Consider 4 distinguishable coins. (a) List the number of microstates for the macrostates '0 heads', '1 head', '2 heads', '3 heads', '4 heads'. (b) Which macrostate is most probable and what is its probability? (c) Relate this to a gas. Solution: (a) Total microstates = 2^4 = 16. Counts: 0 heads: 1 (TTTT); 1 head: 4; 2 heads: 6; 3 heads: 4; 4 heads: 1. (b) '2 heads' with 6/16 = 37.5% probability. (c) Just as 2-heads dominates 4 coins, the equilibrium macrostate of a gas dominates because it owns the overwhelming majority of the 10^(10^23) microstates - probability becomes practical certainty.