I know that for a random walk with two stop values, the expected value of the number of steps needed is $ab$ where the stop values are $-a$ and $b$ and the initial position is at 0.
What about for random walks with probability $1$ of reaching the stop value? For example, I am at position $1$ and I have probability $0.6$ of going left and $0.4$ of going right. I obviously have probability $1$ of reaching $0$, but what is the expected value of number of steps?
Simulations seem to say $E[V] = 5$ which is equal to $\sum_{i=0}^{\infty} C_i * 0.4^i * 0.6^{i+1}$ where $C_i$ is the $i$th Catalan number. However this is equal to of $\sum_{i=0}^{\infty} 0.6^{i+1}*0.4^i*\binom{2i+1}{i}$ which doesn't make intuitive sense. I would have reasoned it to be $\sum_{i=0}^{\infty} 0.6^{i+1}*0.4^i*\binom{2i}{i} = 3$ since the last head is set (the last coin tossed has to be a head). The first equation seems to double count?
Thanks! (Also is there a general formula or approach for random walks with one stopping value?)