Let us have $a<0<b$, and $k$ random walks (games). Given $j\in\{1,\dots,k\}$, let $X_1^j,X_2^j,\dots$ be discrete random variables with identical distributions (but can be different for different $j$), such that $X_i^j$ represents the gain/loss at round $i$ in game $j$. Let $S_i^j=\sum_{k=1}^iX_k^j$. Let $T_n^j$ be a stopping time such that either $S_{T^j}^j\leq a$ or $S_{T^j}^j\geq b$. For all $i,j$ we have the following: $E(T^j)\leq c(b-a)^2$ for some constant $c$ (not depending on $a,b$), $E(X_i^j)=0$, $Var(X_i^j)>0$, and all the variables $X_i^j$ are mutually independent.
Let us now consider the following game: At each time $i$ we choose one of the $k$ games to play, and we play this chosen game once before moving to the next time step. Mathematically speaking, let $X_1,X_2,\dots$ and $G_1,G_2,\dots$ be random variables, s.t. $X_i$ represents the effect at round $i$, and $G_i$ represents the game chosen at step $i$. Then it holds that $X_i=X_i^{G_i}$, and the total gain/loss at step $i$ is $S_i=\sum_{k=1}^i X_k$.
We can now define a stopping time $T$ as $S_{T}\leq -a$ or $S_T\geq b$.
The question is: Does there exist some constant $d$, that does not depend on $a,b$, such that $E(T)\leq d(b-a)^2$ regardless of our strategy?
Intuitively it should hold, but every attempt I made to show this has failed thus far... If $k=1$ them it holds trivially. I can also solve it if each of the $k$ games has only two possible outcomes, where in game $j$ at each step we either gain $a_j$ with probability $p_j$ or lose $\frac{a_jp_j}{1-p_j}$ with probability $1-p_j$ (mean is zero). In such case we can "simulate" each of the $k$ games by a game that gives +1/-1 with equal probabilities (one step of game $j$ is "flipping fair coin until we reach either $a_j$ or $-\frac{a_jp_j}{1-p_j}$"), and thus the choice of which of the $k$ games we chose does not matter as it can be viewed as always flipping the fair coin. Thus in this case the expected number of steps is $E(T)\leq d(b-a)^2$ for some $d$. However when the structure of the individual games is more complicated, for example when $X_i^j$ can take on multiple values (potentially even infinitely many) I have no idea how to show this...
edit: I managed to show this holds if there exists $m$ such that each $|X_i^j|<m$ using martingales and stopping times, however the case with unbounded $|X_i^j|$ still eludes me. In the bounded case we can construct a martingale $M_i=(S_i)^2 - \sum_{j=1}^i Var(X_j\mid F_{j-1})$, then show that $Var(X_j\mid F_{j-1})$ is always bounded from below by a constant to argue that $E(T)<\infty$ in order to apply optional stopping theorem, from which we get the result. The problem with unbounded $|X_i^j|$ is that we cannot bound $S_T$ with a finite value. And the idea of "bounding too large steps" from Expected stopping time in bounded random walk with unbounded steps does not seem usable, as the bounded random variables may satisfy $E(X_i^x)<0$ while $E(X_i^y)>0$, thus we cannot construct the submartingale/supermartingale...