ballot box puzzle I am analyzing the below puzzle presented here on p37

In an election, two candidates, Albert and Benjamin, have in a ballot
  box a  and b votes respectively, a > b, for example, 3 and 2 If
  ballots are randomly  drawn and tallied, what is the chance that at
  least once after the first tally the candidates have the same number
  of tallies?

For a=3 and b=2, the following sequencies are given 


AAABB       *AABBA         *ABBAA 

AABAB        *ABABA         *BABAA

*ABAAB         *BAABA        *BBAAA

*BAAAB 
I don't understand the idea of this puzzle. 
The bolded combinations are not fullfilling the conditions the starred ones are.
I am confused with the way the sequances are traeted. 
Can anybody explain please?
 A: Each letter sequence gives the votes in the order they are drawn from the box.
Here are the tallies for the first vote sequence:

AAABB

1. A | A=1 | B=0
2. A | A=2 | B=0
3. A | A=3 | B=0
4. B | A=3 | B=1
5. B | A=3 | B=2

The sequence is marked in bold to indicate that A never equals B (A $\neq$ B for all steps 1-5).
Here are the tallies for the third letter sequence:

*ABAAB

1. A | A=1 | B=0
2. B | A=1 | B=1 *
3. A | A=2 | B=1
4. A | A=3 | B=1
5. B | A=3 | B=2

The sequence is starred to indicate that at some point the tallies became equal. That occurred for this sequence on the second draw, A=B=1.
Note that the puzzle asks about equality, and the vote tally can only be divided evenly between A and B when the total number of votes are even -- that is, the condition can never be satisfied on an odd counting step.
Calculating the probability that sequences will tie at some point (as the puzzle asks) is part of the Proof by Reflection approach to Bertrand's ballot theorem.
A: Adding this to Jeremy's answer. 
Lets us suppose in a sequence m A's and n B's, such that m > n, the first time their values are equal, we have x A's and x B's, which means at 2xth position. 
Before 2x votes, either A will be more or B. But using the Principle of Symmetric outcomes, we can declare that both will be the same. This means for 2x - 1 votes, y ways A is ahead, and symmetrically B is ahead in those y ways.   
Let's add another slightly independent idea. If b is first, there has got to be an equilibrium somewhere down the line(since n < m), but before that B's are all more in number. Now the probability of these sequences to occur are b/(a + b).
There also sequences where there are more A's than B's before the first equilibrium. Counting them directly is a bit tough. But since we already established the Symmetry principle, we know that they also occur b/(a + b) times. Hence, the total number of ways we get atleast one equilibrium is 2*b/(a + b).  
A: The sequences show the order that the ballots are drawn, left being first and right being last.
For the bolded ones, there is no value $n$ with $1 \leq n \leq 5$ for which the first $n$ ballots have the same number of $A$ and $B$ ballots.  Those bolded cases satisfy the condition.
Contrarily, looking at $AABBA$ (one of the starred ones) we see that for $n=4$ we have $AABB$ drawn, which is the same number of ballots for each.  This case does not satisfy the condition.
