Bankrupt

This is a pretty straightforward Markov chain problem. There are 4 states. The transition graph is given in Figure 1.

The problem starts at state 1. As has been explained in the lessons of this course, we use the following equation:
\begin{equation}
s_1 = \sum_{i=0}^{3}p_{1,i}s_i
\end{equation} \begin{equation}
s_2 = \sum_{i=0}^{3}p_{2,i}s_i
\end{equation} Furthermore, $s_0=0$ and $s_3=1$. Then we have
\begin{equation}
s_1 = \frac{1}{3}*0 + \frac{2}{3} * s_2
\end{equation} \begin{equation}
s_2 = \frac{1}{3}* s_1 + \frac{2}{3} * 1
\end{equation} Solving these equations gives us $s_1=4/7$ and $s_2=6/7$. So, starting with 1 dollar, player A has a 4/7 chance of winning.

Proof
If we substitute Equation 4 in Equation 3, we have

\begin{equation}
s_1 = \frac{1}{3}*0 + \frac{2}{3} * (\frac{1}{3}* s_1 + \frac{2}{3} * 1)
\end{equation} \begin{equation}
s_1 = \frac{2}{9} * s_1 + \frac{4}{9}
\end{equation}  \begin{equation}
\frac{7}{9} * s_1 =  \frac{4}{9}
\end{equation}  \begin{equation}
s_1 =  \frac{4}{9} / \frac{7}{9} = \frac{4}{7}
\end{equation}