(In reply to
solution by Charlie)
Really nice analysis and writeup. Interesting that it is essentially 2^(n+1) tries on average to get n+1 heads in a row -- the same as the number of strings of n+1 heads or tails. I conjecture that the problem can also be solved in closed form when the probability of a head is p. In fact, the expected value should be 1/p + 1/p^2 + ... + 1/p^n and with r=1/p, this sums to
r(r^n - 1)/(r - 1).
Edited on December 13, 2004, 11:00 pm
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Posted by Richard
on 2004-12-13 22:10:02 |