The Best Strategy and Highest Possible Score for the "Threes!" Game.

[There's still the strategy to go. A suitably robust argument that establishes what is statistically the best strategy will be accepted.]

Here's my description of the game:

There's a $4\times 4$ grid with some random, numbered cards on. The numbers are either one, two, or multiples of three. Using up, down, left, and right moves, you add the numbers on adjacent cards to make a new card like so: $$\begin{align}\color{blue}1+\color{red}2&=3\tag{1} \\ n+n&=2n\end{align}$$ for $n=2^k3\ge 3$, where $k\in\{0, 1, . . . , 10\}$, so the highest a card can be is $2^{11}3$. But at each move the "free" cards move too and a random new card appears at a random point along the edge you slide away from. Everything is kept on the grid. The card for the next move is indicated by colour at the top of the screen: blue for $1$, red for $2$, and white for $n\ge 3$ (such that $n$ is attainable using the above process). The white $2^\ell 3$-numbered cards are worth $3^{\ell+1}$ points; the rest give no points. Once there are no more available moves, the points on the remaining cards are summed to give your score for the game.

Here's another description I've found; it's the least promotional. It has the following gif.

So:

What's the best strategy for the game? What's the highest possible score?

Thoughts:

We could model this using some operations on $4\times 4$ matrices over $\mathbb{N}$. A new card would be the addition of $\alpha E_{ij}$ for some appropriate $\alpha$ and standard basis vector $E_{ij}$. That's all I've got . . .

NB: If this is a version of some other game, please let me know so I can avoid giving undue attention to this version :)

The number on each card can be written $n=2^k3^{\varepsilon_k}$, where $$\varepsilon_k=\cases{\color{blue}0\text{ or }1 &: $k=0$ \\ \color{red}0\text{ or }1 &: $k=1$ \\ 1 &: $k\ge 2$;}$$that is, $\varepsilon_k=\cases{0 &:$n<3$ \\ 1 &:$n\ge 3$}$. So we can write $(k, \varepsilon_k)$ instead under $$(k, \varepsilon_k)+(\ell, \varepsilon_\ell)\stackrel{(1)}{=}\cases{(k+1, 1)&: $\varepsilon_k, \varepsilon_\ell, k=\ell > 0$ \\ (0, 1)&: $\color{blue}k=\color{blue}{\varepsilon_k}=\color{red}{\varepsilon_\ell}=0, \color{red}\ell=1$ \\ (0, 1)&: $\color{blue}\ell=\color{red}{\varepsilon_k}=\color{blue}{\varepsilon_\ell}=0, \color{red}k=1$.}$$

Looking at a $2\times 2$ version might help: the moves from different starting positions show up if we work systematically. It fills up quickly.

It'd help to be more precise about what a good strategy might look like. The best strategy might be one that, from an arbitrary $4\times 4$ grid $G_0$ and with the least number of moves, gives the highest score attainable with $G_0$, subject to the random nature of the game. That's still a little vague though . . .

Solution 1:

The strategy I employ is simply to make the move that leaves the most available moves, and to disregard score entirely. As a natural consequence of playing more moves, the score of the board will increase simply because the only way to continue playing is to make combinations, and combinations generate higher scores.

At the beginning of the game, the most important thing to consider is the placement of $1$s and $2$s. They are unique in that nothing can be combined adjacent to them to make a valid combination, they will only combine with their complement, which can only be achieved by board translation (verses, say a $12$, which can be adjacent to a $6$, which combines with another $6$ and then the $12$ can subsequently combine with that $12$. There's no way to make a $1$ or a $2$, it must simply be moved around the board).

Later, with higher scoring tiles on the board, the "bonus" tile (which shows up as a white tile with a $+$ in it at the top) becomes increasingly important, and the best strategy I've found is to attempt to place the bonus tile as near a mixed group of larger tiles as possible. The bonus tile will always be at least $6$, but never the same score as your highest scoring tile in play.

There is also the nature of the tile selection. It's been reverse engineered that the random generator uses a "bag" where $12$ tiles are shuffled. The original board layout uses this method, and $9$ tiles are placed into the board with $3$ remaining in the "bag". Once the bag is exhausted, the tiles are shuffled again. There are always $4$ of each: $1$, $2$, and $3$. Once you reach a high tile of $48$ a "bonus" tile is inserted with a potential value of greater than $3$. This changes the size of the "bag" to $13$ instead of $12$. So, keeping track of where you are in the "bag" and how many of each color you've seen can give you an advantage when looking at future moves.

Curiously, the possibility space for scoring is actually quite sparse. All scores will necessarily be multiples of $3$, but it turns out that only about $\frac38$ of the multiples of $3$ between $0$ and the max score are actually valid. There are a lot that are simply impossible to get, like a $19$ in cribbage.

The lowest one that isn't trivially small is still $39,363$, though, which seems well out of the range of the average player. The next lowest I found is $52,485$. There are lots of gaps at the high end, due to the fact that highest scoring tile is worth over $500$k by itself.