Puzzle about Video Game Taipan

updated below – Weds 8pm

Here’s a question my brother asked: There is a video game called Taipan. It can be played as a 2-player game, where the player with the most money at the end wins. You make money by buying and selling opium in Hong Kong (you start the game with some money). The price of opium fluctuates based on some random process (which, if the player plays a bunch of times, he may be able to approximate). You have a storage facility in Hong Kong — when you feel the price is low, you buy opium and store it. When the price is high, you sell it and free up space in your storage facility (which you’ll use for future purchases).

Question: If you want to maximize your probability of winning, could there ever be a circumstance (and if so, what kind of circumstance?) where the optimal strategy involves selling a portion (greater than 0% but less than 100%) of the opium in your storage facility? Or, alternatively, should you always either sell all or none of your opium?

UPDATE: P gave a solution which looks good. Now, consider this different question:

Instead of playing it as a 2-player game, you play it as a 1-player game, and your opponent plays it as a 1-player game, separately, out of sight from you. Opium prices in your opponent’s game follow the same random process as in your game, although they are not equal to, or even correlated with, the prices in your game. In that case, assuming you want to maximize the probability of beating your friend, should you ever sell only a portion of your opium?

I am told by my brother that Taipan actually is a 1-player game, so this followup question is the more relevant one, and the one I should have asked to begin with!