Let (X, Y) be an ordered pair of real-valued random variables. Say that (X, Y) is fair if E(Y | X) = X a.s. It is shown, for example, that if X has a finite mean and the pair (X, Y) is fair, then X and Y cannot be stochastically ordered unless X = Y a.s. The conclusion is in general false, if X does not have a mean. On the other hand, if X is independent of the increment Y - X, the preceding statement remains in force without any moment restrictions on X. The last assertion, combined with a gambling idea of Dubins and Savage, yields a simple proof of a theorem of Chung and Fuchs on the upper limit of a random walk with mean zero.