[bonanova]

With a nod to

There is a unit cube in 3-space.

A unit sphere (r=1) is placed randomly, so that its center is somewhere inside the unit cube.

Every point inside the unit cube has the same probability of being chosen as the center of the unit sphere.

What is the probability that the cube fits entirely inside the sphere?

Which probability is intuitively greater: the cube fitting inside the sphere, or the square fitting inside the circle?

[k-man]

Just saw this, so without doing any math here is the answer to the intuitive question

The probability in 3-d space will be lower than that on the plane.

[Guest]

Intuitively, the probability calculated for 2-D (31.51%) should be the same for a cylinder of diameter 1 and length 1 inside the sphere.

Since we are here talking about a cube of side 1 the probability should be lower.

Since the ratio of areas of such a cylinder and a cude is 4/pi, the probability should be that much smaller.

The new probability should be 31.51%*pi/4 = 24.74%

Let me know if this is correct pls.

Edited May 9, 2011 by DeeGee

[k-man]

The probability that the cube will fit in the sphere is the volume of the shape near the center of the cube formed by the intersection of 8 unit spheres centered in the corners of the cube. This shape resembles the octahedron, but it has curved edges and faces. Approximating it down to the octahedron we can calculate the length of the edge to be 1-1/sqrt(2) or ≈0.2928932. The volume of the octahedron is then 5/(3*sqrt(2))-7/6 or ≈0.01184463531.

Accounting for the spherical faces and round edges is something I haven't tried yet, but I figure it should amount to no more than 10% of the volume of the octahedron, so the final answer is somewhere around 1.3%.

[superprismatic]

I couldn't see an easy analytical solution to this,

so I wrote a little Monte Carlo program to estimate

the probability. Using 10,000,000,000 trials, my

estimate is 0.0152.

In 4 dimensions, there is only one point in the

hypercube which can be the center of such hypersphere

which contains it. This is easy to see because the

corners of the unit hypercube are at (±.5,±.5,±.5,±.5)

which are hit precisely by the hypersphere of radius

1 and centered at the origin. Any wiggle of the center

away from the origin will miss some corners. So the

probability is 0.

In dimension D>4, there are 2^D corners each of which

are .5×√D away from the origin and hence, outside the

unit hypersphere (because .5×√D>1 when D>4) centered

at the origin. So the probability is 0 here as well.

For completeness, we should include dimension 1 where

the unit cube is a line segment of length 1 and the

unit sphere is a line segment of length 2. Here, the

probability is 1.

So, as the dimension increases from 1, the probabilities

are 1,~.315,~.015,0,0,0,...

[EventHorizon]

After quite a few calculus mistakes, I stumbled on the right answer. (stupid superfluous negations causing negative probabilities...)

First, picture the unit cube right in the center of the sphere. Now extend the faces and cut the sphere into 27 pieces... sort of like the individual parts that make up a Rubik's cube. The answer we are looking for is the sum of the volumes of the 8 corner pieces (think about why... and ask if you can't figure it out).

If you take the whole sphere, subtract three times the volume of the disk that makes up the middle layer (center piece, 4 middle of side pieces, and 4 edge pieces), add three times the volume of the center pillar (center piece and two middle of side pieces), and subtract the center piece... you get the desired volume.

The volume of the sphere is easy... 4*pi/3.

The volume of the disk that makes up the middle layer is the sphere volume minus two caps of height 1/2. = 4pi/3 -2* (5pi/12) = 11*pi/12.

The volume of the center piece is simply the volume of the unit cube.... 1.

This leaves the volume of the pillar in the center... yay for calculus.

Given a circle of radius r, you can calculate it's area by integrating 2*sqrt(r^2-x^2) from -r to r. &random=false]Example using r=e

Instead, integrate that from -.5 to .5. +%2B+2*%281-x^2%29+*+ArcSin[1%2F%282*Sqrt[1-x^2]%29]&random=false]Then we want to integrate that (replacing r with sqrt(1-z^2)) from z=-.5 to z=.5.

The result is that the volume of the pillar is (1/3)*( sqrt(2) + 15*arcsin(1/sqrt(3)) - 4*arctan(5/sqrt(2))).

Putting it all together we get that the desired volume is sqrt(2) + 15*arcsin(1/sqrt(3)) - 4*arctan(5/sqrt(2)) - 1 - 17*pi/12, which is about 0.01520549

On a side note, I would love to see an easier way to solve this one.