$(v_1 \dots v_{M})$ followed by a non-zero element, then this element becomes $a-1$ at $t_{j+1}$ and hence $S(j+1)\le S(j)-1$.

Now let us investigate what happens if the opposite occurs. Then each element $a\ge 2$ in $(v_1 \dots v_{M})$ is followed by a sequence of $0$s and $1s$ such that we observe either $$\dots  a\ 0\ b\dots$$ or $$\dots  a\ 0\ 1\ \underbrace{z_3\ z_4\ \dots \ z_m}_{m-2}\ b\dots,$$ where $m\ge 2$, $b\ge 2$ and $z_i\in\{0,1\}$. Because of Corollary \[cor\_even\], we cannot have an alternating sequence of $0$s and non-zero elements; therefore, we must be able to find somewhere in the sequence of $v$s a chunk which looks either like $$\dots  a\ \underbrace{\underbrace{0\ 1}\ \underbrace{0 \ 1}\ \dots\  \underbrace{0 \ 1}}_{2l\mbox{ elements}}\ 0\ 1\ c\dots
\mbox{ where } c\ge 1 \tag{A}$$ or $$\dots  a\ \underbrace{\underbrace{0\ 1}\ \underbrace{0 \ 1}\ \dots\  \underbrace{0 \ 1}}_{2l\mbox{ elements}}\ \underbrace{0\ \ 0\ \ ?}_{\ i\ \ i+1\ i+2}\dots \tag{B}$$ where $l\ge 0$. Note that a configuration of type (A) at time $t_{j+1}$ with probability $1$ becomes a configuration of type (B). At the same time, in configuration (B), with probability of at least $\frac 13$ the $0$ located at position $i+1$ is chosen before either the $0$ 