Bertrand's Ballot Theorem (Random Walk)
J.Bertrand's Ballet Theorem stated that the probability of p leading q in a random walk is:
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:
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:
How do I reconcile the 2 values?
Thanks for any insight.