Expected value for number

Question

Suppose you have a game with $n$ stages. For every stage $i$, you have $p(i)$ probability to advance to the next stage, and $1-p(i)$ probability to return to stage $1$. You win the game by advancing through all $n$ stages. What is the expected value of the number of times it will take to advance through all $n$ stages?

Answer 1

You will either fall at an earlier hurdle, or finish.
Let $q(i)$ be the probability that you fall at the $i$th stage, and $q(n+1)$ the probability that you pass through.
$q(1)=1-p(1)$
$q(2)=p(1)(1-p(2))$
$q(3)=p(1)p(2)(1-p(3))$
$q(n+1)=p(1)p(2)...p(n)$
Then the expected number of stages is S, where $S=q(1)(1+S)+q(2)(2+S)+...+q(n)(n+S)+q(n+1)n$

Answer 2

This is like a Markov chain with absorbing state $n$. For each $1\le i \le n$ define $h(i)$ to be the expected number of steps you need until you reach state $n$. Thus, for $1 \le i \le n-1$ you have that $$h(i)=1+p(i)\cdot h(i+1)+\left(1-p(i)\right)\cdot h(1)$$ and for $i=n$ you have that $h(n)=0$. Now solve recursively the system of equations to determine $h(1)$.