TITLE: Surface of a sphere and cube
QUESTION [6 upvotes]: I have a sphere $S^2 \in \mathbb{R}^3$ with radius 1 that is painted red on the surface (90% of it), the rest is painted blue.
Now I shall show for every configuration of a  cube that is possible in this sphere, there is always at least one configuration such that a cube with length 2 down each diagonal(that means all corners touch the sphere) has all corners on red painted points. 
My idea was the following: 
If this was wrong than at each position of the cube there would be a point that is painted blue. Since we have 8 corners, the minimum likelihood for blue would necessarily be 12.5%, such that this fails, right?
But somehow I think my argument is not very strong, as this is one that refers to a discrete configuration. But here we are talking about anuncountable number of points. maybe anybody here knows how to make my argument more rigorous, if it is not even wrong after all?
If something is still unclear, please let me know.

REPLY [3 votes]: We compute the expected value of red vertices on the inscribed cube. If this number is greater than $7$, we have solved the problem: there must exist some cube with $8$ red vertices. 
The probability a vertex is red is $.9$. There are $8$ vertices on a cube and expectation is linear, so there is an expected value of $7.2$ vertices. By the above remark, we're done.