How to approximate the expected minimum Hamming distance with $N \gg n$ strings 
If we sample $N \gg n$ binary strings of length $n$, uniformly and
  independently, what is the expected minimum Hamming distance
  between the closest pair?

It seems likely this is hard to compute exactly so a good approximation would be very welcome.
This related question asks specifically about the case where $N = 3$. In a very nice answer @joriki sets out that the mean minimum Hamming distance in this case is approximately:
$$
\boxed{\frac n2-\frac34\sqrt\frac n\pi}\;.
$$
In order to have a concrete value to aim at:


*

*If $N=2^{12}$ and $n=50$ then the mean minimum Hamming distance is approximately $7.3$.

*If $N=2^{12}$ and $n=64$ then the mean minimum Hamming distance is approximately $11.7$.

*If $N=2^{14}$ and $n=50$ then the mean minimum Hamming distance is approximately $5.9$.

*If $N=2^{14}$ and $n=150$ then the mean minimum Hamming distance is approximately $40.2$.

 A: Maybe this helps: from the additive form of the Chernoff bound, for any fixed pair of strings $X,Y\in \{0,1\}^n$ generated uniformly and independently, one has for any $t\geq 0$:
\begin{equation}
\Pr(\text{dist}(X,Y)\leq n/2 -t)\leq \exp\bigg(\frac{-2t^2}{n}\bigg).
\end{equation}
This is just because $X$ and $Y$ are random, so their distance is just a sum of $n$ Bernoulli random variables indicating whether or not the $i$th bit disagrees (and these are all independent for a fixed pair). Applying this with $t=\sqrt{(3/2)n \ln(N/2^{1/3})}=\Theta(\sqrt{n\ln N})$ gives that the probability that a fixed pair deviates from $n/2$ by at least this $t$ is at most $2/N^3$ (I kept the constants inside in case you care about those rather than just the asymptotics). By a union bound over all ${N \choose 2}\leq N^2/2$ pairs, this implies that the probability that any pair deviates by this amount is at most $1/N$.
Note that the maximum deviation from $n/2$ is exactly $n/2$ (if you get the same string twice). So we can conclude that if $X_1,\ldots,X_N\in \{0,1\}^n$ are the sampled strings,
\begin{align*}
\mathbb{E}[\min_{i\neq j} \text{dist}(X_i,X_j)]&\geq (n/2-t)\Pr(\exists i\neq j: \text{dist}(X_i,X_j)-n/2\geq -t)\\
&\geq (n/2-\Theta(\sqrt{n\ln N}))(1-1/N)\\
&=n/2-\Theta(\sqrt{n\ln N})-o(1),
\end{align*}
where we use $N>>n$ to absorb the small negative contribution of the terms that are at most $O(n/N)$. I leave it to you to keep track of the constants if you need them...
On the other hand, this is basically the right scale: here's a sketch. By the central limit theorem,
\begin{equation}
\Pr(\text{dist}(X,Y)\leq n/2 -t\sqrt{n})\sim \frac{1}{\sqrt{2\pi}}\int_t^{\infty} e^{-x^2/2}dx\sim e^{-t^2/2}.
\end{equation}
For $t\sim \sqrt{\ln N}$, this is approximately $1/\sqrt{N}$. Now, by splitting our $N$ random variables into $N/2$ disjoint pairs, we have
\begin{align}
\Pr(\exists i\neq j: \text{dist}(X_i,X_j)\leq n/2-t\sqrt{n})&\geq \Pr(\exists 1\leq k\leq N/2:\text{dist}(X_{2k-1},X_{2k})\leq n/2 -t\sqrt{n})\\
&=1-\Pr(\forall 1\leq k\leq N/2:\text{dist}(X_{2k-1},X_{2k})> n/2 -t\sqrt{n})\\
&= 1-\Pr(\text{dist}(X,Y)> n/2 -t\sqrt{n}))^{N/2},
\end{align}
as we note that by splitting, these events are independent. In particular, this probability is lower bounded as
\begin{equation}
\Pr(\exists i\neq j: \text{dist}(X_i,X_j)\leq n/2-t\sqrt{n})\succeq 1-(1-1/\sqrt{N})^{N/2}\geq 1-\exp(-\sqrt{N}),
\end{equation}
where I am using $\succeq$ to denote where I used some sort of asymptotically equal approximation (i.e. CLT). In particular, with overwhelming probability, there exists a pair whose pairwise distance is at most $n/2-\Theta(\sqrt{n\ln N})$. I leave it to get tight constants and/or make this sketch fully rigorous.
