Boltzmann entropy of a two-box occupancy ($S(k)=k_B\ln \left(\genfrac{}{}{0ex}{}{N}{k}\right)$)

Take $N=20$ distinguishable, labeled particles and two boxes of equal volume. Retain only whether each particle is in the left or right box: use a bit $0$ for left and $1$ for right. Each length-20 string is one coarse microstate of this toy model. Assume the labels make independent side choices with equal probability, so the $2^{20}=1048576$ strings are equiprobable. Equal volumes motivate these weights for noninteracting particles with no preference for either box; equal volumes alone would not establish them for an arbitrary interacting gas.

Now retain only the number $k$ in the left box as the macrostate. Choosing which $k$ labels are left gives a binomial coefficient, soThe factorials count assignments to labels; the probability of a macrostate sums the probabilities of its constituent microstates. This is the binomial distribution with parameters $20$ and $1/2$. The same binomial coefficient appears in the counting of distinguishable two-level systems, although here the macrostate records occupancy rather than energy. ^[ref]

The largest binomial coefficient occurs at $k=10$, the balanced macrostate. All-left means $k=20$ and supplies just one string. ThusThere are $184756$ balanced strings for every all-left string, even though each individual string has the same probability. Exact balance has about $17.6\%$ of the total probability; equilibrium reasoning concerns the many macrostates near balance too, rather than certainty of exactly ten particles on each side. The diagram compares the counts with their natural logarithms.

Four fair, independent coins give a smaller worked example with the same binomial distribution. If $k$ is the number of heads, the five macrostates $k=0,1,2,3,4$ contain respectively $1,4,6,4,1$ microstates. Their probabilities are respectively $1/16,4/16,6/16,4/16,1/16$. In particular, two heads is one macrostate containing six ordered coin sequences, with Boltzmann entropy $k_B\ln 6$ in this counting analogy. A prescribed ordered sequence has one realization. The next diagram enumerates the same sixteen strings as left/right assignments.