Lothaire [@Lothaire]), we are lead to: $$card(\mathcal L_{n}) =
\frac{2^{n}}n(1+O(2^{-n/2}))\mbox{ and }card(\mathcal N_{n})=O(2^{n/2}).$$

In the sequel, we will consider the two following probability spaces:

-   the set $\mathcal A^n$ of words with length $n$, endowed with the uniform probability ${\mathbb P_n}$,

-   the set $\mathcal L_n$ of Lyndon words with length $n$, endowed with the uniform probability ${\tilde{\mathbb P}_n}$.

The probability measure ${\tilde{\mathbb P}_n}$ can be seen as the conditional probability on $\mathcal A^n$, given $\mathcal L_n$. The next lemma is obvious, but it is very useful in our proofs because it allows to transfer results on random words to random Lyndon words by neglecting non-primitive words and using the shepherd’s principle.

\[usarg\] For $A \subset \mathcal L_n$, we have: $$\mid \tilde {\mathbb P_n}(A)- {\mathbb P}_{n}({\rho}^{-1} (A)) \mid \le O(2^{-n/2}).$$

**Proof.** It is sufficient to note that ${\tilde{\mathbb P}_n}(A)={\mathbb P}_{n}({\rho}^{-1} (A)|\mathcal P_n)$ and ${\mathbb P_n}(\mathcal P_n)=1- O \left(2^{-n/2} \right).$

Number of runs and length of the longest run {#run}
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This section deals with the number of runs and the length of the longest run in a random Lyndon word. The results exposed in this section are not new, but are presented in a convenient way for our proofs. The 