We randomly select four points independently on the surface of a sphere. What is the probability that the tetrahedron defined by them contains the center of the sphere?
I. solution. Let us mark a point $A$ on the sphere, which we will consider fixed. Then let us select three points on the sphere's surface independently and randomly. The probability we are looking for is the same as the probability that the tetrahedron determined by these three points and $A$ contains the center of the sphere. Since the probability that any two of the four points coincide is 0, we can disregard these cases - just as we can disregard the cases where the plane defined by the three selected points passes through the center of the sphere. If $X, Y, Z$ are three points selected in this way, then the tetrahedron $A X Y Z$ contains the center $O$ of the sphere precisely when the segment $O A^{\prime}$ intersects the triangle $X Y Z$, where $A^{\prime}$ is the reflection of point $A$ over $O$.

The triples of points to be selected can be grouped into octets in the following way. Let us take three pairs of antipodal points - let these be $\left(B, B^{\prime}\right),\left(C, C^{\prime}\right),\left(D, D^{\prime}\right)$ - such that no point coincides with $A$, and the plane of $B C D$ does not pass through the center of the sphere. The convex hull of the six points thus obtained is an octahedron symmetric about the center of the sphere, with parallel pairs of faces being $\left(B C D, B^{\prime} C^{\prime} D^{\prime}\right),\left(B^{\prime} C D, B C^{\prime} D^{\prime}\right),\left(B C^{\prime} D, B^{\prime} C D^{\prime}\right)$ and $\left(B C D^{\prime}, B^{\prime} C^{\prime} D\right)$. Any of the vertices of these faces can serve as the three points to be selected, and every admissible triple of points corresponds precisely to one face of such an octahedron. The idea of the solution is based on the following observation: the random selection of the three points can be done by first randomly selecting three pairs of antipodal points, and then from each pair, selecting one of the points with a $1 / 2-1 / 2$ probability. In other words, we first randomly select a suitable octahedron, and then one of its faces, each with an equal, individual $1 / 8$ probability, whose vertices form the selected triple of points.

Consider a suitable octahedron. The line connecting point $A$ to the center $O$ of the sphere intersects two of the octahedron's parallel faces, avoiding the others. By symmetry, exactly one of these two parallel faces is intersected by the segment $O A^{\prime}$, and the other is not. Therefore, of the eight triples of points associated with the given octahedron, exactly one will be suitable. Since this can be said of any octahedron, based on our reasoning, the probability we are looking for is $1 / 8$.

Remark. The justification for the two-step selection of the three vertices may seem intuitively clear, but it actually requires proof, which goes beyond the tools of high school mathematics - hence we do not detail it (and we did not expect it from our solvers either). In this strict sense, the solution below is also not complete; however, like the previous one, it conceptually shows why the probability we are looking for is precisely $1 / 8$. It might also be worth pondering while reading how (un)different this solution is from the first.

II. solution. Using the notation from the previous solution: let the reflection of a point $X$ on the sphere over the center be denoted by $X^{\prime}$. From the points $A, B, C, D$, let us first select the first three, $A$, $B$, and $C$; the tetrahedron $A B C D$ contains (in its interior) the center of the sphere precisely when the last point to be selected, $D$, falls within the spherical triangle $A^{\prime} B^{\prime} C^{\prime}$. The (conditional) probability that we select the point $D$ correctly (given that $A, B, C$ have already been selected in some way) is thus the ratio of the surface area of the spherical triangle $A^{\prime} B^{\prime} C^{\prime}$ to the surface area of the sphere. The probability of selecting the four points correctly is then the average (expected value) of this ratio (the fraction of the surface area of any spherical triangle to the surface area of the sphere). Instead of directly calculating this, notice that the spherical triangles $A B C, A^{\prime} B C, A B^{\prime} C, A B C^{\prime}, A^{\prime} B^{\prime} C, A^{\prime} B C^{\prime}, A B^{\prime} C^{\prime}, A^{\prime} B^{\prime} C^{\prime}$ cover the surface of the sphere exactly once, so the sum of their surface areas is exactly the surface area of the sphere. Since each spherical triangle has the same expected surface area, this common value is $1 / 8$ of the surface area of the sphere.