2048 - Scoring (Part 2)

In Part 1 we talk about how to estimate the score of your game based on a single snapshot. However, we hadn't yet corrected for the fact that $4$'s sometimes spawn. So, to compensate for the number of $4$'s that spawn, we need to know how often they spawn! The online, open-source version has a $10\%$ chance of spawning a $4$, but based on nearly $38,000$ (No, I didn't count them all, I'll show you how I came about this in a little bit.) spawns on my iPhone app version, there is about a $15.3\%$ chance that a $4$ will spawn (leading me to believe that it's supposed to be a $15\%$ chance).

With that in mind, we need to have an estimate for the total number of tiles spawned. And under our old assumption (that only $2$'s spawn), we can compute that very easily. Consider the following. Every $2$ tile that spawns counts as a spawn. Therefore, every $4$ tile that we have is really two total tiles that spawned (the two $2$ tiles). The $8$'s are 4 spawns (two $4$ tiles each with two $2$ tiles spawning). It doesn't take long to figure out that each tiles has $$c_a = a/2$$ where $a$ is the face value of the tile. Thus we just sum this for each of the $16$ tiles on the board and we have a good estimate for the number of total tiles that spawned, $C_e$ $$C_e = \sum_{16\;Tiles}c_a$$ Well, we just assumed that all the spawns were number $2$ tiles. To compensate for the fact that $4$ tiles can spawn at a probability of $p_4$ we know the following two relationships: $$p_4 = \frac{C_4}{C_a}$$ $$C_a = C_e - C_4$$ where $C_4, C_a,$ and $C_e$ are the number of $4$ tiles spawned, the actual number of total tiles spawned, and the estimated total tiles spawned, respectively. The first relationship is fairly straight forward; it's how the probability of a four-spawn is defined. However, the second relationship involves a little more thought. We guessed, based on only $2$ tiles spawning that there are $C_e$ total spawns. If we actually know that there are $C_4$ four-spawns, then we have to account for the number of two-spawns that would've smashed into these $4$'s. So, since there are two $2$'s that make a $4$ we have $C_2 = C_e - 2C_4$ two-spawns. To get the actual number of spawns we add the two-spawns to the four-spawns $$C_a = C_2 + C_4 = (C_e - 2C_4) + C_4 = C_e - C_4$$ Plugging the second equation into the first and solving for $C_4$ we find out that $$C_4 = \frac{p_4C_e}{1+p_4}$$ Lastly, since each of these $4$'s that spawned does not give us 4 points (we assumed we got the points for all $4$ tiles), our expected score ($S_x$) is the estimate score ($S_e$) minus four times the number of $4$'s that spawned: $$S_x = S_e - 4C_4 = S_e - 4\frac{p_4C_e}{1+p_4}$$ Since we know $p_4$ and we know how to find $S_e$ (See Part 1) and $C_e$, we can compute our expected score. I played approximately 30 games with this and the average relative error for these guesses is about $0.2\%$. The results were even better once you scored above $10,000$. Only averaging the games with scores higher than $10,000$ (which was all but four of them), the average relative error was about $0.1\%$. Pretty cool results!

I said I would tell you how I came about the probability of a four-spawn of $15.3\%$. So I played a bunch of games and compared the actual score of the game to what I estimated it would be if there were only $2$ tiles that spawned. The difference between these two divided by four is the actual number of $4$ tiles that spawned. With that I used the two equations above and calculated what $p_4$ was!

There are a few more questions about 2048 that I still have. What is the highest score one can get? What is the lowest score? These questions (and others if I think of more) will be discussed in Part 3.