Tricolor columns-only Hackenbush boards: must adding the "rainbow" board eventually stabilize? This is a question about "columns-only three-color Hackenbush." Formally, a board is a finite formal sum (or multiset if one prefers) of finite strings from $\{1,2,3\}$. For $i\in\{1,2,3\}$ and a board $B$, a legal move in $B$ for player $i$ consists of replacing one occurrence of a string $\sigma\in B$ with one occurrence of a string $\tau$ such that $\tau i\preccurlyeq\sigma$. (Intuitively, a string $\sigma$ occurring in $B$ represents a "stack of edges" labelled according to $\sigma$ from the ground up.) As in an earlier MO question of mine, define the "basic type" of a board $B$ to be the set $bt(B)$ of pairs $(x,A)$ such that $x\in \{1,2,3\}$, $A$ is a nonempty proper subset of $\{1,2,3\}$, and there are strategies $\pi_a$ for $a\in A$ which when followed ensure that the first player to be unable to move will not be in $A$ assuming player $x$ starts and move ordered proceeds in the obvious cyclic order from then on. (Intuitively, $(x,A)\in bt(B)$ means "team $A$ can avoid losing in $B$ if $x$ goes first.")
Now let ${\bf z}=\langle 1\rangle+\langle 2\rangle+\langle 3\rangle$ be the "rainbow" board. Interestingly, and in contrast with the two-color version, adding ${\bf z}$ does not preserve basic type. For example, we have $(1,\{3\})\in bt(\langle 123\rangle)$ but $(1,\{3\})\not\in bt(\langle 123\rangle+{\bf z}$). Essentially, adding a copy of ${\bf z}$ lets players $1$ and $2$ swap roles for one round, and that helps them beat $3$ if $1$ goes first.
Despite this, I'm curious whether adding ${\bf z}$ does eventually preserve basic type:

For a board $B$, is there necessarily a natural number $k$ such that $$bt(B+m{\bf z})=bt(B+n{\bf z})$$ whenever $m,n\ge k$?

Here $aC=C+C+C+...+C$ ($a$ times) for $a\in\mathbb{N}$ and $C$ a board. I would also be interested in the answer to the analogous question for $p>3$ players.
 A: Yes, the basic type of $B+n{\bf z}$ always stabilizes.
For teams $A$ of size $2$, adding a copy of ${\bf z}$ is always weakly beneficial to $A$. To be precise, for all $x\in\{1,2,3\}$, if $(x,A)\in bt(B)$, then $(x,A)\in bt(B+{\bf z})$. This is because team $A$ can ignore the ${\bf z}$ component, using their winning strategy for $B$ part, and respond to the third player playing in ${\bf z}$ by playing in ${\bf z}$ for the next two turns.
For the same reason, for teams of size $1$, adding a copy of  is never helpful, since the opponent is a team of size $2$.
In either case, we have shown that the membership status of $(x,A)$
in $bt(B+n{\bf z})$ as $n\rightarrow\infty$ must converge.

In fact, this is true for any number of players:
Let $p$ be the number of players, let $x\in\{1,\dots,p\}$, and let $A$ be a nonempty proper team. Divide the game into rounds, where each round is a sequence of $p$ consecutive moves starting with the initial player, $x$. For a game played on $B+n{\bf z}$, call a round boring if all $p$ moves of that round are played in one of the ${\bf z}$ components. Under optimal play, you can assume that all boring rounds happen at the end of the game. Indeed, suppose there is an optimal line of play in which round number $i$ is boring, but round number $i+1$ is interesting. Suppose team $A$ was the first team in round $i+1$ to make a move in the $B$ component; then team $A$ could have just made that same move in round $i$. If you repeat this logic, you can move all boring rounds to the end.
Now, let $e$ be the total number of edges in $B$. There can be at most $e$ interesting rounds in any game of the form $B+n{\bf z}$. Therefore, for any $n>0$, optimal play on $B+(e+n){\bf z}$ is identical to optimal play on $B+e{\bf z}$, followed by $n$ boring rounds. This means they have the same outcome, so that adding $e$ copies of ${\bf z}$ is sufficient to stabilize the basic type.

Finally, even when $p=3$ there do exist boards such that the time you have to wait for stabilization is arbitrarily high. For any $n,k\in \mathbb N$, consider the board
$$
\langle1^n2^n\rangle+\langle3^{n-1}\rangle+k{\bf z}
$$
with players $1$ and $2$ teamed up, with $1$ going first. In order to avoid losing, player $2$ has to make $n$ separate moves in the $\langle1^n2^n\rangle$ component. Since any move by player $1$ in $\langle1^n2^n\rangle$ would destroy all of $2$'s potential moves, player $1$ needs to make $n$ moves in the ${\bf z}$ components. That is, the team can win if and only if $k\ge n$.
