Consider a 1d random walk starting from the origin $0$ and moving $2n$ steps where each step is of size 1 (i.e. $\pm 1$).
We want to calculate the number of paths that do not pass $0$ during the walk. By considering different ending position $\pm 2, \pm 4, \cdots, \pm 2n$, we can write out the formula: $$ \sum_{k=1}^n (N^{\hat{0}}_{2n} (0,2k)+ N^{\hat{0}}_{2n} (0,-2k))$$
where $N^{\hat{0}}_{2n} (0,m)$ denotes the number of paths from $0$ to $m$ with $2n$ steps that do not pass $0$ during the walk.
After some calculations we can verify the result is equal to the number of paths from $0$ to $0$ with $2n$ steps without any restriction, i.e. $N_{2n} (0,0)$.
Since these two numbers are equal, there must be some bijection that maps valid walks for one problem to paths for the other. How can we construct such a bijection? (If we can do so, it will be much easier to solve the original problem) I have tried some using reflection but failed.
Note: thinking graphically, i.e. taking the step(or time) and the position as x and y axis respectively, we will naturally have a map between the following
But how should we map the following lines to broken lines connecting $(0,0)$ and $(6, 0)$
I tried to find patterns starting from $n=1,2$, but when $n=3$ things become complex. Please help.