Is the 2048 game zero sum? I have been asked to consider whether The 2048 game is a zero sum game.
According to wikipedia:

each participant's gain or loss of utility is exactly balanced by the
  losses or gains of the utility of the other participants. If the total
  gains of the participants are added up and the total losses are
  subtracted, they will sum to zero

Assuming so called 'player 2' is adversarial (always placing the new '2' and '4' tiles in the worst possible position) what factors must be considered to prove or disprove the game as zero-sum?
Is it as simple as win/lose(i.e. 1 + -1 = 0)?
What about the points that player 1 accumulates?
Perhaps the utility of player 2 is the inverse of the points player 1 accumulates? (e.g. 100 points for player 1 is -100 for player 2. 100 + -100 = 0)
Questions: 


*

*Is it zero sum?

*How might we draw this conclusion?


The question is ill-posed. A zero-sum game requires two players, with payoff functions. So the answer boils down to establishing whether you want to assume that the algorithm choosing tiles is a player with payoffs opposed to yours, or not. 
I suspect that you have been asked something different; that is, how to model "2048" as a zero-sum game. Start by assuming what the payoffs for Player~1 should be: one possibility is to assign $+1$ when he reaches 2048 and $0$ when it fails; another possibility is to assign a payoff equal to his final score. Now assume an opponent that chooses tiles and has payoffs opposite to yours. It is a whole different matter to write down strategies (theoretically possible, but extremely cumbersome, I suspect). 
