method is to get results for random words, and to transfer them to random Lyndon words via lemma \[usarg\].

Let $w$ be a word in $\mathcal A^n$. We denote by $N_n(w)$ the number of runs in $w$, by $X_1(w),X_2(w), \dots,
X_{N_n}(w)$ their lengths, by $M_n(w)=\max \{ X_i(w), \; 1 \leq i
\leq  N_n(w)\}$ the length of the longest run in $w$ and by $M_n^a(w)$ the length of the longest run of “a” in $w$.

$n=9$

                         $N_6 $   $(X_i)$   $M_6$   $M_6^a$
  --------------------- -------- --------- ------- ---------
      $w=aabbbbaaa$       $3$     $2,4,3$    $4$      $3$
   $\rho(w)=aaaaabbbb$    $2$      $5,4$     $5$      $5$

\[Nn\] For every $\gamma>0$, there exists a constant $C_\gamma>0$ such that for any $\varepsilon>0$, $$\begin{aligned}
{\tilde{\mathbb P}_n}\left( \left| N_n-\frac{n}2 \right| \ge \gamma n^{-1/2+\varepsilon}
\right)
& \le &  O \left( \exp \left( -C_\gamma n^{2\varepsilon}
  \right) \right).\end{aligned}$$

**Proof.** First, we prove the above inequality for the probability measure ${\mathbb P_n}$ on the set $\mathcal A^n$ of words with length $n$. The cardinal of the event $\{N_{n}=k\}$ corresponds to the number of compositions of the integer $n$ with $k$ parts (see Andrews [@Andrews] and Pitman [@Pitman]): $$\forall k \in \{1,2, \dots, n\}, \; {\mathbb P_n}(N_{n}=k)= \frac{1}{2^{n-1}}
\binom{n-1}{k-1}.$$ Thus $N_{n}-1$ is a binomial random variable with parameters $(n-1,1/2)$, whose 