# Thread: Bertrand's Ballot Theorem (Random Walk)

1. ## Bertrand's Ballot Theorem (Random Walk)

J.Bertrand's Ballet Theorem stated that the probability of p leading q in a random walk is:
${{p-q} \over{p+q}}$ 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: ${{p-q} \over{p+q}}={2\over6}={1\over3}$

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

How do I reconcile the 2 values?
Thanks for any insight.

2. Ok. I got this figured out.

The first probability measures the percentage of paths reaching (6,2) which are strictly positive.

The second probability measures the probability of paths which reach (6,2).

So the total probability of strictly positive paths which reach (6,2) is given by:
$({1\over3})({15\over64})={5\over64}$