I had an argument about the following statement, which I think is wrong:
One third of odd numbers (33.333%) will be divisible by 3.
Let $\Bbb{N}$ refer to the positive integers.
Let $\Bbb{O} := \{o | o = 2n-1; n \in \Bbb{N}\}$, the set of odd numbers.
$2n-1$ is a (trivial?) bijective function to $\Bbb{N}$.
Thus $\Bbb{O}$ is countably infinite.
Thus $|\Bbb{O}| = |\Bbb{N}| = \aleph_{0}$.
Let $\Bbb{M}(m) := \{o | o = 2n-1 \land o\mod m = 0; m,n \in \Bbb{N}\}$, the set of odd numbers which are divisible by $m$ without remainder.
The additional condition $ o\mod m = 0$ does not remove the bijective property of the odd numbers, as I think we can do a kind of Hilbert Hotel here: For $\Bbb{O}$ we have $1 \to 1$, $2 \to 3$, $3 \to 5$, $4 \to 7$, $5 \to 9$, $6 \to 11$, $7 \to 13$, $8 \to 15$, $7 \to 17$, .. For e.g. $\Bbb{M}(3)$ we have $2 \to 3$, $5 \to 9$, $8 \to 15$, .. which can be renumbered as $1 \to 3$, $2 \to 9$, $3 \to 15$, ..
Thus $\Bbb{M}(m)$ is countably infinite.
Thus $|\Bbb{M}(m)| = |\Bbb{O}| = \aleph_{0}$.
The quoted statement is now in those terms $\frac {|\Bbb{M}(3)|}{|\Bbb{O}|} = \frac {\aleph_{0}}{\aleph_{0}} = 1 \neq \frac {1}{3}$, which is a contradiction.
Where am I wrong? Or where is more precision in my argumentation needed?
In step 6, allowing "$m,n\in\mathbb N$" is too permissive. Consider that $\mathbb M(2)$ is the empty set! It would be better to require that $m$ is an odd prime in the definition, or at least that $m$ is an odd number. Then you can reason about general $\mathbb M(m)$ by calling on the Chinese remainder theorem.
In step 7, you're right that we can "do a kind of Hilbert Hotel", but you might want to write down the explicit bijection $\mathbb N\to\mathbb M(m)$ instead of hand-waving it. It's a simple enough formula, and proving that it works will help to show what kind of condition you need on $m$.
In step 10, as the comments say, that sort of division isn't going to work. Cardinal arithmetic is tricky; for example, $3\aleph_0=\aleph_0$.