I'm currently solving a similar problem to this one: Cashier has no change... catalan numbers.. probability question. In my case, you have an equal number of "votes/steps" for A and an equal number of "votes/steps" for B.
The question is: given $2n$ steps, where there are $n$ "-1" steps and $n$ "+1" steps, what's the probability you always remain above 0. The solution is $\frac{1}{n+1}$, which I understand.
The issue is my approach. I applied conditional probability: P(X) = P(X | last element = -1)*P(last element = -1) + P(X | last element = 1) * P(last element = 1).
But P(last element = -1) = P(last element = 1) = 1/2. Also, P(X|last element = 1) = 0 because if the last appearance is a +1 and the final result sums to 0, the 2nd to last result must be -1. Hence, P(X) = 1/2*P(X | last element = -1). With this form, we have n-1 "-1" values and n "+1" values, allowing us to use Bertrand's theorem here https://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem#Bertrand's_and_Andr%C3%A9's_proofs which states that "Suppose that candidates A and B are in an election. A receives a votes and B receives b votes, with a > b. The probability that A is always ahead of B is "$(a-b)/(a+b) * {a+b \choose a}$"
So the end result I get is $\frac{n - (n-1)}{(n+n-1)} * {2n-1 \choose n} = \frac{1}{2n-1}*{2n \choose n}*1/2$. Then dividing by the total sample space, we get $P(X) = \frac{1}{4n-2}$. Where did I go wrong?
$(a-b)/(a+b)$ is the probability that $A$ stays strictly ahead of $B$ throughout the vote count. The probability that $A$ stays weakly above $B$, meaning ties are OK, is $(a-b+1)/(a+1)$. As a sanity check, when $a=b=n$, we get $(n-n+1)/(n+1)=1/(n+1)$. The weak formula is discussed in the last section of the wikipedia article you linked.
If we use the corrected formula, your argument becomes $$ P(X)=P(X|\text{last element is $-$1})\cdot \frac12=\color{#0a0}{\frac{n-(n-1)+1}{n+1}}\cdot \frac12=\frac2{n+1}\cdot \frac12=\frac1{n+1}\;\;\color{#0f0}{\checkmark} $$