I'm working on a gambler's ruin problem where a player starts out with $i$ money, and 'winning' is when their total money reaches $N$ (ie they will keep playing until they reach N or run out of money, at N they will stop). The game has probability $p$ of the player winning each time they play, and $1-p=q$ of losing.
I know that the probability of a player obtaining total $N$ with initial capital $i$ is given by $$P_iN=\frac{1-( \frac{q}{p})^i}{1-(\frac{q}{p})^N}$$ However I'm trying to look at the number of turns taken to either win or reach gambler's ruin, and I don't quite know how to get that number or an approximation of it. Instinctively I want to model this using Poisson but I can't quite see how to make the jump. Can anyone offer any suggestions? Thanks!
Edit: I can compute the expected value of N but as far as I understand this is not the same. Can I use this to get the answer I want?
---- you can get the average duration of a game by considering the cash flow- if odds of a coin toss is p to win, q= (1-p) to lose, in a single toss, you gain an average of (p-q).
If the average duration of a game is T flips, in Z games, you will gain $ Z * T * (p-q) $.
This equals the Number of Won games * the amount won in a game (which is (n-i) ), minus the Number of Lost games * the amount lost in a game (which is i ). You know the probability of winning= P(i,N,p).
So the number of won games = P * Z,
the number of lost games = (1-P)* Z
so Z * T * (p-q) = P * Z * (N-i) - (1-P) * Z * i.
and you get the average duration $$T=\frac{P (N -i) + (1-P) i }{ ( p - q)}$$ or $$T = \frac{ (PN - i)}{ (p-q) } $$ with P given by the quantity you have.