided to see where the cut point (i.e. value of u) would be if we wanted to cut the sphere in half. We know that it should be 0

(u - (1/3)u^3)pi + (2/3)pi = (4pi/3) x (1/2)
(u - (1/3)u^3)pi + (2/3)pi = 2pi/3
(u - (1/3)u^3) + (2/3) = 2/3
u - (1/3)u^3 = 2/3 - 2/3
u - (1/3)u^3 = 0
u(1 - (1/3)u^2) = 0
Notice that there is only one solution between -1 and 1: u=0 (the other solutions are +/- sqrt(3) )
#15
01-08-2006, 12:21 AM
 Saint Cad Guest Join Date: Jul 2005
Quote:
 Originally Posted by SaintCad the formula thus would be i.e. (u - (1/3)u^3)pi + (2/3)pi = 2pi/15 the (4/3)pi takes into account that the lower bound is -1 and not 0

Make that (2/3)pi
#16
01-08-2006, 05:44 AM
 Capt. Ridley's Shooting Party Guest Join Date: Jul 2003
Can someone explain what a Lebesgue measure is and it's relevance to this problem? I've never heard of that before.
#17
01-08-2006, 06:29 AM
 Saint Cad Guest Join Date: Jul 2005
Quote:
 Originally Posted by Dominic Mulligan Can someone explain what a Lebesgue measure is and it's relevance to this problem? I've never heard of that before.
It is a way to measure the total length of a set of points.

What does that have to do with probability? Look at it this way, the probability of an event happening is:
(number of sucessful events)/(number of events)
If these are whole numbers then there's no problem - you get a fraction between 0 and 1 inclusive.

But what happens if I have an infinite number of total events? Pick a number between 0 and 1; there's an infinite number of those so we need another way of calculating probability. The obvious way is to look at intervals or lengths. For example, pick a number from 0 to 1 - what is the probability that it is bigger than 0.6? Well, look at the length of the interval 0.6 to 1 (0.4) and the total length from 0 to 1 (1.0). The probability is 0.4/1.0 or 40%.

So far so good . . . but what if I want a weird set of points, like the probability of randomly picking a rational number from the interval 0 to 1? What is the length of all of the rational points put together? It is here that Lebesgue measure is needed. The Lebesgue measure of the rational numbers is 0. What's nice is that the Lebesgue measure of an interval is just its length, so the Lebesgue measure of the interval (0, 1) is 1. So the probability of picking a rational number between 0 and 1 is 0/1 = 0
#18
01-08-2006, 06:34 AM
 Capt. Ridley's Shooting Party Guest Join Date: Jul 2003
Thanks.

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