Re: The Ballot Box
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
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