Low entropy forces a heavy fiber

Let $Z : \Omega \to \beta$ be any function from a finite sample space $\Omega$ (nonempty) into a finite set $\beta$ (nonempty), and let $H(Z)$ denote its base-2 Shannon entropy under the uniform distribution on $\Omega$ (shannonEntropy, as introduced in Definitions.DiscreteEntropy). Then some value $b \in \beta$ is attained on a fraction of $\Omega$ at least $2^{-H(Z)}$:

This is the elementary pigeonhole/counting consequence of low entropy that the entropy method in combinatorics (Spencer's partial coloring lemma and its relatives) exploits: a random variable of small entropy cannot spread its probability mass thinly across many outcomes, so at least one fiber must carry an exponentially large share of $\Omega$.

The proof is a direct one-line comparison per term: writing $p_b$ for the empirical probability of $b$ and $p^\ast = \max_b p_b$, the inequality $\log_2(1/p_b) \ge \log_2(1/p^\ast)$ holds for every $b$ with $p_b > 0$ (since $p_b \le p^\ast$), so summing $p_b \log_2(1/p_b) \ge p_b \log_2(1/p^\ast)$ against the total probability $\sum_b p_b = 1$ gives $H(Z) \ge -\log_2(p^\ast)$, i.e. $p^\ast \ge 2^{-H(Z)}$.

Formalization Note The conclusion is stated as a lower bound on the raw fiber cardinality (cast to $\mathbb{R}$) rather than on $p_b$ directly, since that is the form the entropy method's pigeonhole step is applied in.