Suppose $X_n$ is the fortune of a gambler after $n$ th game. Then the game is called fair (Breiman 1968) if
$$E[X_{n+1} \mid X_1, \dots, X_n] = X_n \forall n$$
My question is why a fair game is not defined as the following
$$E[X_{n+1}] = E[X_n] \forall n$$ i.e. $$E[X_{n+1}- X_n]=0$$. This should be the proper definition as a fair game is where avg. gain is zero. Nothing conditioning should be there.
$$\operatorname{E}[X_{n+1}\mid X_1,X_2,\ldots, X_n] = X_n\;, \forall n$$
That's an ... interesting definition. It says the game's odds can be adjusted given knowledge of the gambler's past fortune, but will be considered fair if that adjustment always means that the gambler's fortune can be expected not to change after each game (whatever it may be at the time).
Of course, since $\operatorname{E}[Y] = \operatorname{E}\!\left[\operatorname{E}[Y\mid X]\right]$ then the familiar definition follows:
$$\begin{align}\operatorname{E}[X_{n+1}] & = \operatorname{E}\left[\operatorname{E}[X_{n+1}\mid X_1,X_2,\ldots, X_n]\right] \\ & = \operatorname{E}[X_{n}]\;, \forall n \end{align}$$
Which is an equivalent, but somewhat weaker, statement that a game is fair if the gambler's expected fortune does not depend on how many games are played.