Here's another "paradox":
Game A: You lose a dollar every time.
Game B: If the last game you played was game A, you win a million dollars. Otherwise, you lose a dollar.
AAAAAAA... loses, BBBBBBB... loses, but ABABABABA... makes you rich!
Suddenly it doesn't seem so paradoxical to me.
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.
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
\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
\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.
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.
Your actual balance is ridiculous to include in a game's calculations.