, 2019

Solution:

Let the unit cube be placed on a $x y z$-coordinate system, with edges parallel to the $x$, $y$, $z$ axes. Suppose the three points are labeled $A$, $B$, $C$. If there exists a cube with side length $\frac{1}{2}$ and edges parallel to the edges of the unit cube that contain all three points, then there must exist a segment of length $\frac{1}{2}$ that contains all three projections of $A$, $B$, $C$ onto the $x$-axis. The same is true for the $y$- and $z$-axes. Likewise, if there exist segments of length $\frac{1}{2}$ that contain each of the projections of $A$, $B$, $C$ onto the $x$, $y$, and $z$ axes, then there must exist a unit cube of side length $\frac{1}{2}$ that contains $A$, $B$, $C$.

It is easy to see that the projection of a point onto the $x$-axis is uniform across a segment of length $1$, and that each of the dimensions are independent. The problem is therefore equivalent to finding the cube of the probability that a segment of length $\frac{1}{2}$ can cover three points chosen randomly on a segment of length $1$.

Note that selecting three numbers $p < q < r$ uniformly and independently at random from $0$ to $1$ splits the number line into four intervals. That is, we can equivalently sample four positive numbers $a, b, c, d$ uniformly satisfying $a + b + c + d = 1$ (here, we set $a = p$, $b = q - p$, $c = r - q$, $d = 1 - r$). The probability that the points $p, q, r$ all lie on a segment of length $\frac{1}{2}$ is the probability that $r - q \leq \frac{1}{2}$, or $b + c \leq \frac{1}{2}$. Since $a + d$ and $b + c$ are symmetric, we have that this probability is $\frac{1}{2}$ and our final answer is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.