TITLE: Can we approximate the most likely event's probability if we know the distribution's entropy?
QUESTION [6 upvotes]: Suppose that we know that a distribution $p$ over small $n$ elements has entropy $H(p)=0.01$. I intentionally chose a small number. Is there some kind of inequality or aproximation to the probability $p_i$ for $i = \arg\max_j p_j$?
What I came up with so far is the following inequality:
$H(p) = \sum _j -p_j \log(p_j) \geq -p_i \log(p_i)$. This gives us in principle an inequality for $p_i$, because if we assume that the most likely event has probability larger than $0.5$ (a consequence of the entropy being sufficiently small for a given $n$), then this inequality $H(p) \geq -p_i \log(p_i)$ should give us a bound on $p_i$ in terms of the entropy. I'm not sure how to solve that inequality for $p_i$ though, or if it's even possible. Maybe an approximation would help as well, that works for small entropy.
Is there a bound or approximation on the maximum probability $p_i$ in terms of the entropy, that works for sufficiently small entropy?

REPLY [3 votes]: Notice that $$H(X) = h(p_j) + (1-p_j) H(X \backslash x_j)  \tag 1$$
where $h$ is binary entropy function and $H(X \backslash x_j)$ is the entropy of the remaining values (or equivalenty, $H(X \mid X \ne x_j)$  (proof).
This gives a lower and upper bound of the entropy in terms of $p_j$:
$$ h(p_j) \le H(X) \le  h(p_j) + (1 - p_j) \log(n-1) \tag 2 $$
where $n=|{\cal X}|$ is the cardinality of the variable $X$ support. (Of course, the logarithm should be in base $2$ if the entropy is expressed in bits).

The graph displays this lower bound  (eq $(2)$, in green) and the upper bound for $n=3$ (blue) and $n=9$ (red), for $p_j \to 1$.  The entropy, in the vertical axis, in bits.
The permisible pairs $(H, p_j)$ lie between the green and red/blue lines.
Hence $(2)$ give lower and upper bounds for $p_j$, which should be computed numerically. See also here.
A semiempirical approximation for the lower bound, for the range $0.01 \le H < 0.4$ :
$$ p_j \approx 1- \frac{a H}{b - \log(H)}$$
where  $a=0.73$, $b = 2.3$.