The proportion of H's to total flips is simply p: Expected H's per run = p + p^2 + p^3 + ... = p/(1-p) = p/q Expected T's per run = 1 Expected flips per run = p/q + 1 Proportion of H's to total flips = (p/q)/(p/q + 1) = p/(p + q) = p Tom
Okay, now suppose the same question but suppose for given p+q = 1, a large number of people flip their own biased coin, with
Prob(H) = p and Prob(T) = q = 1-p,
until reaching the first T.
What will be the asymptotic proportion of H's and T's flipped, in simplest form?
--Dan
P.S. I had to solve this problem before I appreciated the reasoning behind Gene's beautiful solution, which didn't register with me at first.
<< PUZZLE: Suppose that in a very large population, each family has children until the first boy is born, and then has no more. What is the percentage of girls in the population?
(Of course, make all the usual simplifying assumptions in this kind of problem.)
I sleep as fast as possible so I can get more rest in the same amount of time.
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