Suppose we have a sphere and more than a half of its surface is red. Prove or disprove that we can place all vertices of a regular hexagon on this sphere and at least four vertices of it will be placed on red surface.
It depends on what you mean by "half". Probably the simplest definition is that P(A random point turns out to be red) > 50%.
Consider the expected number of red vertices on a random hexagon on the sphere. By properties of expectation - in particular, expected value is additive, even for nonindependent events - the expected number of red vertices is
P(vertex one is red) + P(vertex two is red) + ... + P(vertex six is red) > 50% + 50% + ... + 50% > 3.
If every hexagon had three or less red vertices, the expected value would be less than three. However, it's more than three, so there must indeed exist a hexagon with four red vertices.