Variation on Bertrand's ballot theorem, chance that gambler is never worse off than at beginning and never better off than at end If a man playing for a constant stake wins $2n$ games and loses $n$ games, how do I see that the chance that he is never worse off than at the beginning and never better off than at the end is$${{n^2 + n + 2}\over{4n^2 + 6n + 2}}?$$This seems to be a variation off of Bertrand's ballot theorem: https://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem. I've been able to come up with a "brute force" solution through writing out all the routes and summing them up, but I'm wondering if there is a quick way to see this via mimicking the reflection proof of Bertrand's ballot theorem.
Edit: Showing my work: I got a large sum counting all the routes that simplified to$$\binom{3n}{n} - 2\binom{3n}{n - 1} + \binom{3n}{n - 2},$$and then dividing this by $\binom{3n}{n}$ gets us the desired answer. But I wonder if there is a quicker way via reflection that's less calculation intensive to get $\binom{3n}{n} - 2\binom{3n}{n - 1} + \binom{3n}{n - 2}$.
 A: Hint: Use the reflection theorem twice, and the inclusion-exclusion principle.
Let us start from point $(0,0)$. Once the man wins, goes from $(x,y)$ to $(x+1,y)$. Otherwise goes to $(x,y+1)$.
Then the number of all trials should be the number of paths from $(0,0)$ to $(2n,n)$, which is $\binom{3n}{n}$.
Recall how we calculate the number of paths that do not touch line $y=x$: once touch, we do the reflection that maps $(0,0)$ to $(-1,1)$, and there are $\binom{3n}{n-1}$ paths from $(-1,1)$ to $(2n,n)$. I skip some details to prove the equivalence. Please refer to the wiki page you link for details.
Similarly, we do the reflection that maps $(2n,n)$ to $(2n+1,n-1)$ and get the number of paths that do not touch $y=x-n$, which is also $\binom{3n}{n-1}$.
However, there are some paths that touch both $y=x$ and $y=x-n$, which is equivalent to the paths from $(-1,1)$ to $(2n+1,n-1)$, and there are $\binom{3n}{n-2}$ paths.
Using the inclusion-exclusion principle, we have $\binom{3n}{n}-2\binom{3n}{n-1}+\binom{3n}{n-2}$ paths that satisfy our constraint.
