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 only half in the market, your expected value grows at 1.125^n.
But if you have a log utility function, the decreased variance is helpful because then your expected utility grows each period rather than staying flat. (With 100% exposure, your EV of one period is .5*log(2)+.5*log(.5), which is obviously 0, but with 50% exposure your EV of one period is .5*log(1.5)+.5*log(.75), which is positive.) If you have a log utility function, 50% exposure happens to maximize your growth in expected return.
(I do agree with you that the link saying that the offer is a "wash" without bringing in the log utility function, or the tradeoff between variance and expected return, is bad, but those are somewhat subtle issues that they might not want to introduce along with the game.)
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.
That illustrative example highly depends on the number of heads being exactly equal. If the number of heads and the number of tails differed even slightly, the result would not be the same amount that you started with, and the fact that the ratio of heads to tails was close to 50% would not affect that. If you had 100 heads and 101 tails, you'd end up with half as much as ...