We define the $s$-sparse hypercube in $\mathbb{R}^d$ as \begin{align} \mathbb{H}_s = \bigl \{ {\bf{v}} \in \{ -1, 0 , 1\}^d \colon \| {\bf{v}} \|_0 = s \bigr\}, \end{align} where $ \| {\bf v} \|_0 $ is the number of nonzeros of vector ${\bf v}$. Note that $\mathbb{H}_s$ contains ${d \choose s}\cdot 2^s$ elements. I was wondering what is the structure of $ |\langle {\bf v}_1 , {\bf v}_2 \rangle| $ for $ {\bf v}_1 , {\bf v}_2 \in \mathbb{H}_s$.

For any fixed ${\bf v}$, define $$ M_k = \#\bigl \{ {\bf v}_1 \in \mathbb{H}_s\colon | \langle {\bf v}_1 , {\bf v} \rangle| = k \bigr\} $$ as the number of vectors in $\mathbb{H}_s$ whose inner product with ${\bf v}$ is $k$ or $-k$. Here $k \in \{ 0, 1,\ldots, s\}$.

I was wondering if we can derive some upper bound on $M_k/ M_{k-1}$ for all $k$. Or can we directly compute $M_k$ or derive some simple upper bound that can tell us the growth of this number?

From a probabilistic point of view, if ${\bf v}$ is uniformly taken over $\mathbb{H}_s$, what is the distribution of $M_k$ like?