Say we have a sphere with 10% of it's surface area painted in blue and the remaining 90% red. We want to show that, no matter how the sphere is painted, we can always inscribe a cube inside such that it's 8 corners (vertices) are all touching red.

Formalizing this in terms of probabilities, we operate over the space $\Omega$ of all possible sphere paintings. We know that if we randomly placed a vertex $i$, the chance of it being red is $\mathbb{P}(R_i) = 0.9$ and similarly $\mathbb{P}(B_i) = 0.1$. Thus we can state our problem as: $$ \begin{align} \mathbb{P}\left(\bigcap^8_{i=1} R_i \right) &= 1 - \mathbb{P}\left(\bigcup^8_{i=1} B_i \right) \\ &\geq 1- \sum_{i=1}^8 \mathbb{P}(B_i) \\ &= 1 - 8 \cdot 0.1 \\ &= 0.2 \\ &\neq 0 \iff \bigcap^8_{i=1} R_i \in \Omega \end{align} $$ In other words, we showed that some event (with the property of having all 8 vertices be red) is non-zero. This works because to show that some specific element $\omega \in \Omega$, it is sufficient to show that there exists a probability distribution $p$ over $\Omega$ where $p(\omega) > 0$. This follows from the fact that if $\omega \not\in \Omega$ then $p(\omega) = p(\varnothing) = 0$. This proof technique is called The Probabilistic Method and notably non-constructive: given the sphere's blue coloring, it does not show how to orient a cube, rather, it shows that there exists some orientation regardless of the exact blue locations.

Talking around, Aidan (cowboy hat) came up with another proof. Essentially, each of the eight points correspond to an octant with $100/8 = 12.5\%$ surface area. And as any $c\%$ is painted in any single octant, the other seven will be similarly constrained: there will be a corresponding $c\%$ 'implicitly painted' in all others such that placing a vertex in this area will result in at least one other vertex to be blue. Thus we can view the problem as an issue of covering an octant's $12.5\%$ surface area given $10\%$ blue -- impossible! The proof is a tad handwavy, but gives a sense of how we could collapse the octants to reveal the (minimally $2.5\%$) remaining red area and thus construct a valid orientation.