For each $n \in \mathbb{N}$ let $X\_n$ be a random variable taking its values in a finite set $E\_n$ with $P(X\_n=x\_n)>0$ for all $x\_n \in E\_n$. Say that $X\_n$ is asymptotically degenerate if $\min\_{x\_n \in E\_n}\Pr(X\_n \neq x\_n) \to 0$.
Denoting by $h(X)$ the entropy of (the law of) a discrete random variable $X$, assume that $\frac{h(X\_n)}{\log k\_n}\to0$ with $k\_n = \sharp E\_n$. Does it imply that $X\_n$ is asymptotically degenerate ?



asymptotic behaviour of the entropy and degeneracy 
Of course, not. Your "asymptotic degeneracy" condition means that the weight of the maximal atom of the distribution of $X\_n$ goes to 1, which is a pretty strong condition. If the sizes of your sets $E\_n$ go to infinity, take a sequence $\epsilon\_n\to 0$. Then the sequence of measures $\nu\_n$ with two atoms of weight $(1-\epsilon\_n)/2$ and the remaining mass $\epsilon\_n$ equidistributed among the other points of $E\_n$ is a counterexample. On the other hand, if the sizes of $E\_n$ are uniformly bounded, then $h(X\_n)\to 0$ indeed implies asymptotic degeneracy in your sense.


