Counting/Probability problem involving runs of a given length Suppose we have a row of $c\,W$ elements. We select half of them, $\displaystyle \frac{c\,W}{2}$. What is the probability that we have at least one run of at least $W$ selected elements, in terms of $c$ and $W$?
EDIT: $c$ and $W$ are both positive integers.
 A: The following assumes that $W$ and $a$ are positive integers, that one chooses uniformly at random $aW$ integers between $1$ and $2aW$, and that one asks for the probability of the event $A$ that the resulting set $S$ contains at least $W$ consecutive integers.
Let us compute the probability of the event $A_i$ that $S$ contains the $W$ consecutive integers starting from $i$. Then $A_i=\emptyset$ if $i\ge(2a-1)W+2$, otherwise $S$ is determined by its $(a-1)W$ elements not between $i$ and $i+W-1$. Hence
$$
P(A_i)=\frac{{(2a-1)W\choose (a-1)W}}{{2aW\choose aW}}=p(W).
$$
Similar formulas can be written for $P(A_i\cap A_j)$ for example, depending of $|i-j|$.
Since $A$ is the union of the events $A_i$, an exact formula for $P(A)$ is given by the exclusion-inclusion formula, and an easy upper bound is
$$
P(A)\le(2a-1)Wp(W).
$$
When $W\to+\infty$, Stirling formula yields
$$
p(W)=\sqrt{\frac{2a-1}{2a-2}}(C_a)^W(1+o(1)),\quad C_a=(2a-1)^{2a-1}2^{-2a}a^{-a}(a-1)^{1-a}.
$$
Roughly speaking, the error made in the uper bound is to count several times the segments of length greater than $W$ included in $S$. This remark and some heuristic arguments suggest that a discount of $\displaystyle\frac{a}{2a-1}$ should be applied to the upper bound, hence
$$
P(A)=a\sqrt{\frac{2a-1}{2a-2}}W(C_a)^W(1+o(1)).
$$
