Bertrand's ballots puzzle - Proof by reflection I am looking for a some clarification to the proof of the Bertrand's ballot problem. 
 
I understood the the trick with swapping B(A) for A(B) in every position up to the first tie.

What I don't quite get is why the probability of beginning a sequence with B expressed as $\frac{b}{a+b}$  represents the probability of getting at least one tie when starting with B. 

It seems those are two different things. Secondly how does it account for many combinations giving the sequencies with a tie.

Can anybody explain this link please?
 A: The argument given at your link shows that there is a bijection between sequences that reach a tie at some point and sequences that begin with $B$. Thus, there are exactly as many sequences that reach a tie at some point as there are sequences that begin with $B$. Let $n$ be the number of sequences that reach a tie at some point (and hence also the number that begin with $B$). There are $\binom{a+b}a$ sequences altogether, since a sequence is completely determined once you know the $a$ positions of the $A$s, so the probability that a randomly chosen sequence reaches a tie at some point is $\frac{n}{\binom{a+b}a}$, and this is also the probability that a randomly chosen sequence begins with $B$. But we can directly evaluate that probability as $\frac{b}{a+b}$, since each sequence has $b$ $B$s and $a$ $A$s. Thus
$$\begin{align*}
\operatorname{Pr}(\text{there is a tie at some point})&=\frac{n}{\binom{a+b}a}\\
&=\operatorname{Pr}(\text{sequence starts with }B)\\
&=\frac{b}{a+b}\;.
\end{align*}$$
A: The key point in answer to your specific question is:


*

*The probability that a vote counting sequence starts with a vote for B is exactly the probability that the first vote counted is for candidate B, which is $\dfrac{b}{a+b}$. 


The related points are 


*

*$a \gt b$ so eventually Candidate A must win

*So any vote counting sequence which starts with a vote for B must have at least one tie point, since candidate B starts ahead but candidate A wins in the end

*The reflection between the start and the first tie, leading to a bijection between ties where the first vote is for B and ties where the first vote is for A, means that the probability of a tie at some stage is therefore $2\dfrac{b}{a+b}$

*The probability candidate A is always ahead in the count is the probability there are no ties at any stage, which is then $1-2\dfrac{b}{a+b}=\dfrac{a-b}{a+b}$
