This question is inspired by Bertrand's ballot theorem. I want to check if I understood the counting method correctly. I changed the initial problem of the ballot theorem a bit.

Suppose we have two candidates, A and B. After counting the votes, we have a tie. How many paths exist in which candidate A is never? $ geq2 $ votes ahead?

This is my approach:

I imagine the election as a path in the $ x $-axis that begins in $ (0, 0) $ and ends in $ (2p, 0) $, where $ 2p $ denotes the total number of votes. In the test of the *voting theorem* they use a second path that is constructed reflecting in part the original paths on the $ x $-axis. I will try to follow this idea.

First, I change the beginning of all my paths to $ (- 2.2) $ and the first two votes should always be in favor of B. Second, I will extend the way to $ (2p + 2, 2) $ while the last two votes will count for A. Of those paths I will only consider those that begin with two consecutive votes for B and end with two consecutive votes for A (otherwise, we would include paths that are definitely not allowed). Denote $ M $ as the set of those paths. Thus, $ | M | = {2p choose p} $.

Then i build *assistant* roads as follows:

Allow $ P $ a path of $ M $. While $ P $ does not touch the horizontal line that passes $ (0.2) $ (see red line in the image) Reflect your values on the horizontal line that crosses $ (0.2) $. Those values are the first points of the *assistant* routes. When $ P $ play the $ (0.2) $-line the *assistant* paths will follow the rest of $ P $. Building those *assistant* roads is a bijection in the set of those roads of $ M $ that touch or cross the $ (0, 2) $ line. So I just have to subtract everything *assistant* paths of $ | M | $.

Now I count all *assistant* paths (I will do this with a little more detail):

I have added $ 4 $ votes for $ 2p $ votes from the beginning (see green lines). $ 4 $ votes of each *assistant* the route always has the same meaning as the first two votes and the last two votes always count for A. Therefore, all *assistant* the roads add up: $ {2p choose p-2} $. The total number of roads where A is never $ geq 2 $ votes ahead is $ {2p choose p} – {2p choose p-2} = frac {2} {p + 2} {2p choose p} $.

Is this correct?

I appreciate any comments or suggestions and let me know if I should be clearer at some step.