# 2048 - Maximum score that doesn't rely on luck

Inspired by this thread: Largest possible tile in 2048, I've always wondered with such games to what extent you NEED luck to be able to get to a high score. For example, the maximum tile of 131072 requires the luck to spawn a '4' tile in the last free square on the board.

What often seems to screw you over, might just imply that your strategy is sub optimal however.

So here's the questions, both assuming you don't deliberately try to fail:
1) Assuming you don't know what will spawn where, what is the first combined tile-value you cannot be guaranteed to achieve?(so sum of the 16 tile values on the board).
2) If you'd know the spawn tile numbers series ahead of time, but not their places, would that knowledge affect the results?

To point out the obvious:

• No cheating
• No 'Undo'
• Any step involving 'luck' (a 2 spawn where a 4 spawn would mean game over, or a 4 spawn where a 2 spawn would mean game over, or a spawn of 2 or 4 where that spawn in any other free square would mean game over) is not allowed. There is only said to be a 10% chance of a 4 spawning instead of a 2, but if it can happen, assume Murphy hates you.

This thread: Lowest score in 2048 theorises it's possible to get as low as 48 combined. You'd really need to not make any sensible step for this, so assume the answer to the question is higher then 48.

Note to readers: I have no exact answer to this myself. It's just a brainteaser to see what logic people can build around this.

• Have you seen sztupy.github.io/2048-Hard/index.html ? There should be no luck and you can get an idea how luck is important in 2048). Oct 3, 2014 at 11:28
• I'll look into that. If only there was a highscore page for it.. (also: it largely depends on how the 'worst position' is calculated. A set mechanic may actually be exploited somehow) Oct 6, 2014 at 6:45
• stackoverflow.com/questions/22342854/… Oct 23, 2014 at 22:45
• Actually it depends on how you play the game. If you play it foolishly, then there is no definitive answer, but if there is a perfect player, an answer may be possible. Jan 13, 2015 at 18:36