Difference between Shannon entropy and min-entropy I would like to construct a sequence of discrete random variable $X_2, X_3,...,X_n,...$, where $X_n \in\{0,1,2,...,n-1\}$. Given any $\epsilon \in (0,1)$, its Shannon entropy and min-entropy should satisfy the following relationships
\begin{cases} 
      H(X_n)\geq(1-\epsilon)\log_2(n)\\
      H_{min}(X_n)=const
\end{cases}
for all $n\geq\mathbb{N}_{\epsilon}$ and some $const > 0$.
My understanding is that the Shannon entropy indicates the underlying distribution should be approximately uniform. And the min-entropy suggests that the largest possibility of $X_n$ should be $2^{-const}$. But I am stuck with coming up with such a distribution. Is there anyone who could provide some hints?
 A: Since the OP may be interested in what $\epsilon$ are achievable I am providing this alternative to the other answer, where it is correctly stated:

If you fix the maximal atom (say, $p$) of a distribution $\mu$ supported by $n$ points, then its entropy is maximal when all the remaining atoms have the same weight $(1-p)/(n-1)$.

Also note that
$$
p\geq \frac{1-p}{n-1} \iff np\geq 1\iff p\geq \frac{1}{n}
$$
so that $p$ is indeed the maximal atom of $\mu.$
This means that one actually obtains the equality below for the Shannon entropy:
$$
H(\mu) = -p\log p - (1-p)\log(1-p) + (1-p)\log(n-1),  
$$
when $p$ is fixed which gives
$$
H(\mu) = H_2(p) + \left(1-p\right)\log(n-1)   
$$
or
$$
H(\mu) \geq \left(1-p\right)\log(n-1)\sim \left(1-p\right)\log n \quad (1)
$$
where $f(n)\sim g(n)$ denotes that $\lim_{n\rightarrow \infty} \frac{f(n)}{g(n)}=1.$ This means that for $n$ large enough you can pick any $\epsilon \geq p$ for which
$$
H(\mu)\geq (1-\epsilon) \log n
$$
is indeed satisfied.
Depending on what application you have in mind, this may suffice.
A: Such a family doesn't exist. If you fix the maximal atom (say, $p$) of a distribution $\mu$ supported by $n$ points, then its entropy is maximal when all the remaining atoms have the same weight $(1-p)/(n-1)$. Whence
$$
\begin{aligned}
H(\mu) &\le -p\log p - (1-p)\log\frac{1-p}{n-1} \\
&= -p\log p - (1-p)\log(1-p) + (1-p)\log(n-1) \;,
\end{aligned} 
$$
which grows slower than $(1-\epsilon)\log n$ for sufficiently small $\epsilon$.
