Consider the first $n$ natural numbers and randomly pick $m$ numbers out of these. Define S as the set of these $m$ numbers, having density $d=m/n$. Consider all arithmic progressions of length 3 in S. What is the expectation of the proportion of these triples that crosses $n/2$? That is, the progressions that start at < $n/2$ and have endpoint > $n/2$
Simulation seems to indicate that for sufficiently large $n$, the ratio is 0.5 when $d$ is close to 1 and 0.25 when $d$ is close to 0. Does that make sense?
Let's say $d=1$, then $S=\{1,2,3,...n\}$.
For an arithmetic progression $s_1,s_2,s_3$, assume that $s_1<s_2<s_3$. Now, we want to pick $s_1<n/2$ and $s_3>n/2$ in S. We can ignore $s_2$ here because there will definitely be a $s_2$ between $s_1$ and $s_3$. (of course we assume that $s_3-s_1$ is even. This will not matter since we will be comparing tuples containing arithmetic progression among themselves)
If both of these possibilities occur ($0.25$ possibility), this is a situation we do not want.
Likewise, we do not want these two possibilities to occur together ($0.25$ possibility). Now, the sum of the possibilities we do not want is $0.5$. Then the situation we want is $0.5$.
So far everything is normal. You already stated that this ratio is 0.5 when d=1. If the set S shows a uniform distribution, $d$ will not affect the ratio you mentioned in any way. Because we can define S as a fuzzy set whose membership function is equal to a constant $d$:
$$S=\{1,2,3,...n\}, \ f(s_i)=d$$
Here $f$ is membership function of S. Thus, expected value of the number of elements of the set S will be $n.d=m$. And the solution above applies exactly the same here. From this perspective, it is clear that d will not change this ratio.
I think there must be something missing in your simulation. Because to be sure of my answer, I wrote a small code for this in Matlab and tried it. I saw that for d's close to 0, it was still 0.5.