A Dyck word is a sequence of open and closed brackets such that the brackets come in correctly matched pairs. For example $(()(()))()$ is a Dyck word, while $())(()$ is not. A Dyck path is a visual representation of a Dyck word, where for every open bracket the path goes up one step and for every closed bracket the path goes down one step. For example the Dyck path to $(()(()))()$ is
[![enter image description here][1]][1]
It is easily seen that the Dyck path uniquely determines the Dyck word and that any such path, which stays above or at the starting height and ends at the starting height, is a Dyck path. A peak in a Dyck word is an occurrence of $()$.

The Narayana numbers $N(n,k) = \frac{1}{n} {n \choose k} {n \choose k-1}$ count the number of Dyck paths/words of length $2n$ with $k$ many peaks.

**Question:**
Given $q,k,n \in \mathbb{N}$ with $k \leq n$ and $q < 2n$ and a Dyck word $W$ pulled randomly (uniformly) from the set of Dyck words of length $2n$ with $k$ many peaks, what is the probability that the positions $q$ and $q+1$ form a peak in $W$?

Trivial examples: For $k=n$, the only possible Dyck word is $()()\cdots()$ and the probability is $1$, if $q$ is odd, and zero otherwise. For $k=1$, the only possible Dyck word is $((\cdots()\cdots))$ and the probability is $1$ for $q=n$ and zero otherwise.


  [1]: https://i.sstatic.net/RjmXu.jpg