# Probability of microstate in Einstein solid

This question from my course got me confused.

Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed.

(c) Assuming that this system is in thermal equilibrium, what is the probability of finding all the energy in solid A?

I know the answer already: it's the amount of microstates where everything is in A divided by the total amount of microstates. 20 over 29 / 20 over 39

Intuitively, I would say it's (1/2)^20 because every unit has .5 chance to be in A as well as in B. And the position of every unit is independent of other units's positions. Where am I wrong?

The problem comes down to distinguishability. If the units of energy were distinguishable, then indeed it would behave as $\frac{1}{2^{20}}$. Unit $p$ being in oscillator $A_1$ with unit $q$ being in oscillator $A_2$ would be counted as distinct from unit $p$ being in oscillator $A_2$ with unit $q$ being in oscillator $A_1$. This is what happens when their probabilities are truly independent. However, with indistinguishable units of energy, the two states described above only count once, showing that the probabilities of the two units are not totally independent. Now, the question is really, how many ways can we distribute 20 units of energy across 10 oscillators compared to across 20 oscillators? As the number of units increases, the number of microstates is reduced more and more, deviating farther and farther from the assumption of independent probabilities. Because of these reductions, the number of microstates with 19 units in $A$ and 1 unit in $B$ decreases relatively more than, say, with 18 units in $A$ and 2 units in $B$. So, the probabilities wind up being dependent on how many units are already in a given region. This is basically the source of the exchange interaction for otherwise non-interacting bosons.