Doubling strategy Let {X_n} be simple symmetric random walk. Consider a game where in
case of X_n = +1 the gambler wins and in case of X_n = -1 the gambler
loses his stake on game n. The stake on the first game is 1 euro. According
to the player's strategy, he doubles the stake after each loss and quits the
game after the first win (or if there is not enough money for doubling).
Show that a player A with finite initial capital will earn 0 (in average),
however a player B with infinite initial capital will win 1 euro.
How to define stopping times for the players A and B ?
I was actually thinking of using the Doob's Optimal Stopping theorem.
 A: Suppose the game ends at a finite time $N$ due to $X_{1},X_{2},\ldots,X_{N-1}=-1$
and $X_{N}=1$. Then, the payoff is
$$
\sum_{n=1}^{N}X_{n}2^{n-1}+X_{N}2^{N}=2^{N}-\sum_{n=1}^{N}2^{n-1}=1.
$$
Since the player B continues to play until $X_{n}=1$ and the probability
of the event $(X_{1},X_{2},\ldots)=(-1,-1,\ldots)$ occuring is zero,
the claim about player B follows immediately.

Let us now focus on claim about player A. If the player is unable
to play the $(N+1)$-th game, the payoff is
$$
\sum_{n=1}^{N}X_{n}2^{n-1}=-\sum_{n=1}^{N}2^{n-1}=1-2^{N}.
$$
If player A does not have enough money to play the first game, the
claim is trivial. Therefore, suppose that their wealth $S\geq1$.
In this case, the payoff can be written
$$
\mathbb{E}\left[1_{\{X_{\tau}=1\}}+\left(1-2^{\tau}\right)1_{\{X_{\tau}\neq1\}}\right]=1-\mathbb{E}\left[2^{\tau}1_{\{X_{\tau}\neq1\}}\right]
$$
where
$$
\tau=\inf\left\{ N\geq1\colon X_{N}=1\text{ or }S<1+2^{N}\right\} .
$$
$\tau$ is the stopping time you are looking for. You can revisit
the claim about player B from this perspective by noting that if $S=\infty$,
$\tau<\infty$ (along with $X_{\tau}=1$) occurs with probability
one.
