Problem involving a probability distribution and random variables A speculator invests in the currency markets. Let $X_n$ denote the amount of money in her account at the end of day $n$. Her daily gains, $J_n = X_n - X_{n-1}$, are independent random variables, each having distribution as shown below:
$$ P(J_n = j) = 0.18, \; \; if \; \; j = - £200.$$
$$ P(J_n = j) = 0.30, \; \; if \; \; j = - £100.$$
$$ P(J_n = j) = 0.12, \; \; if \; \; j = £0.$$
$$ P(J_n = j) = 0.40, \; \; if \; \; j = + £100.$$
The speculator's aim is to increase her capital to £4000, when she will stop trading. She will also stop if the account falls to zero. 
Let $h_k$ denote the probability that the speculator's capital ultimately reaches £4000, given that her account currently holds $k \times £100$.
Now, how do I calculate the values of $h_0$, $h_{-1}$ and $h_{40}$ ?
 A: Since the expected return is -26 per round, it is hard to imagine that there is much hope to turn 100 into 4000, but it is, of course possible.
One way to approach this is to use a Markov chain.  If the states are 0, 100, 200, ... 4,000, with 0 and 4000 being absorptive, then the transition matrix looks something like:
\begin{bmatrix}
\space\space\space 1 \space\space\space\space  0 \space\space\space\space  0 \space\space\space 0 \space\space\space 0 \space\space\space 0 \space\space\space 0 \space...\space 0 \\
 .48 \space\space .12 \space\space.4 \space\space\space 0 \space\space\space 0 \space\space\space 0 \space\space\space 0 \space... 0 \\
  .18 \space\space .3 \space\space .12 \space\space .4 \space\space\space 0 \space\space\space 0 \space\space\space 0 \space... 0 \\
  0  \space\space .18 \space\space .3 \space\space .12 \space\space .4 \space\space\space 0\space\space\space 0 \space... 0 \\
...\\
  0  \space\space\space 0  \space\space\space ... \space\space 0  \space\space\space .18 \space\space .3 \space\space .12 \space\space .4  \\
  0  \space\space\space 0  \space\space\space ... \space\space\space 0 \space\space\space 0  \space\space\space 0 \space\space\space 0 \space\space\space 0 \space\space 1  \\
\end{bmatrix}
Of course I'm assuming that losing 200, when you only have 100 is the same as losing 100 (you end up with nothing and no chance for recovery).
If you multiply your starting state [ 0 1 0 0 0 0  .... 0] (state holding 100) by the above transition matrix raised to a fairly high power, you get a vector with the probabilities of each state being final.  States 2 to 40 slowly go away, and the 41st element goes to about $5.2 \cdot 10^{-8}$ and the 1st is 1 minus that.
For higher starting amounts just adjust the row vector above [ 0 0 1 0 ... 0 ] for starting with 200, etc.
