This question is inspired by Bertrand's ballot theorem. I want to check if I have understood the counting method correctly. I changed the initial problem of the ballot theorem a little bit.
Suppose we have two candidates, A and B. After having counted the votes we have a tie. How many paths exist on which candidate A is never $\geq2$ votes ahead?
This is my approach:
I envision the election as a path on the $x$-axis which starts at $(0, 0)$ and ends at $(2p, 0)$, where $2p$ denotes the total number of votes. In the proof of the ballot-theorem they use a second path which is constructed by partly reflecting the original paths over the $x$-axis. I will try to follow this idea.
Firstly, I shift the start of all my paths to $(-2,2)$ and the first two votes must always be in favor of B. Secondly, I will extend the path until $(2p+2, 2)$ wheras the last two votes will count for A. Out of those paths I will only consider those ones which start with two consecutive votes for B and end with two consecutive votes for A (otherwise we would include paths that are definitively not allowed). Let's denote $M$ as the set of those paths. Hence, $|M|={2p \choose p}$.
Then I construct auxiliary paths as follows:
Let be $P$ a path of $M$. As long as $P$ doesn't touch the horizontal line which goes through $(0,2)$ (see red line in the picture) I reflect its values over the horizontal line which goes through $(0,2)$. Those values are the first points of the auxiliary paths. When $P$ touches the $(0,2)$-line the auxiliary paths will follow the rest of $P$. The construction of those auxiliary paths is a bijection into the set of those paths of $M$ which touch or cross the $(0, 2)$ line. So I simply have to subtract all auxiliary paths from $|M|$.
Now I count all the auxiliary paths (I will do this in a bit more detail):
I have added $4$ votes to the $2p$ votes from the beginning (see green lines). $4$ votes of every auxiliary path are always the same meaning that the first two votes and the last two votes always count for A. Hence, all auxiliary paths sum up to: ${2p \choose p-2}$. The total number of paths where A is never $\geq 2$ votes ahead is ${2p \choose p}-{2p \choose p-2}=\frac{4p+2}{(p+1)(p+2)}{2p \choose p}$.
Is this correct?
I appreciate any comment or suggestion and please let me know if I should be clearer in any step.
Best Answer
You have the correct answer, but there is a simpler explanation which does not require augmenting paths. We are counting paths from $(0,0)$ to $(2p,0)$ whose steps are all of the form $(1,\pm 1)$, and whose $y$ coordinate is never equal to $-2$. We take all $\binom{2p}p$ paths, and subtract the bad paths which at some point hit $y=-2$. If we reflect such a bad path after the first time it hits $y=-2$, we end up with an arbitrary path to $(2p,-4)$. Therefore, the number of bad paths is the number of paths to $(2p,-4)$, which is $\binom{2p}{p-2}$.