A random walk question: what is the given probability? Let $\{X_n\}_{n\in\Bbb N_0}$ be a simple random walk, given $n\in \Bbb N$ 
what is the probability 
$$
\mathbb P(X_1\ge0,X_2\ge0,\ldots, X_{2n-1}\ge0,X_{2n}=0)
$$
I think that I should benefit from reflection principle but I cannot use it. Please help me. Thank you 
 A: Let $\mathbb{P}(X_{n+1}=X_{n}+1)=p$ and $\mathbb{P}(X_{n+1}=X_{n}-1)=1-p=q$.
Since $X_{2n}=0$, we must have taken $n$ upward and $n$ downward steps, as $\#\{$steps up$\}+\#\{$steps down$\}=2n$ and $\#\{$steps up$\}-\#\{$steps down$\}=0$. So the probability is $C_{n}p^nq^n$ where $C_n$ is the number of ways it can be done.
We will prove that $\displaystyle C_n=\frac{1}{n+1}\binom{2n}{n}$. (Catalan's number). Take the connected curve $(0,0)-(1,X_1)-\ldots-(2n,X_{2n})$, and rotate it by $45^\circ$ degrees and scale it back by $\frac{1}{\sqrt{2}}$, so you'll get a movement on the lattice points, and the line $x=0$ becomes $x=y$, and $(2n,0)$ becomes $(n,n)$, if $(X_{n+1}=X_{n}+1)$, then you take a step in the direction of the $y$ axis (up), if $(X_{n+1}=X_{n}-1)$, you take a step in the direction of the $x$ axis (right). So you have to get from the left bottom corner of a lattice square to the right top without going under the diagonal.
There are $\binom{2n}{n}$ total paths that go from $(0,0)$ to $(n,n)$ with $n$ steps up and $n$ steps right. Calculate the number of paths that hit $x-1=y$ (bad paths). Take the first such event, and reflect the rest of the path on $y=x-1$ - this will keep distinct paths distinct, so in the end you'll get to the point $(n+1,n-1)$, still taking $2n$ steps, so the path has $n-1$ up and $n+1$ right steps, so we have $\binom{2n}{n+1}$ such paths. Thus:
$$C_n=\binom{2n}{n}-\binom{2n}{n+1}=\frac{1}{n+1}\binom{2n}{n}$$
And the probability is:
$$\frac{p^nq^n}{n+1}\binom{2n}{n}$$
