Even if you don't want the games to be able to refer to the last game you played, it looks like this is greatly convoluted. Why not just consider something simpler like the following?

Game A: If you have an even number of chips, gain one. Otherwise, lose two.

Game B: If you have an odd number or chips, gain one. Otherwise, lose two.

The point being that game A (or B) always leaves you with an odd (or even) number of chips, so that if you keep playing it you lose two every turn, but if you alternate, you win one every turn (after the first, possibly). For simplicity, I'm ignoring the behavior near zero chips, but this still seems to capture the essential properties of the "paradox" in a much simpler fashion.

Exactly - most of the effort was making a game that hid that simple mechanic so that it took some (admittedly interesting) math to figure out where the dependence was.

The author does say that the Parrondo's paradox only works if the games are not independent. It still is paradoxal that "A combination of losing strategies becomes a winning strategy".

It's only 'paradoxical' when the details of the games are sufficiently obscured.

If you word it as "a combination of losing strategies becomes a winning strategy" many people will be surprised and ask you to explain.

If you word it as "losing in A adds to the prize in B, so playing both beats the house" people aren't going to be impressed. note: used a simpler A/B mechanic than the blog post for illustration purposes

That's what "paradox" is: when a simple model or explanation seems to show a contradiction, and a more sophisticated model is needed to understand the situation.

But in this case the 'simple model' is less 'simple' and more 'misleading'. Something should be a paradox without trickery. Someone hiding invalid division to make 1=2 is not a paradox. "Set of sets that don't contain themselves" is a paradox.

If they were independent, it would be impossible: E(X + Y) = E(X) + E(Y) if X and Y are independent. (where E(X) = E(x | X=x) is the expected value of a random variable). This is easily provable: E(x | X = x) = \integral_{-\inf}^{inf} x f_X(x) dx, and by integration by parts:

  \integral_{-\inf}^{inf} \integral_{-\inf}^{inf} (x + y) f_X(x)f_Y(y) dx dy = \integral_{-\inf}^{inf} f_Y(x) \integral_{-\inf}^{inf} x f_X(x) dx dy + \integral_{-\inf}^{inf} f_Y(y) \integral_{-\inf}^{inf} x f_X(x) dy dx = \integral_{-\inf}^{inf} f_Y(y) dy \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} f_X(x) dx \integral_{-\inf}^{inf} y f_Y(y) dy

Since p.d.f.s f_X(x) integrate to 1 over their domain,

\integral_{-\inf}^{inf} f_Y(y) dy \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} f_X(x) dx \integral_{-\inf}^{inf} y f_Y(y) dy = \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} y f_Y(y) dy = E(X) + E(Y)

Therefore, if two games are independent, the expected loss is the sum of the expected losses. For real-valued expected losses, it is not possible to add two real numbers of the same sign and get a real number of the opposite sign, and so the 'paradox' is therefore impossible for independent games.

But you are right, for non-independent games, it doesn't seem that surprising, so it doesn't really meet the definition of a paradox.

Linearity of expectation applies to any two random variables, not just ones that are independent. The issue is that these games are not random variables, they are random processes (sequences of random variables) with state affected by the choices you make and previous outcomes. Linearity of expectation is irrelevant.

I was thinking the same thing. Is there a version that makes it a paradox?

No.

Your balance isn't really outside information. In a game like Texas Holdem Poker, your balance is a key factor in how much you bet.

But if we change the game so that "if you flip more than two heads in a row, your chances of flipping a third head are only 10%; if you flip two tails in a row, your chances of flipping a third tail are 90%", the result is the same, the odds turn more negative over time than Game A and you should switch to A after two consecutive flips.

It's not your balance that matters there, it's the amount of money you choose to bring into the game at the start. If I could boost my odds by bringing in only $498 dollars of my five hundred then I would do so every time.

Your actual balance is ridiculous to include in a game's calculations.

In texas hold'em your actual balance is a very significant factor in your optimal behaviour in a particular hand. Short-stack tactics are different to big stack tactics.

Okay, I didn't make myself clear enough. Let's say you have a thousand dollar bill in your hand. You decide to convert $600 into chips because that's a multiple of three, and then you put $400 into your room safe. Your real balance is 1000, but your balance for the game is 600. Your stack tactics are based on the 600. Your chance of winning is based on the 600 you chose to bring. Anything that counted the 400 in your room vault would be ridiculous. Yet the game in the blog would.

Yes you can; you just lose a dollar each time.