Question. Suppose we have iid random variables $\{X_k\}_{k\geq 1}$ where $$P(X_k = 1) = \frac{2}{3} \quad\text{and}\quad P(X_k = -1) = \frac{1}{3}.$$
We have the random walk $\{S_t\}_{t \geq 1}$ where
$$S_0 = 4, \quad S_t := S_0 + \sum_{k=1}^t X_k.$$
I'm trying to find the probability that we arrive at the value of $6$ before time $t = 8$. Formally, we want to find the probability that at any time, $1 \leq t \leq 7$, we have $S_t = 6$.
What I have. I think we can solve this using the recurrence relations $$p_{t,i} = \frac{2}{3}p_{t-1,i+1} + \frac{1}{3}p_{t-1, i-1},$$
where $p_{t,i}$ is the conditional probability that at time $t$ we have value $i$. The obvious case is $p_{t,0} = 0$.
I'm unsure of how we do this. My intuition says we just find the sum of all possible instances that we can hit a value of 6 before time $t = 8$, i.e. $\sum_{t=2}^8 p_{t, 6}$?