enough meaning that the conditioning only changes the probability by a factor at most $2$. Then, we can conclude by noticing that $\ell_N=\xi_1= N-\sum_{i=1}^k \chi_k$.

The equilibrium measure in the small-bias case {#Subsec:eqSmallBias}
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We aim at showing that with large probability the density of particles everywhere is of order $k^{1+o(1)}/N$. Given $\xi\in {\Omega}^0_{N,k}$ we let $Q_1(\xi)$, resp. $Q_2(\xi)$, denote the largest gap between two consecutive particles, resp. between two consecutive empty sites. $$\begin{split}
Q_1(\xi)&:=\max\{ n\ge 1 \ : \ \exists i\in {\llbracket}0,N-n{\rrbracket}, \ \forall x\in {\llbracket}i+1,i+n{\rrbracket}, \xi(x)=0 \},\\  
 Q_2(\xi)&:=\max\{ n\ge 1 \ : \ \exists i\in {\llbracket}0,N-n{\rrbracket}, \ \forall x\in {\llbracket}i+1,i+n{\rrbracket}, \xi(x)=1 \},
 \end{split}$$ and $Q(\xi)=\max(Q_1(\xi),Q_2(\xi))$.

\[lem:dens\] For all $x\in {\llbracket}1,N{\rrbracket}$, we have $$\label{encadr}
 \frac{k}{N} {\lambda}^{x-N} \le \pi_{N,k}(\xi(x) = 1) \le \frac{k}{N} {\lambda}^{x-1}\;.$$ Furthermore, there exists a constant $c>0$ such that for all choices of $N\ge 1$, $u>1$ and $p_N\in (1/2,1]$ and all $k\le N/2$ $$\label{splam}
 \pi_{N,k}\left(Q(\xi)\ge \frac{{\lambda}^{N}  N u}{k} \right) \le  2k e^{-cu}\;.$$

We set $A_x:= \{\xi(x) = 1\}$, we first prove that for all $y\in {\llbracket}1,N-1{\rrbracket}$ we have $$\label{swiz}
  \pi_{N,k}(A_y) \le  \pi_{N,k}(A_{y+1}) \le  {\lambda}\pi_{N,k}(A_y).$$ We observe that the map $\xi \mapsto \xi^y$ defined in induces a bijection from $A_y$ to $A_{y+1}$ and that for every $\xi\in A^y$, 