Solvability of jenga

Question

I was playing Jenga, and a thought occurred to me: Assuming that layers with only the center piece missing and layers with only the center piece remaining are effectively "dead", as in they can no longer be tampered with, and that all players are capable of perfectly pulling out pieces from non "dead" layers, is the game of Jenga solvable for, say, $2$ players?

(Also assuming that the top $3$ layers cannot be tampered with, and that the Jenga tower is the standard $18$ layers)

I've been thinking about this for a bit and I would appreciate any insight or help, since it seems that both players have the ability to make layers "dead" layers up to a certain turn

edit: I neglected to mention that as the blocks are being pulled, the players stack the pieces on top with them having to put their piece on the newest uncompleted layer, creating new layers, as per normal Jenga rules

Answer 1

Answer: This idealized version of Jenga is a win for the second player.

Let us denote a position as an ordered pair $(f,h)$, where $f$ is the number of full layers, and $h$ is the number of half layers (meaning layers with a center and one side piece remaining). However, we adopt the convention that $f$ is a fraction whose fractional part is either $0,\frac13$ or $\frac23$, where the fractional part denotes the proportion of the top layer which is complete. A typical position is $(5\frac23,3)$, meaning there are $5$ full layers, $3$ half layers, and two pieces in the top layer. If you move on one of the half layers, this position becomes $(6,2)$, since a half layer gets removed, and the top layer gets an extra piece, so there are now six full layers.

Since you cannot touch the top three layers, at any point in the game, the second and third layers from the top will always be full. Therefore, any legal position has $f\ge 2$. In fact, it is always true $f\ge 2\frac13$, since in the case where $f$ is an integer, there will be three full layers on top.

The available options from a position $(f,h)$ are as follows:

$$ \begin{cases} (f-\frac23,h+1) & \text{if $f>3$ (remove end piece from full layer)} \\ (f-\frac23,h) & \text{if $f>3$ (remove center piece from full layer)} \\ (f+\frac13,h-1) & \text{if $h\ge 1$ (remove piece from half layer)} \end{cases} $$ Note that if $f=3$, then the only full layers are the top three layers, so taking from a full layer is illegal.

At this point, it is a simple matter to compute the outcome of the initial position $(18,0)$ using a recursive program. Call a position winning if the player whose turn it is can win. A position is winning if and only if it has at least one losing option, where there outcome of the three options are computed recursively. It turns out that $(18,0)$ is a losing position. Moreover, I observed a simple pattern to the outcomes:

The outcome of $(f,h)$ is the same as the outcome of $(f+3,h)$ and $(f,h+3)$, for all $f\ge 2\frac23$ and $h\ge 0$.

This should be tediously provable by induction. This means that the table of outcomes is a repeating pattern of the $27$ outcomes for $2\frac23 \le f\le 5\frac13$ and $0\le h\le 2$, which are as follows:

$h$ \ $f$	$2\frac23$	$3$	$3\frac13$	$3\frac23$	$4$	$4\frac13$	$4\frac23$	$5$	$5\frac13$
$0$	L	L	W	W	W	L	L	W	W
$1$	W	L	L	W	W	W	W	W	W
$2$	W	W	W	W	L	L	W	W	W

This completely determines the outcome of any position of Jenga, except for the three positions $(2\frac13,0),(2\frac13,1)$ and $(2\frac13,2)$, which are respectively losing, winning, and losing.