Boltzmann distribution - why does distinguishability increase likelihood? I am looking through derivations of the Boltzmann distribution. The method I've seen uses an argument that involves counting distinguishable microstates of a system with fixed energy, and then assuming that these distinguishable microstates are equally likely to occur.
A first assumption is that from an experimental perspective, it is not possible to distinguish certain configurations. This seems reasonable. However, the derivations I've seen never explicitly say whether the (possibly indistinguishable) rearrangements of particles are still physically meaningful/realizable or not. Are they? Here's why I think this is important to know.
By example, consider the derivation of Eisberg & Resnick, Appendix C. Assume a four particle system of total energy $3\Delta E$, with energy divisions $\{0, 1\Delta E, 2\Delta E, ... \}$. Let's just consider two of the possible valid macrostates to avoid getting bogged down.

*

*One particle at energy $3\Delta E$, three particles at $0$ energy. In principle there are $4!$ rearrangements, but $3!$ are irrelevant due to indistinguishability, giving only $4$ distinguishable microstates.


*One particle at energy $2\Delta E$, one particle at energy $\Delta E$ and two particles at energy $0$. There are again $4!$ rearrangements in principle, but $2!$ are irrelevant due to indistinguishability. This leaves $12$ microstates.
In macrostates 1 and 2 both, there are $4!$ possible orderings, but they are not all distinguishable. However, for a moment, let's suppose that each of these orderings, despite not being distinguishable to an experimenter, do correspond to a valid and physically realizable configuration. If all $4!$ rearrangements of a macrostate are physically realizable, would it not be more reasonable to then assume that (a) "each possible rearrangement (distinguishable or not) is equally likely," not that (b) "each distinguishable rearrangement is equally likely?"
To see the difference in practice, suppose I have a lab notebook, and every $T$ seconds I observe this system to find it in one of the two configurations above, i.e. macro state 1 or 2. Also assume that I write the distingushable microstate that I observe. That means that, I write one of 1.1-1.4 for macro 1, and for macro 2, I write 2.1-2.12. Suppose I do this for a long time.
Under the assumption (a) above, my entries tend toward an equal number of 1's and 2's, but. But as for microstates, I would have them in differing frequencies. This also seems to agree with a statement the book makes: "all possible divisions of the energy of the system occur with the same probability." It is tempting to interpret this as saying that all macrostates are equally likely (and thus distinguishable microstates should not be).
Under the assumption (b), in contrast, my entries would have 1.1-1.4 and 2.1-2.12 occurring in equal amounts -- all distinguishable microstates equally likely. Overall, macrostate 2 would be happening much more often than macrostate 1, and this is obviously reflected in the standard derivation.
Have I deeply confused myself? How do I justify the assumption (b) without drawing a strange relationship between distinguishability and likelihood?
thanks.
 A: I think that there are some basic concepts to be reviewed in your question.
First, a macrostate is defined by macroscopic variables, in your example of the four particles, fixing the macrostate is equal to fixing the total energy of the system. A microstate is defined by a configuration compatible with a macrostate. So the usual question is to ask: How many micro-states are compatible with total energy $3\Delta E$? Your points (1.) and (2.) are instances of such microstates.
Second, the phrase "but 3! irrelevant due to indistinguishability, giving only 4 distinguishable microstates", are the particles indistinguishable or distinguishable?
This might seem irrelevant, but I think that it would help to answer your doubts if you rephrase it in the usual framework of statistical physics.
Example: Say that the total energy is $\Delta E$ and there are 4 particles. The energy of these particles is quantized, so particle $i$ can only have energy values equal to $\epsilon_i=n\Delta E$ with $n=0,1,2,3,\dots$.  Then, we can count the number of microstates with distinguishable and indistinguishable statistics.
Indistinguishable: There is only one possible microstate. One particle with energy $\Delta E$, the rest with energy 0.
Distinguishable: There are four possible microstates. These are:($\epsilon_1$,$\epsilon_2$,$\epsilon_3$,$\epsilon_4$)=($\Delta E$,0,0,0); (0,$\Delta E$,0,0); (0,0,$\Delta E$,0) and (0,0,0,$\Delta E$). In this case, you can count the number of microstates as the ways of sorting 1 energy pack among 4 distinguishable particles (see Wikipedia: combinatorics). This is, $\Omega(E=\Delta E)=\binom{1+4-1}{1}=4$. This result has nothing to do with indistinguishability.
A: 
A first assumption is that from an experimental perspective, it is not possible to distinguish certain configurations.

The point here is not whether they can be distinguished in a realistic experiment, but whether they can be distinguished in principle, in any conceivable experiment - that is whether the configurations correspond to different microstates or not. In classical physics every particle can be characterized by its position and momentum - these may be hard to measure and trace, but in principle it could be done. In quantum mechanics particles are indistinguishable, because there is no way of knowing which is which or even separating them - this is hard-coded in QM formalism, as Slater determinants for fermions or permanents for bosons.
