[PseudoPserious]

Hey irchans,

After a decent night's sleep and a half hour commute, I think I realized what I did wrong -- I think the idea was good, I just didn't implement it correctly. Here goes:

Call P(n) the probability of being tied for the first time on the nth draw. Call p(n) the probability of being tied on the nth draw. Then:

P(n)=Product[(1-p(i)),(i=1 to (n-1))]*p(n)

This looks like a straight application of conditional probability (a probability 'tree'), so it should handle properly handle independence.

Then by using:

p(x)=C(a,x/2)C(b,x/2)/C(a+b,x), or
p(x)=C(x,x/2)(.5)^x

for the first and second cases respectively, you can sum up all of the P(n) to get the probability.

Better?
PP