Suppose we play a coin game: we flip a coin, upon tails you get one dollar from me, upon heads I get one dollar from you. We repeat this until one of us is broke.

Now, at the beginning of the game, I have $4 and you have $6.

1. What is the chance of me winning the game? (thus ending up with $10, leaving you broke)

2. What is the expected number of turns we have to take until the game finishes? (i.e. until one of us is broke)

Intuitively I feel the answer to Q1 should be $\displaystyle \frac{4}{10}$, but I can't seem to calculate this properly. No idea how to approach Q2. I can see that $\displaystyle \mathbb{E}(T_0) = \mathbb{E}(T_{10}) = 0$ and $\displaystyle \mathbb{E}(T_4) = 1 + \frac{1}{2}\mathbb{E}(T_3) + \frac{1}{2}\mathbb{E}(T_5)$ (with $\displaystyle T_n$ being the number of turns from a game situation where I have n dollar) but I don't know how to resolve this to a definite result.