Title: Optimal algorithm to lose game 2048
Tags: algorithm
Question: By now, especially after this post and other similar internet resources, I guess most people have figured out how to easily win at Gabriele Cirulli's game 2048: even manually, by observing simple rules, reaching 2048 is not a problem.
However, losing at this game seems far more challenging than winning! As much as I try, the minimum tile I got so far was 16. It seems to me that losing depends on chance much more than winning. Is there any strategy that can guarantee to lose with no tile more than 8?
(Of course, some of the hints suggested here might help, such as calculating all possible moves for n steps and choosing the combination of moves that maximises the probability to get the tiles stuck and end the game. But is there a more logical principle to obtain that?)
In the luckiest case, you would alternate 2s and 4s; alternating 2s, 4s and 8s should be easier. Actually I have just made it with five 8s, seven 4s and four 2s.
Comment: It seems rather unlikely that you can avoid getting an 8 unless you allow for ideal tile placement (rather than random). Otherwise it seems like it's far more likely that random tile placement will restrict your movements to two movement directions, at which point it becomes likely that the final two directions will be cutoff as well.

Comment: This question appears to be off-topic because it's not about programming, per se. You could try asking on [programmers.se], which is more about algorithms and conceptual questions that don't actually contain code.

Comment: @lostsock [codegolf.se] is for programming contests / challenges, not for questions about programming. I suggest you read [our Help Center article about what's on topic](http://codegolf.stackexchange.com/help/on-topic).

Comment: Intuitively I don't think you can guarantee losing under that limit because 2s and 4s come at random and there are too few 4s (1 in 10 if I remember correctly.)



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