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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

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    It seems like a Bayes' rule problem essentially: if $W_1$ is "you won the first toss" and $W_\infty$ is "you won the game" then $P(W_1 \mid W_\infty)=\frac{P(W_\infty \mid W_1) P(W_1)}{P(W_\infty)}$. Now $P(W_\infty \mid W_1)$ is just $1$ because of the betting strategy, $P(W_1)=1/2$, so I think your problem boils down to showing that $P(W_\infty)=0.8$. I think you can infer that from the optional stopping theorem (what is $E[X_\infty]$, and what states is $X_\infty$ concentrated on?) – 2017-02-15
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    I see, thanks! But how would I apply this to describing the distribution of the of the whole sequence $X_1, X_2, ...,X_n$ conditioned on the fact that the gambler reached 10 dollars/won the game? Would I populate the transition matrix similar to the one I provided above with percentages $5/8$ and $3/8$ – 2017-02-15

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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: $E[X_\infty]=E[X_0]=8=0 \cdot P(W_\infty^c) + 10 \cdot P(W_\infty)$.

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    I see, thanks! But how would I apply this to describing the distribution of the of the whole sequence $X_1, X_2, ...,X_n$ conditioned on the fact that the gambler reached 10 dollars/won the game? Would I populate the transition matrix similar to the one I provided above with percentages $5/8$ and $3/8$ – 2017-02-15
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    @zhou8910 The win rates conditional on winning in the end will not be homogeneous; try computing them the same way that we computed the probability of winning the first game given that you win the game in the end and start with \$8. (Things may get a bit complicated in the case where you begin with \$4 and the case where you begin with \$2 because then $P(W_\infty \mid W_1)$ is not $1$.) – 2017-02-15
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    I'm not exactly sure about the inhomogeneity of the probabilities. How would one even calculate the conditional when you begin with $4, $6, or $2? And what would the transition matrix look like? I thought the transition matrix described moving from one state to another at each time step -- how could you condition this based on the even we calculated earlier? – 2017-02-15
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    At \$6 it's easy, you just do the same thing over again, because if you win the first game starting from \$6 then you win the game. Try doing the \$4 case. You'll still want to use Bayes' rule but the difficulty will be that $P(W_\infty \mid W_1)$ will not be $1$. (But it might be something you've calculated already...) – 2017-02-15
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    @zhou8910 By the way, there is an important conceptual point here, which you should at least convince yourself about, if not actually prove: the process conditioned on winning in the end is still Markov. This might be surprising given that there is some "dependence on the future" built into how we describe it, so you should think about this. – 2017-02-15