Let us assume a set $\mathcal{A}$ of $K$ unique integers: $\mathcal{A} = (X_{1}, \ldots, X_{K})$ and randomly pick an integer $X_{j}$ from the set $\mathcal{A}$.
Now, let us draw a random subset $\mathcal{B} \subset \mathcal{A} \setminus (X_{j})$ of $M < K-1$ integers (excluding the $X_{j}$ integer from the draw).
My problem is to find that the probability that the integer $X_{j}$ is less than the minimum integer in subset $\mathcal{B}$? That is, find/estimate the probability:
$\mathbb{P}(X_{j} < \min(X_{i}): X_{i} \in \mathcal{B})$.
Many thanks for your help!
Example: Let $\mathcal{A} = (2,3,4,5,6,7)$ and $X_{j} = 4$. Let also $\mathcal{B1} = (3,6,7)$ and $\mathcal{B2} = (6,7,5)$. In $\mathcal{B1}$, the integer $X_{j} = 4$ is not less that 3 = min(3,6,7), thus, does not satisfy the criterion, while in $\mathcal{B2}$, the integer $X_{j} = 4$ is less than 5 = min(5,6,7), which satisfies the criterion.
We order the element elements of $A$ from the min to max $x_{1},\dots,x_{k}$. Fix the chosen $x_{j}$ on the first step, then $\min{B} > x_{j}$ if and only if the set $B$ is a subset of $\{ x_{j+1}, \dots, x_{k}\}$, there are ${k-j}\choose{m}$ subsets of $m$ elements of this set, and of course, $j\leq k-m$ otherwise we can't pick $m$ elements greater than $x_{j}$; so $$P(X_j\leq \min (B))=\sum\limits_{s=1}^{k-m}\frac{1}{k}\frac{{k-s}\choose{m}}{{k-1}\choose{m}}=\sum\limits_{i=m}^{k-1}\frac{1}{k}\frac{{i}\choose{m}}{{k-1}\choose{m}}$$
Now we use this identity to get $$P(X_j\leq \min (B))=\frac{1}{k}\frac{{k}\choose{m+1}}{{k-1}\choose{m}}=\frac{1}{m+1}$$