Give a simple symmetric random walk, denote by $\mathbb P_0(T_1 < \infty)$ the probability of starting at $0$ and for the first time reaching level $1$ within finite discrete time steps. What is $\mathbb P_0(T_1 < \infty)$?
So I know by the law of total probability $$ \begin{split} \mathbb P_0(T_1 < \infty) &= \mathbb P_0(T_1 < \infty|S_1 = 1) \mathbb P_0(S_1 = 1) + \mathbb P_0(T_1 < \infty|S_1 = -1) \mathbb P_0(S_1 = -1) \\ &= \frac12 + \frac12 \mathbb P_0(T_1 < \infty|S_1 = -1) \end{split} $$
Also, $\mathbb P_0(T_2 < \infty)$ = $\mathbb P_0(T_1 < \infty)$$\mathbb P_1(T_2 < \infty)$.
In order to solve the system of the above two equations, I need to find the relationship between $P_0(T_1 < \infty|S_1 = -1)$ and $\mathbb P_0(T_2 < \infty)$. I guess they are equal since from $-1$ to $1$ is the same as from $0$ to $2$, but I'm not sure if that's correct.