[argv_empty]

On top of that, the argument rests on the kind of error you expect to see freshmen making in a discrete math course:

> The basic argument is as follows. For simplicity but without loss of generality, assume there are n past price changes, each of which is either UP (1) or DOWN (0). How many possible trading strategies are there? Suppose we allow a strategy to either be long, short, or neutral to the market. Then the list of all possible strategies includes the strategy which would have been always neutral, the ones that would have been always neutral but for the last day when it would have been either long or short, and so on. In other words, there are three possibilities for each of the n past price changes; there are 3n possible strategies.

There are 2^n possible histories of length n. If a strategy maps each history to one of three positions, there are 3^(2^n) strategies that consider n bits of history.

[kizer]

Why wouldn't it be 3^n? Edit: wait, it's because bitstring -> strategy is a function, right?

[argv_empty]

Yes. Mapping from 2^n bitstrings to 3 market positions.

[kizer]

Thanks.