After thinking about it a little, looks like the argument works in situations like this: if a stock you have sells for $x today and tomorrow it could become 2x or x/2 with equal probability, then it is better to wait tomorrow to sell (assuming you must sell today or tomorrow) and you are risk neutral. However, for the problem that was posed, say for instance X is a random variable sampled from distribution with positive support and PDF f(.). Then you are revealed either X or 2X (you do not know which one). If you observe a quantity "z" on your first attempt, assuming my calculations are right, the probability that X was revealed will be f(z)/(f(z)+0.5f(z/2)). So the expected value of the unrevealed one will depend on z.