So here is the problem. Let’s assume that I am playing a game, I start with 100 money units I lose 1 money unit with probability p = 2/3 and I win one with probability q = 1/3. I am not stopping the game until either I lose all of my money or double them, what is the probability for each of those two outcomes.
At start I’ve tried it to solve it with Binomial distribution and I get that sum (or some variation of it).
$\sum_{n=0}^{\infty} {2n -100 \choose n}p^{n}q^{n-100} $
After trying a loooooot of time grinding out something from that sum (even thought I wasn’t sure if it is correct to begin with) I decided that the problem doesn’t solved with binomial distribution.
Then I decided to treat it as a Markov Chain so:
$E= \{ 0,1,2,3,..,100,..,200\} \;\; and \;\; p(n,n+1)=1/3, \;\; p(n,n-1)=2/3$
for $ \pi(n)$ I get
$ \pi(0)= \frac{2\pi(1)}{3}, \pi(1)= \frac{2\pi(2)}{3} $
$ \pi (2)= \frac{\pi(1)}{3} + \frac{2\pi(3)}{3} $ etc…
if I calculate them correctly I have:
$\pi(1) = \frac{3\pi(0)}{2},\pi(2) = \frac{9\pi(0)}{4}, \pi(3) = \frac{21\pi(0)}{8},\pi(4) = \frac{45\pi(0)}{16}$
so the rule is $\pi(n) = \frac{a_n}{2^n}\pi(0)$ with $a_0=0, a_1=3, a_{n+1}=2a_n+3 $
At that point I calculate also (in order to be sure) $\pi(5)=\frac{93\pi(0)}{32}$ which verifies the above.
I am not sure if the same rule still applies to ( $\pi(199)$ and $\pi(200)$), but I can’t figure out a way to check so I assume that it applies.
So I have that sum: $\sum_{n=1}^{\ 200} {\frac{a_n\pi(0)}{2^n}} \ = 1 \Rightarrow \pi(0)\sum_{n=1}^{\ 200} {\frac{a_n}{2^n}} \ = 1$
and I am stuck on figuring out how from there I can calculate $\pi(0)$ (if that’s possible) for 2 days or so…
Any assistance you can provide would be greatly appreciated.
PS. Sorry for my English it’s not mother tongue