I was wondering if anything is known about this problem. Fix $0\leq m\leq k$. We are given a graded poset and we fix an element $x$ of rank $k$. Is it possible to estimate the number of elements $y$ of rank $k$ such that $x\wedge y$ has rank $m$? (I suppose the poset must be a meet-semilattice for this to be well-defined.)

I have kept the formulation general, but I'd be happy to hear any results with extra hypotheses on the poset. In fact, my use case is a lattice of certain subsets of a finite ground set. I think another formulation would be in terms of shadows in a set system. Is there a bound on the cardinality of the $(k-m)$th upper shadow of the $(k-m)$th lower shadow of a singleton in $X^{(k)}$? I apologise in advance if this question turns out to be trivial.

Edit. Based on the answer below, it seems there isn't something one can really say in general, so I'll just post my motivating example. Consider the poset of all subsets of $\{1,2,\ldots,n\}$ that are a (nonempty) arithmetic progression and fix a progression $P$ of length $k$. How many progressions of length $k$ have $m$ elements in common with $P$, for $0\leq m\leq k$?

I'd also be interested in any references that deal with any interesting properties of this poset. For example, the size of this poset seems to be $$1+n+\sum_{m=1}^n\sum_{k=1}^m \bigg\lfloor{m\over k}\bigg\rfloor,$$ according to the OEIS.