Suppose that $A \subset \mathbb{R}^n$ is a random linear subspace of dimension $k < n$. I am interested in the event that $A$ intersects the hypercube $[-1,\ 1]^n$ at a specific face.
In other words, the event that there is some vector $a\in A$ with particular configuration of $r$ entries equal to $1$ and $s$ entries equal to $-1$ (the remaining entries are all in the range $(-1, 1)$) with $r+s\leq k$.
In particular, I am interested in calculating or bounding the probability of such an event. The distribution of $A$ can be chosen as any continuous distribution that makes this tractable/easy.
I would appreciate any pointers on this as I'm unsure where to start. thanks :)
You can skip ahead, past the break, for the condensed argument and conclusion. I'm only focusing on the case when the faces are as large as possible (facets). I suspect (but haven't thought much about confirming) that the full story can be obtained, fairly straightforwardly, once enough is known about the case of facets, together with the combinatorial aspects of a cube.
I'll explain my viewpoint on the problem, focusing on the standard cube in $\Bbb R^3$. We'll abstract from there.
For the moment, let's focus on $1$-dimensional subspaces. I'm going to assume we're choosing a unit vector, uniformly at random. Thus, we have a unit sphere's worth (not entirely true; but we'll correct soon) of choices of directions for our subspace. And, instead of working with the cube, we'll work with a spherical cube (by scaling each point on the cube to have length $1$).
The $6$ faces of spherical cube partition the unit sphere into $6$ regions of equal area; thus, the probability that a unit vector lies in a particular region (corresponding to a particular face) is $$\dfrac{\text{area of region}}{\text{area of sphere}} = \frac{1}{6}.$$
But wait, there's more! This is only the probability that a particular point on the unit sphere is in a certain region. Each $1$-dimensional subspace determines two antipodal points on the sphere, so our probability is really $\frac{1}{3}$, essentially by identifying opposite faces of the cube.
Since our formula for probability involves areas, the probability of a $1$-dimensional subspace going precisely through an edge or vertex is $0$.
I'll leave serious thought about $2$-dimensional subspaces to you. Essentially each subspace determines a "belt" of four faces and there are three such belts, with each face contained in two belts; this leads to a probability of $\frac{2}{3}$. Thinking hard about this step lead naturally to thinking about higher dimensions and generalizing the preceding thoughts.
We may ignore subspaces going through "small dimensional" faces (vertices, edges, etc) of a cube, as the probability of such a thing happening with a randomly chosen subspace is $0$.
Any$^1$ subspace of dimension $d$ intersects $[-1, 1]^n$ in exactly $2d$ facets (faces of dimension $n - 1$), these $2d$ facets splitting into $d$ pairs of opposite facets (this was the hardest part of the argument for me to figure out). Thus, any subspace of dimension $d$ uniquely determines $d$ pairs of opposite facets, and there are $n$ such facet-pairs.
Thus the probability that a random $d$-dimensional subspace intersects a particular facet of the cube is exactly the probability that a given facet-pair is contained in a set of $d$ facet-pairs.
There are ${{n - 1} \choose {d - 1}}$ ways to choose $d$ facet-pairs of an $n$-dimensional cube containing a given facet-pair, and ${n \choose d}$ total ways to choose $d$ facet pairs. The probability of including a particular facet, then, simplifies quite nicely to $\frac{d}{n}$, which suggests to me there's a more streamlined argument!
$^1$ "Any" here means any subspace that intersects the interior of a facet whenever it intersects a face of that facet; we're ignoring the subspaces like those in $\Bbb R^3$ that go through two vertices, or two opposite edges, of a cube, without intersecting the interior of a facet.