Entropy and Probabilities: does not "particle indistinguishability" alter probability values? This issue has been bothering me for a bit but I fail to find a convincing answer.
It all has to do with particles indistinguishability and how they alter the concept of probability.
As I understood it, in the microcanonical ensemble all configurations are equiprobable: we find the macroscopic equilibrium configuration by selecting the one which occurs with higher probability. 
But indistinguishability must somehow alter what we compute as probabilities.
Let me clarify directly with an example.
I have a system consisting of three boxes, and three particles. Each particle has to go in one box. There is no energy, only configurational entropy.
All it matters to me, macroscopic observer, is how many boxes are left empty, $1$, $2$ or $3$.
An observed unable to distiguish particles could get one of the following configurations:
$$ (3, 0, 0) \\
(0,3,0) \\
(0,0,3) \\
(1,1,1) \\
(1,2,0) \\
(2,1,0) \\
(0,1,2) \\
(0,2,1) \\
(1,0,2) \\
(2,0,1) $$
Now I wonder, what is the probability (in the mathematical sense) of having a configuration whereby one box is empty?
Looking at the above, one might be tempted to fallaciously reply such probability equals $ \frac{6}{10} $.
An easy calculation shows the probability of having one empty box equals $ \frac{6}{9}$.
Indeed, treating particles as indistinguishable does not delete the fact that the configuration say $(1,2,0)$ can occur in $3$ different ways, and so on. Hence, the six configurations above with one empty box can occur in $18$ ways: adding the $(3,0,0)$ which can occur in $3$ ways, and the $(1,1,1)$, which can occur in $6$ ways, one gets $27$ total possibilities, not $10$ as one would if considered indistibguishability.
If I wanted to find the most probable configuration in the microcanonical ensemble, I would need to look at probabilities.  How does indistinguishability fit in all this?
I studied Gibbs’ paradox, and I see how indistinguishability solves it: but is not the microcanonical intuitive picture somehow altered??
Yes for an observer maybe $(A, BC, 0)$ is indistinguishable from $ (B, AC, 0)$, yet these occurences must count if we look at mathematical probabilities.
Thanks a lot for your help
 A: Indistinguishability is more subtle than that. As a simple example, consider two particles ($A$ and $B$) and two states. If the particles are distinguishable, there are four equally probable configurations,
$$(A, B), (B, A), (AB, 0), (0, AB)$$
so, e.g. the probability of having both states occupied is $1/2$. If you had another observer that couldn't distinguish between the particles (suppose they're red and green, and the observer is colorblind), the probabilities remain exactly the same, even though the particles are now "indistinguishable". The microcanonical ensemble doesn't depend on the observer.
True indistinguishability is deeper: it means that the separate labels $A$ and $B$ are completely meaningless, so that the states could be, e.g.
$$(1, 1), (2, 0), (0, 2)$$
if the particles are bosons, so that the probability of having both states be occupied is now $1/3$. This is sometimes expressed by saying that bosons have a tendency to 'clump up'. 
You can try putting labels on -- for instance, maybe we can think of $A$ as the "first particle we put in" and $B$ as the second. Then formally the states $(A, B)$ and $(B, A)$ are
$$a_0^\dagger a_1^\dagger |0 \rangle, \quad a_1^\dagger a_0^\dagger |0 \rangle$$
but these are the exact same state because $a_0^\dagger$ and $a_1^\dagger$ commute, so you are double counting the state. It's the same mistake as saying that when you flip a coin, you can get heads, not-tails, and tails, so the chance of getting tails must be $1/3$. 
