Consider a particle that moves according to a simple random walk. Denote by $X_n$ the position of the particle immediately after step $n$. Assume that $X_0 = 0$ and that, at each step, the position of the particle goes up (by one) with probability $p$ and down (by one) with probability $1-p$. Find the probability that the particle visits $0$ for the first time after $4$ steps.
Working:
Let $P(\varepsilon_i = 1) = p$, $P(\varepsilon_i = -1) = 1-p$ for $i= 1, 2, 3, 4$. Then $X_4 = \varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_4$. We need to find $P(X_1 \neq 0, X_2 \neq 0, X_3 \neq 0, X_4 = 0)$. So we know that $$P(X_1 \neq 0, X_2 \neq 0, X_3 \neq 0, X_4 = 0) =P(X_2 \neq 0, X_4 = 0 | X_1 \neq 0, X_3 \neq 0)P(X_1 \neq 0, X_3 \neq 0)$$
But where do I go from here? I know that $P(X_1 \neq 0) = P(X_3 \neq 0) = 1$ since $1$ and $3$ are odd, and I can intuitively argue that the events $\{X_1 \neq 0\}$ and $\{X_3 \neq 0\}$ are independent, but how can I show this rigorously through probability formulas? Furthermore, I also think $\{X_2 \neq 0, X_4 = 0\}$ and $\{X_1 \neq 0, X_3 \neq 0\}$ are independent events, but again, how do I show this properly?
Consider the possible paths of the particle so that "the particle visits $0$ for the first time after $4$ steps".
Case $X_1=1$: If $X_2=0$ then the statement is false. So $X_2=2$. Then the only way for the particle to return to $0$ from here is two downward steps. So probability of this occurring because they're independent events is $p^2(1-p)^2$. Note this is the only possible path in this case.
Case $X_1=-1$: More or less the same, and the path is down, down, up, up, so $(1-p)^2p^2$. This is the only path in this case.
So we have $P(X_1≠0,X_2≠0,X_3≠0,X_4=0)=2p^2(1-p)^2$.