For a <discrete random variable> $X$ with <probability mass function> $p_x$, using base-two <logarithms> for both <Shannon entropy> and <min-entropy>,
$$
H(X)\geq H_\infty(X)\geq0.
$$
The first inequality says that average <information content> cannot be below the least <information content> of a possible outcome. For a finite <alphabet>, equality $H(X)=H_\infty(X)$ holds exactly when the <probability mass function> is uniform on its positive-probability outcomes. Impossible outcomes do not need to share that positive <probability>. In particular, $(1/2,1/2,0)$ has both quantities equal to $1$ <bit>, while $(1,0,0)$ has both zero. The same equality characterization holds for a countable <alphabet>: equality forces only finitely many positive-probability outcomes, because all must have the same positive mass. The uniform maximum of <Shannon entropy> is discussed in Shannon's original treatment.<A Mathematical Theory of Communication by Claude Shannon>[Shannon, A Mathematical Theory of Communication, Section 6, definition of H and properties 1–2]{ref}

For the proof, $0<p_x\leq p_{\max}$ gives $-\log_2p_x\geq-\log_2p_{\max}$; taking the <expected value> proves <information entropy dominates min-entropy>. More precisely, the nonnegative gap is
$$
H(X)-H_\infty(X)
=\sum_{x:p_x>0}p_x\log_2\frac{p_{\max}}{p_x}\geq0.
$$
It vanishes precisely when every positive term vanishes, namely $p_x=p_{\max}$ for every possible outcome. With $m$ such outcomes, normalization forces $p_{\max}=1/m$. If <Shannon entropy> is infinite, the inequality still holds and is strict, because <min-entropy> of a countable <probability mass function> is finite. No subtraction of two infinite quantities is needed.

An analytical family makes the gap visible. Give $A$ <probability> $q$ and each of $B,C,D,E$ <probability> $(1-q)/4$, with $1/5\leq q\leq1$. Then $A$ is a most likely outcome, so its <probability> is also the best single-guess success <probability>, and
$$
H_\infty(q)=-\log_2q,\qquad H(q)=h_2(q)+2(1-q),
$$
where $h_2$ is <binary entropy>. The <entropy bound with one prescribed probability> explains the second expression: it separates whether $A$ occurred from which of four equally likely alternatives occurred. At $q=1/5$, all five outcomes are uniform and both quantities are $\log_2 5\approx2.3219$ <bits>. At $q=1/2$, <Shannon entropy> is $2$ <bits> while <min-entropy> is $1$ <bit>, reproducing the biased example in <min-entropy>. At $q=1$, both are zero; the four impossible alternatives contribute nothing. For every $1/5<q<1$, the <probability mass function> is nonuniform on its possible outcomes and the gap is strictly positive.

\Image[/min-entropy-guessing-inequality-family.png]
{title=An analytical five-outcome model separates average surprise from the surprise of the most likely outcome}

For finite <Shannon entropy>, exponentiating the inequality gives the useful but potentially loose bound $p_{\mathrm{guess}}(X)\geq2^{-H(X)}$. Equality holds exactly for a <discrete uniform distribution> on the possible outcomes. Thus <Shannon entropy> alone supplies a lower bound on optimal success, not the upper bound needed to promise that every guess is unlikely. <Shannon's source coding theorem> concerns average compression under its hypotheses, while <min-entropy> answers the single-guess question directly. The two quantities serve different tasks even when both are expressed in <bits>.


 Information entropy dominates min-entropy (source code)