A monkey is sitting on 0 on the real line in period 0. In every period $t$ $\in${$0,1,2,...$} it moves $1$ to the right with probability $p$ and $1$ to the left with probability $1-p$. $p\in[1/2,1]$. What is the probability that the monkey will reach positive integer N in some period $t>0$.
My working: The total number of $N$ steps is the sum of k forward steps and N-k backward steps. The probability reaching a point N is therefore
$\sum_{k=1}^N{N \choose k}p^{k}(1-p)^{n-k}$. I'm stuck here, if k ran from 0 to N then, the sum of the probabilities would have been 1 since it represents the sum of a binomial PMF. It can't be that k = 0, for that would mean that the monkey was at N since the beginning, however the monkey here starts at zero. What should I do ? Please help.
This could be solved by conditioning on the first step.
Denote by $r$ - the probability to reach $1$, starting at $0$. Note that the whole process is shift invariant, i.e. this is the probability to reach $4$, starting from $3$. And this is a Markov process, i.e. no memory involved - only the current state matters.
Then $$ r = p + (1-p) \cdot r^2.$$
How do you reach $+1$. You either go straight to $(+1)$ - this is $p$, or first go to $(-1)$, but then, due to shift invariance, you need to first get back to $0$ (which has probability exactly $r$) and proceed from $0$ to $(+1)$ - also $r$, by definition.
What's left is to solve. In your range only one solution is possible, for $p < \frac{1}{2}$ things become more involved, as one solution should be dropped (namely $r = 1$).