Today we will continue our discussion with Ballot theorem. In the previous post we mentioned the theorem. In this post, we mention the conditional probabilities. Out of an urn with n balls the sampling followed a random walk. The probability that given S0 = 0 and Sn = y  > 0, the random walk never becomes zero up to time n was given by
Po ( S1  > 0 , S2 > 0 , ... Sn > 0  | Sn  = y )  = y/n
Now we say that if n,y are both positive or zero, and are both even or both odd then,
then the same probability is (y + 1) / 1/2 ( n + y ) + 1
In particular, if n is even,
            P0 (S1 >= 0, ... Sn >= 0 )  = 1 / (n/2) + 1
We use the corollary we discussed earlier to prove this :
The number of paths from (0,0) to (n,y)  that remain above the horizontal axis is given by
n Choose (n+y) / 2 - n Choose ((n+y)/2 + 1)
The number of paths from (0,0) to (n,y) = n Choose (n+y)/2
so the probability we are interested can be expressed as a conditional P0 (Sn >= 0, ..., Sn >= 0, Sn = y) / P0 (Sn = y)  =   The number of paths from (0,0) to (n,y) that remain above the horizontal axis   /   The total number of paths from (0,0) to (n,y)
 and by substituting the expressions from above for the number of paths and eliminating the common terms we prove the probability above.