Bertrand's Ballot Theorem (Random Walk)

J.Bertrand's Ballet Theorem stated that the probability of p leading q in a random walk is:

$\displaystyle {{p-q} \over{p+q}}$ given that p>q.

Bertrand's ballot theorem - Wikipedia, the free encyclopedia

Let's give values to work the problem.

Let p=4, q=2.

On a graph, we can recast the problem as the probability of a path starting at (0,0) reaching (6,2) without touching the x-axis and is given by the Ballet Theorem. (6 because 4+2, 2 because 4 of '+1' - 2 of '-1')

BUT with that one condition: p>q

The probability worked out is:$\displaystyle {{p-q} \over{p+q}}={2\over6}={1\over3}$

My problem is this:

What if we remove the condition p>q? What happens to the actual probability:the probability that p leads q without any conditions attached?

The probability of a path starting at (0,0) to reach (6,2) is given by:

$\displaystyle {6\choose2}2^{-6}= {15 \over64}$

How do I reconcile the 2 values?

Thanks for any insight.