# Worst case scenario in 2048

The game 2048 can be thought as a two player game, where one player shifts the table around to get to the 2048 tile, while the other player tries to hinder them by putting down 2s and 4s on the table. While the original game will put down the tiles randomly there are variants, such as some where this is not random, and also some where you are the one putting down the tiles, and the computer will shift the table around (not to mention the general one, where you can play both sides).

Theoretically, the worst situation is the following table, but for this the player shifting the table has to play really badly.

 2 | 4 | 2 | 4
---+---+---+---
4 | 2 | 4 | 2
---+---+---+---
2 | 4 | 2 | 4
---+---+---+---
4 | 2 | 4 | 2


If however both players play flawlessly against each other, then what is the highest tile that the shifting player can achieve?

• Hard to define "perfect" here... – kBisla May 15 '14 at 0:30
• @BlueFlame I don't think it's really ambiguous; ideal strategy, like any other game. – Kevin May 15 '14 at 1:26
• How many 2s and 4s are allowed to be put in a turn? Your last link allows the table to be filled as per your worst case example. – SQB May 15 '14 at 8:11
• @SQB: only one per turn. The last example is just a sandbox – SztupY May 16 '14 at 13:13

## 1 Answer

When playing deterministic 2048, where a new tile is placed in the first available spot, you can get a pattern like this using left-down-left-down:

The worst I have been able to get in regular 2048 is this:

It is possible to get that worst-case scenario if your tile generator and human are working together to do so.

• That a really large images :O – martijnn2008 May 15 '14 at 22:53
• Sorry. I took them on a netbook, so i was expecting them to be smaller. – TheDoctor May 15 '14 at 23:56