During the process of counting ballots in an Election of candidates, Show that $Pr$ ($A$ is ahead during the counting process ) = $\frac{a-b}{a+b}$ 
So this question is apparently from my past papers. I am practicing these for my upcoming exam, but I need a bit of guidance in it. So here is my working I have done so far and the problem I am facing in solving it.
My Working:
$1)$ So for the first part of this question I have gone as follows:
The total number of ways of choosing ballot randomly among $a+b$ ballots is: $(a+b)!$
Now if the last ballots to be chosen is from $A$ then there are $a$ possibilities for it and for each $a$ possibilities there are $(a+b-1)!$ possibilities of choosing remaining ballots. Hence:
$Pr$ (the last ballot is vote for $A$) = $\frac{a.(a+b-1)!}{(a+b)!}=\frac{a}{a+b}$
Is my way of arguing and proving part $a$ correct?
Also I am unable to even begin to prove part $b$. I need help in it.Kindly anyone guide and help me. I will really appreciate it.
 A: Extending the comment of Ritam Dasgupta, the following related problem is #22 from  "fifty challenging problems in probability" (F. Mosteller).  For the record, no way would I have thought of the following solution on my own.
Assume that $a > b$, and let $P$ denote the chance that at any point in the counting of the votes, there is a tie.  For each sequence of $(2n)$ votes, where the first tie recorded is on the $(2n)$-th vote, if you reverse the balloting of the $(2n)$ votes so that each vote for $a$ becomes a vote for $b$, and vice-versa, then you will have a mirror-image sequence of votes that also results in the first tie being recorded on the $(2n)$-th vote.
For each such pair of sequences, exactly one of the two will have the first vote being for $B$, rather than $A$.  Further, for each sequence of votes where the first vote is for $B$, since $a$ is assumed $> b$, such a sequence must result in a tie at some point.
Thus, the chance that the 1st vote is for $B$ is $\frac{b}{a+b}$, and so $P = $the chance that the sequence of votes results in a tie at some point, which equals $2 \times \frac{b}{a+b}.$
Therefore, the chance that $A$ is always ahead is 
$\displaystyle 1 - P = 1 - \frac{2b}{a+b} = \frac{a-b}{a+b}.$
Edit
By the way: 
In theory, I am supposed to wait until you show work on part(b), before handing you the answer.  To me, this is similar to asking you to prove Fermat's Last Theorem, and then having Andrew Wiles say that he will have to wait for you to show work before giving you the answer.
If I get downvoted, so be it.  At times, I regard the strict interpretation of the mathSE protocol, as detailed here as problematic, to describe it generously.
A: Here is how to solve the problem using the given hint. Let $E$ be the event that $A$ is always ahead, and let $L_A$ be the event that the last ballot is for $L_A$, similarly for $L_B$. Then as long as $a>b$ (note the strict inequality),
\begin{align}
P(E)
&=P(E|L_A)P(L_A)+P(E|L_B)P(L_B)
\\&=\color{blue}{\frac{(a-1)-b}{(a-1)+b}}\cdot \frac{a}{a+b}+\color{blue}{\frac{a-(b-1)}{a+(b-1)}}\cdot \frac{b}{a+b}
\end{align}
For each of the $\color{blue}{\text{blue}}$ parts, we are applying the induction hypothesis. Given the last ballot is for $A$, the first $a+b-1$ ballots are a mix of $a-1$ ballots for $A$ and $b$ ballots for $B$, and the event that $A$ is ahead during the entire time during these first $a+b-1$ ballots is a smaller version of the original problem. Since we still have $a-1\ge b$, the formula $((a-1)-b)/((a-1)+b)$ applies. The same logic applies to $P(E|L_B)$.
Simplifying that fraction gives the result you want. You also have to deal with the case $a\le b$, but $P(E)$ is obviously $0$ in that case, as desired.
