If we define a set $[A,B]$ and a set $[a,b]$ which is contained inside the first set, what are the odds when picking a number from set $[A,B]$ that it is also contained inside the set $[a,b]$ if all numbers are equally likely to be picked?
If elements in our sets are $\in\mathbb N$, then we can simply divide the number of all elements we can pick from and the number of elements in our target set, making our odds:
$$p({\mathbb N})= \frac{b-a+1}{B-A+1}$$
For $\in\mathbb Z$, we simply watch out for the negatives:
$$p({\mathbb Z})= \frac{|b-a|+1}{|B-A|+1}$$
But what about when our sets are rational, $\in\mathbb Q$?
If we define our sets like here, we have:
$$p({\mathbb Q})=\frac{|b-a|+1+(|b-a|)\times n}{|B-A|+1+(|B-A|)\times n}, n\to\infty$$
If we take the limit as $n$ approaches $\infty$, we finally get:
$$p({\mathbb Q})=\frac{|b-a|}{|B-A|}$$
Is this a valid solution?
Also, I'm wondering how could one define these sets for $\in\mathbb R$ and calculate $p(\mathbb R)$?
What about $p(\mathbb C)$?