Imagine that the bank has the money $M_1$ and the player has the money $M_2$. The rules are the following: You win with a chance of $\frac{17}{36}$ and lose with $\frac{19}{36}$ each round. Now you are allowed to bet as much as you want. If you win, you get twice your money back and otherwise your money is gone. The question is: What is the probability of going bankrupt in this game?
I heard that the probability is surprisingly small to win this game, so I wanted to have a mathematical answer.
An answer depends also on your ultimate objective as well as your bet sizing and they are dependent. The objective could be to play until you have doubled your money, play until your adversary is bankrupt, etc.
For comparison, if you bet one $1$ unit each round you can use the Gambler's Ruin solution where $M_1+M_2$ is your target:
$$P_{ruin} = \frac{(q/p)^{M_1+M_2}-(q/p)^{M_2}}{(q/p)^{M_1+M_2}-1}$$
Here $p = 17/36$ and $q = 19/36$. If you bet $x$ units each round you can use the same formula by replacing $M_1$ and $M_2$ with $M_1/x$ and $M_2/x$, respectively.
Unless your capital is comparable to that of the bank you will find that the probability of ruin is effectively $1$. The optimal strategy to double your money is to bet everything at once with unfavorable odds.
With favorable odds look at the Kelly criterion.