That blog post describing the scheme starts out
For a player who gambles his entire bankroll each round, it appears to be a wash. No matter the order of returns, if there are an equal number of heads and tails, the player ends up having exactly as much as he did at the start.
If the sequence of coin flips has an equal number of heads and tails, you wouldn't need any complicated scheme to win--you could just bet $0 on everything except the last flip, and you would know with 100% certainty what the result of the last flip would have to be to produce equal numbers, so you'd bet everything on it. This would even work if the win and loss payoffs are equal numerically instead of equal in percent.
I don't see why anyone would postulate that the sequence of coin flips contains an equal number of heads and tails unless they are confusing "as you flip a lot of coins, it gets closer to 50% heads and 50% tails" (true) with "as you flip a lot of coins, the number of heads gets closer to the number of tails" (not true).
This doesn't give me high confidence for the rest of that link. (My first suspicion is that the whole thing actually amounts to a proof that this type of random walk is nonexistent.)
If the sequence of coin flips has an equal number of heads and tails
It only does in expectation; the underlying process is a martingale. They're using an illustrative example to show you that investing everything in that random walk leads to a modal expectation of having the same at the end as you do at the beginning.
But that's an expected value of 0 in log terms; the expected value in linear terms of course follows 1.25^n, where n is the number of flips. Shannon's Demon reduces the variance in return at the price of reducing the mean return. If you're ...