Trying to understand the solution of the simple problem on probability

[Feynman's fan]

The problem statement:

A sphere is colored in two colors: 10% of its surface is white, the remaining part is black. Prove that there is a cube inscribed in the sphere such that all its 8 vertices are black.

And here is the textbook solution:

Choose a random inscribed cube. Then the probability that and one corner is white is 0.1 so the probability that at least one corner is white is at most 0.8. Thus the probability that none are is at least 0.2 so there must be such inscribed cubes.

Note that the the simple calculation makes use of the Boole's inequality and the reasoning itself is the Probabilistic method

And my questions are:
Where do they use that those are cubes? What breaks down if we just go through the same line of reasoning with any 8 point figure? Say, with a parallelepiped which is not as demanding as a cube? How do we know that the 0.2 is about cubes?

Any help is highly appreciated.

 

[haruspex]

Where do they use that those are cubes?

Why not say they are cubes? You're right, any object with 8 vertices that lie in the same spherical shell would do. So they might as well specify cubes.

 

[Ray Vickson]

Why not say they are cubes? You're right, any object with 8 vertices that lie in the same spherical shell would do. So they might as well specify cubes.

Isn't it possible to have a coloring scheme and a non-cubic paralellepepid in which the result is false? Say we have 5% white surfaces at the north and south poles. We can construct a long, narrow paralellepipid having all 8 corners in the white regions. Perhaps the shape can be more general than a cube, but some type of restriction need apply.

 

[PeroK]

Isn't it possible to have a coloring scheme and a non-cubic paralellepepid in which the result is false? Say we have 5% white surfaces at the north and south poles. We can construct a long, narrow paralellepipid having all 8 corners in the white regions. Perhaps the shape can be more general than a cube, but some type of restriction need apply.

As I understand it, you would simply rotate the paralellepipid so that none of its vertices were near the poles. The same argument must apply for any 8 points in a fixed configuration. The only constraint is that any of the points can be on any part of the sphere.

It's an amazing argument!

 

[haruspex]

Isn't it possible to have a coloring scheme and a non-cubic paralellepepid in which the result is false? Say we have 5% white surfaces at the north and south poles. We can construct a long, narrow paralellepipid having all 8 corners in the white regions. Perhaps the shape can be more general than a cube, but some type of restriction need apply.

You're not trying to show all positions give the object at least one black corner. You're trying to show there are some positions where all the corners are black.
The probabilistic argument makes it seem a bit magical, but it can be phrased without mentioning probability.

 

[PeroK]

I guess the full probabilistic argument goes like this:

Imagine you try all orientations of the cube. Every vertex will land once on every point of the sphere. So, each vertex will be on a black point 90% of the time, hence the probability that any given vertex will be black is 90%. So, 90% of all vertices taken over all orientations are black.

Now suppose that there are no orientations where all 8 points are black. Then, for every orientation, the probability that a given vertex is black is at most 7/8. So, taken over all orientations and all vertices the probability that a vertex is black must be at most 7/8, which is less than 90%.

I guess you could use a similar argument with the sphere divided into small areas with 10% of them white and take the limit as the size of the area goes to 0.

Can you colour the sphere in such a way that no 10-pointed configuration is all black?

Or, with 12.5% white, so that no cube is all black?

 

[haruspex]

Imagine you try all orientations of the cube. Every vertex will land once on every point of the sphere.

No, it will land on each point infinitely often.

Can you colour the sphere in such a way that no 10-pointed configuration is all black?

If you did, it would follow that, except on a set of measure zero, there's always exactly one white vertex.

 

[PeroK]

No, it will land on each point infinitely often.

Yes, of course, "land on each point with equal probablity"

 

[Ray Vickson]

As I understand it, you would simply rotate the paralellepipid so that none of its vertices were near the poles. The same argument must apply for any 8 points in a fixed configuration. The only constraint is that any of the points can be on any part of the sphere.

It's an amazing argument!

Of course: stupid of me. What was I thinking (or not).