A gambler has 8 dollars and wants to increase it to 10 dollars in a hurry. He can repeatedly stake money on the toss of a fair coin; when the coin comes down on tails, he loses his stake and when it comes down on heads, he wins an amount equal to his stake along with his original stake being returned.
The gambler decided to take on a strategy in which he stakes all his money if he has less than 5 dollars and otherwise stakes just enough to increase his capital to 10 dollars if he wins.
Question: Show that the probability that he wins the first coin toss, given that he eventually reaches 10 is 5/8 and extend this to describe the distribution of the whole sequence $X_0, X_1 , X_2,...$ where $X_n$ is his money after the nth coin toss.
I have already solved for the transition matrix as the following:
$$ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 & 0 \\ 1/2 & 0 & 0 & 0 & 1/2 & 0 \\ 0 & 1/2 & 0 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 1/2 & 0 & 1/2 \\ 0& 0 & 0 & 0 & 0 & 1\\ \end{matrix} $$
where the state space is $I = {0,2,4,6,8,10}$ but do not know how to proceed from there
Let $W_1$ be "you win the first toss" and $W_\infty$ be "you win the game". By Bayes' rule:
$$P(W_1 \mid W_\infty)=\frac{P(W_\infty \mid W_1) P(W_1)}{P(W_\infty)}.$$
Because of the choice of betting strategy and the initial condition, $P(W_\infty \mid W_1)=1$. Fairness gives $P(W_1)=1/2$. So the desired result is equivalent to showing $P(W_\infty)=4/5$. This can be shown directly by computing the limiting distribution of the chain (find all invariant distributions, then select the one corresponding to your initial distribution). It can be more easily shown using the optional stopping theorem: $E[X_\infty]=E[X_0]=8=0 \cdot P(W_\infty^c) + 10 \cdot P(W_\infty)$.