Election between $2$ candidates ends in a tie: probability one candidate leads until the penultimate vote 
Assume there are two candidate $C_1$ and $C_2$. At the end of the election both candidates receive the same amount of votes. What is the probability $P$ that candidate $C_1$ leads during the whole election process until the penultimate vote? (The last vote must always be in favor of candidate $C_2$)

This question was presented in our lecture in the context of the ballot-theorem. So one should think of paths which start at $(0, 0)$ along the $x$-axis and end at some point $(n,s)$, where $n,s \in \mathbb{Z}$.
My approach:
My sample space $\Omega$ includes all possible paths along the $x$-axis. If the path is above the $x$-axis then candidate $C_1$ has more votes and if the path is below then $C_2$ has more votes . If the paths touches the $x$-axis then both candidates have the same amount of votes. Hence, $|\Omega|={2p \choose p}$, where $p \in \mathbb{N}$ is the number of votes of each candidate.
Firstly, I count all paths which start at $(1,1)$ and end at $(2p,0)$. These are ${2p-1 \choose p-1}$ many. Now I subtract all paths that touch the $x$-axis, these are ${2p-2 \choose p-2}$ many. So in total I count ${2p-1 \choose p-1}-{2p-2 \choose p-2}$ paths which do not touch the $x$-axis. One can interpret all these paths as desired outcomes, i.e. where candidate $C_1$ leads until the penultimate vote. As all paths are equally probable I get the solution just by dividing by $|\Omega|={2p \choose p}$. Hence, $P = \frac{{2p-1 \choose p-1}-{2p-2 \choose p-2}}{{2p \choose p}}$.
I am not sure if this is correct. May be someone can check it or comment on it.

Let's shift gears and use the ballot theorem instead of reinventing it for the case of ties.  Other than the one appeal to the ballot theorem, this will be a conditional probability problem.  ^_^
Our sample space will be all cases where each candidate received $p$ votes.  Let $A$ be the event that $C_1$ lead all the way until the moment before the final vote was read, and let $B$ be the event that $C_2$ received the final vote.  
We know that  


*

*$P(B)=P(\overline B)=\frac12$ 

*$P(A\mid B)=\frac{p-(p-1)}{p+(p-1)}=\frac1{2p-1}\quad$ This is where we are using the ballot theorem.  

*$P(A\mid \overline B)=0\quad$ Obviously, $C_1$ could not have lead throughout the counting and received the final vote, because the count ended in a tie.  
Using all this and the law of total probability, 
$$P(A)=P(A\mid B)\cdot P(B)+P(A\mid \overline B)\cdot P(\overline B)\\=\frac{1}{2p-1}\cdot\frac12+0\cdot\frac12=\frac1{4p-2}$$
Note that this formula agrees with almagest's calculation that the probability was $\frac1{10}$ when $p=3$.
