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Is it possible to reach a score of 50,000 in the game 2048 with only having a highest tile of 512(Only one 512 was present) and one 256 tile and other small numbered tiles such as 2,4,8,16,...,64.

Because my friend has done so in her phone and I am quite skeptical about it so I decided to ask.

Please do show a mathematical explanation instead of giving a direct answer

Jack M
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user3436836
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2 Answers2

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As far as I can tell, you get $n$ points for forming an $2^n$-tile (ie. $32$ points for forming a tile with $32$ on it, etc). But that doesn't mean that if you have a $2^n$-tile, you'll have a score of $2^n$. That's forgetting the amount of points you got from all the intermediate tiles you would have had to form to get that tile.

If you have a $2^n$ tile, you get $2^n$ score from that tile. But that tile was formed from two $2^{n-1}$ tiles, which would have been formed from four $2^{n-2}$ tiles, and so on. So if you see a $2^n$ tiles on the board, the total score from it and everything that went into making it is at most:

$$2^n+2\cdot2^{n-1}+2^22^{n-2}+...=\sum_{k=0}^{n-2}2^{n-k}2^k=(n-1)2^n$$

I'm summing up to $n-2$ because I'm assuming the best-case scenario where all the basic tiles (given to you by the game) were $2$s, which will maximize your score.

So the score for a $512=2^9$ is at most $8\cdot512=4096$. In the best case scenario where you end with one $512$, one $256$, and the other $14$ tiles $128$s, your score would be $16640$. So $50$ thousand certainly seems unlikely, unless there's some aspect of the scoring that I didn't take into account.

Jack M
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  • A couple typos: In your first line, you should say you get $2^n$ points for forming a $2^n$ tile. The second $n-1$ exponent should be $n-2$. Then there might be multiple $512$'s and unfortunately a board of $512$'s except with $256$'s on a main diagonal scores $124096+47*256=56320$. We need a (slightly) finer analysis to show you can't get more than $10\ \ 512$'s without having a chance to make a $1024$. You are very close. – Ross Millikan Apr 21 '14 at 02:23
  • @RossMillikan I was simply taking into account the OP's information, which said the board contained one $512$, one $256$, and "other smaller tiles". The highest possible score with only $512$s would be an interesting, but distinct problem. – Jack M Apr 21 '14 at 10:05
  • Good point. I was just thinking of the highest allowable with nothing higher than $512$ – Ross Millikan Apr 21 '14 at 13:50
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No, not possible I think. Lets count how many points we get by getting to a certain power of $2$. By my rough estimates, if you just have a bunch of $2's$, your score will be around $256$ when you get to $64$. From there, you get $512$ for two $64's$ and $512+128 < 650$ for a $128$. Then you get $650*2+256 < 1600$ for a $256$ and less than $3200+500 = 3700$ for a $512$. Then if your whole board was just filled with $512's$, your score would be less than $3700*16 = 59, 200$. Now, this is disregarding the difficulty of arranging everything...

Sandeep Silwal
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