This is a standard Markov chain problem. The transition matrix of the problem is $$ T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 1\\ \end{bmatrix} $$ Let us denote the state where Player A has £$i$ as $a_i$ Recall from the Markov chain section the formula for the probability of absorption into a particular state. $$ a_i = \sum_{k} a_{k} p_{ik} $$ This gives us four simultaneous equations. $$ a_3 = 1, $$ $$ a_2 = \frac{2}{3} a_3 + \frac{1}{3} a_1, $$ $$ a_1 = \frac{2}{3} a_2 + \frac{1}{3} a_4, $$ $$ a_4 = 0. $$ We need to solve these equations for $a_1$, our starting position. Solving gives $$ a_1 = \frac{4}{7} $$ The probability of Player A winning is $\frac{4}{7}$.