Let $S_n = \sum^n_0 X_k$ be random walk (with steps not necessarily of probability $\frac 12$). I want to show that $$ P(\text{hitting $2$ | hitting $1$ $\cap$ $S_0=0$}) = P(\text{hitting $2$ | hitting $1$})$$ where "hitting $x$" is the event $\bigcup_{n=0}^\infty \{S_n=x\}$.
I conceptually can conceive that the starting point does not affect this probability so it can be deleted in the probability but I struggle to show it properly and rigorously, which makes me wonder if the equality is false.
I write this as $$P(\text{hitting $2$ | hitting $1$ $\cap$ $S_0=0$})=\frac{P(\text{hitting $2$ $\cap$ hitting $1$ $\cap$ $S_0=0$})}{P(\text{hitting $1$ $\cap$ $S_0=0$})}$$
I can deal with the denominator $$P(\text{hitting $1$ $\cap$ $S_0=0$})=P\bigg(\bigcup_{n=0}^\infty\{S_n=1\} \cap S_0=0\bigg)$$ $$=\sum_{n=0}^\infty P(S_n=1 \cap S_0=0)=\sum_{n=0}^\infty P(\sum^n_{k=0} X_k=1 \cap X_0=0)=\sum_{n=0}^\infty P(\sum^n_{k=1} X_k=1 \cap X_0=0)$$ $$=\sum_{n=0}^\infty P(\sum^n_{k=1} X_k=1)P( X_0=0)=P( X_0=0)\sum_{n=0}^\infty P(\sum^{n-1}_{k=0} X_k=1)$$
there will be a problem for $n=0$ but the last quantity is equal to $P(S_0=0)P(\text{hitting} 1)$, which is good news but I don't know if what I did is legal and I don't feel that the numerator can be dealt with similarly.