What does conservation of probability mean in Classical Mechanics and why is it true?

In the context of the Liouville equation, regularly the conservation of probability is invoked. (Of course, the overall probability is always conserved but this is a truism and not what is meant here. See also the discussion here.)

But what concretely does this means in the context of Classical Mechanics and why is it true?

Let's say that for some reason only three initial states are possible $A$, $B$, and $C$ and we can describe our system using the probability distribution: \begin{align} \rho(t=0,A) &= 0.7 \notag \\ \rho(t=0,B) &= 0.2 \notag \\ \rho(t=0,C) &= 0.1 \end{align} The total probability to find our system in a region $R$ which contains $A$ and $B$ but not $C$ is therefore $P(R,t=0)=90$%.

Now, as time passes on ($t\to t=t_1$) $R$ gets dragged by the Hamiltonian flow and becomes $\tilde R$. Moreover, our phase space points $A$, $B$, $C$ also get dragged and become $\tilde A$, $\tilde B$, $\tilde C$. The statement regularly used in the derivation of the Liouville equation is $$ P(\tilde R, t= t_1) = P(R,t=0)=90 \% \, . $$

Is this correct because the time-evolution of the phase space points $A$, $B$, $C$ and the time-evolution of the region $R$ are both described by Hamilton's equation? (Formulated simpler: because we move them around equally as time passes on?)

And secondly, what does it imply for our concrete probability distribution? Liouville's theorem tells us the phase space volume is constant. And this in combination with the conservation of probability tells us that $\frac{d \rho}{dt}=0$. But does this tells us that while the probabilities at our original locations can be wildly different ($\frac{\partial \rho }{\partial t}\neq 0$), e.g.

\begin{align} \rho(t=t_1,A) &= 0.3 \notag \\ \rho(t=t_1,B) &= 0.5 \notag \\ \rho(t=t_1,C) &= 0.2 \end{align}

we certainly have

\begin{align} \rho(t=t_1,\tilde A) &= 0.7 \notag \\ \rho(t=t_1,\tilde B) &= 0.2 \notag \\ \rho(t=t_1,\tilde C) &= 0.1 \quad ? \end{align}