Isn't it a little sketchy to have one of the "games" be a game that takes into account outside information like your balanace? 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.
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…
Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby
#71Linearity 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.