A and B have 10 dollars each. They bet 1 dollar each time. A wins the final game iff B has no money left. 
*

*For each bet, A has prob = 0.5 to win the 1 dollar from B. and prob = 0.5 to lose 1 dollar to B. How to calculate the probability that A wins the final game?

*How about for the initial state A has 20 dollars and B has 10 dollars , what's probability that A wins the final game?

*How about A and B still have 10 dollars each, and A has prob = 0.6 to win each bet, what's probability that A wins the final game?
Thanks!

By symmetry $A$ wins with probability $0.5$ for scenario (1). For (2), consider the probability that $A$ wins given that, at some point in time, $A$ has $k$ dollars and $B$ has $30 - k$ dollars. Define $p_k$ by
$$
p_k = P(A \text{ wins } | \: A\text{ has } k \text{ and } B \text{ has } 30-k).
$$
Conditioning on whether $A$ wins, we get
$$
p_k = 0.5 p_{k+1} + 0.5 p_{k-1}.
$$
Intuitively, if $A$ wins, $A$ will have $k+1$ dollars and $B$ will have $30 - (k+1)$ dollars and if $A$ loses, $A$ will have $k-1$ dollars and $B$ will have $30 - (k-1)$ dollars. We also have
$$
p_0 = 0, \: p_{30} = 1.
$$
Since once $A$ has $30$ dollars he wins and when $A$ has $0$ dollars they lose. This problem of solving for $p_k$ then becomes a problem of solving a homogeneous linear recurrence relation. The details for how to solve such a system can be found here. In the end, we obtain the formula
$$
p_k = \frac{k}{30}.
$$
Thus the probability $A$ wins is $p_{20} = 2/3$. For part (3), you have to instead solve
$$
p_k = 0.6 p_{k+1} + 0.4 p_{k-1}, \: p_0 = 0, \: p_{20} = 1.
$$
The details are much messier in this case and can again be found here.
