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 is 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 (as opposed to a constructive proof, i.e. this hat puzzle where events 10 events of 10% likelihood are constructed in such a way to ensure mutually exclusivity, implying a zero chance of all events not occurring. .)

Talking around, My bud Aidan 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.