The classic example of Bertrand paradox deals with the case where we count the uncountable set of chords in a circle in different ways and ends up getting different probability each time. The wikipedia example shows how we got $\frac{1}{2},\frac{1}{3},\frac{1}{4} $ respectively in different cases.
My question from here is can we always find some, way of choosing the chords so that the probability ends up spanning the whole rational set or even the whole real number set.
What I mean by spanning the whole rational number set $\mathbb{Q}$ is as follows:
$\forall q \in \mathbb{Q} \text{(}or \mathbb{R}\text{)} \cap [0,1]$ there exist some way of choosing the chord in the circle so that probability in that space becomes $q$?
If you are okay with models not very "meaningful" as those on wikipedia, there is always a way: adding a selection step to tilt the weight. Say we have a $\frac 12$-scheme as that described on wiki.
If you want any $p<\frac 12$, you can choose a chord from the $\frac 12$-scheme, and keep it with probability $2p$ while discard it with probability $(1-2p)$. This gives you a scheme with success rate $p$.
If you want $p>\frac 12$, you can let $q=2-\frac 1p\in(0,1)$ and do the following. If you get a successful chord, keep it. Otherwise you discard the unsuccessful chord with probability $q$ and select again. The overall success rate is now
$$\frac 12+\frac 12\cdot q\cdot\frac 12+\frac 12\cdot q\cdot\frac 12\cdot q\cdot\frac 12+\cdots=\frac 12\cdot\frac{1}{1-q/2}=p,$$
as desired.