The floor-probability bounds conjecture for convex sets

Let $K$ be a compact convex set with a flat floor over the segment $[0,1]$, and let $Q_K(n)$ denote the corresponding probability that $n$ independent uniformly random points, together with the two endpoints of the floor, are in convex position. Set

$$t_n=\frac{2^n}{n!(n+1)!},\qquad q_n=\frac{1}{n!(n+1)!}{2n\choose n}.$$

Floor-probability bounds conjecture. For all $n\geq 3$ and all $K\in\operatorname{CCSF}([0,1])$,

$$t_n\leq Q_K(n)\leq q_n.$$

The lower and upper bounds generalize the known $n=2$ bounds, with the triangle and square giving the respective benchmark values. The supplied text presents these inequalities as a belief and does not provide a resolution.