 Solution

The Hoeffding inequality states that if $X_1,\dots ,X_n$ are independent random variables with $a_i\le X_i\le b_i$ almost surely, then for $t>0$The same bound holds for the lower tail. Consequently the two-sided tail is at most twice this bound. When all ranges have length zero, the sum is deterministic and every positive tail probability is zero.

We construct an exponential packing of a Hamming cube. Choose an inclusion-maximal separated set $S$ in $\{-1,1{\}}^n$ with separation at least $r=n/8$ for the Hamming distance. It exists by a finite greedy procedure: keep adding any vertex at distance at least $r$ from all selected vertices until none remains. By maximality, every vertex is within distance strictly less than $r$ of some member of $S$. This is the principle that maximal separated sets give covers.

For a uniformly random vertex $U$ and any fixed vertex $s$, the mismatch indicators in different coordinates are independent Bernoulli random variables with success probability $1/2$. Hence $\rho (U,s)$ has the binomial distribution with parameters $n,1/2$. The lower-tail Hoeffding inequality givesThus the Hoeffding lower-tail bound for Hamming balls shows that every strict-radius ball $B_{<r}(s)$ has at most $2^n{e}^{-9n/32}$ vertices. Using a strict ball handles noninteger $n/8$ without changing the required separation.

The strict balls centered at $S$ cover the entire Hamming cube. Counting their union by the sum of their sizes yieldsThereforeIn particular this proves the assertion for every $n\ge 8$. Maximality here means that no further vertex can be added; finding a largest separated set is unnecessary.