A gambler is playing a game which have a probability 0.6 to wins with \$1 and probability 0.4 to lose with \$1 The gambler will stops to play it when he have won \$4 or lost all the money.
Let $p_x$ be the probability that the gambler end it with \$4 if the gambler start the game with \$x, where $x = 1,2,3$. Using the conditional probability given the result of the first game, express $p_2$ in terms of $p_1$ and $p_3$.hence use the $p_x$ to write out the three equation three unknowns to calculate the probabilities.
I though it was suitable to use tree diagram to solve it but its not really helpful. And it seems to be a recursive probability. The question was found in a chapter of conditional but I have read some books which have some similar example but its about Finite Markov Chains. However I haven't learnt it and don't know how to do this.
Let $X_n$ be the gambler's fortune at time $n$, then $\{X_n:n\in\mathbb N_0\}$ is a Markov chain on $\{0,1,2,3,4\}$ with transition matrix $$P=\begin{pmatrix} 1&0&0&0&0\\ \frac25&0&\frac35&0&0\\ 0&\frac25&0&\frac35&0\\ 0&0&\frac25&0&\frac35\\ 0&0&0&0&1 \end{pmatrix}. $$ This is an absorbing Markov chain, since $P_{00}=P_{44}=1$. For each transient state $i$ and absorbing state $j$, let $$\tau_{ij} = \inf\{n>0:X_n=j\mid X_0=i\} $$ and set $q_{ij} = \mathbb P(\tau_{ij}<\infty)$. Then we have the system of linear equations \begin{align} q_{34} &= \frac35 + \frac25q_{24}\\ q_{24} &= \frac35q_{34} + \frac25q_{14}\\ q_{14} &= \frac35q_{24} \end{align} with solution $$\left(q_{14}, q_{24}, q_{34}\right) = \left(\frac{27}{65}, \frac9{13},\frac{57}{65}\right). $$ Since the probability of absorption is one (for intuition, try computing $\lim_{n\to\infty} P^n$), it follows that $q_{i1}+q_{i4}=1$ for each $i$. Hence $$\left(q_{11}, q_{21}, q_{31}\right) = \left(\frac{38}{65}, \frac4{13},\frac8{65}\right). $$ Let $\alpha_i$ be the expected time until absorption (that is, $\alpha_i = \mathbb E[\tau_{i1}\wedge\tau_{i4}]$), then we have \begin{align} \alpha_2 &= 1+\frac35\alpha_3\\ \alpha_3 &= 1+\frac25\alpha_2+\frac35\alpha_4\\ \alpha_4 &= 1+\frac25\alpha_3, \end{align} and hence $$\left(\alpha_2,\alpha_3,\alpha_4\right) = \left(\frac{200}{83}, \frac{280}{83}, \frac{195}{83} \right). $$