$\P(S_{X,\ell}=1)=\frac{1}{{\mathsf{N}}}\geq \frac{1}{{\mathsf{N}}^2}$.
\item otherwise, $h(X)=\ell$, then
$\P(S_{X,\ell}=1)=0$.
-But in this case, intutively, it is possible to move
+But in this case, intuitively, it is possible to move
from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{N}$). And in
$\ov{h}^{-1}(X)$ the $l$-th bit can be switched.
More formally,
At position $X$ with $i-1$ fair bits,
we do not obtain a new fair if $Z$ is one of the $i-1$ already fair bits
or if $Z$ is a new fair bit but $h(X)$ is $Z$.
-This occures with probability
+This occurs with probability
$p
= \frac{i-1}{{\mathsf{N}}} + \frac{n-i+1}{\mathsf{N}}.\frac{1}{\mathsf{N}}
=\frac{i(\mathsf{N}-1) +1}{\mathsf{N^2}}
Notice that the calculus of the stationary time upper bound is obtained
under the following constraint: for each vertex in the $\mathsf{N}$-cube
there are one ongoing arc and one outgoing arc that are removed.
-The calculus does not consider (balanced) hamiltonian cycles, which
+The calculus does not consider (balanced) Hamiltonian cycles, which
are more regular and more binding than this constraint.
In this later context, we claim that the upper bound for the stopping time
should be reduced.
