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#21




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My original point still stands; start simpler and see what you can learn. Then extend it out to see if what you found is general enough or if you need to work to refine it. The big key here is to not take the monster problem head on. As it is, there is a clear recursion pattern that can be seen by looking at this simpler problem. If P(n, k) = Probability of finding no more than k consecutive zeros in a string of n (binary) digits, then we have the following recursion: P(n, k) = 0.5*P(n1, k) + 0.5*P(n1, k1) And the answer to the OP would be 1  P(2k, 8)  P(2k, 7)  ...  P(2k, 1)  P(2k, 0).
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#22




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P(4,3) = 2/16 P(3,3) = 1/8 P(3,2) = 2/8 Your first term is the outcome is a 1, but the second outcome being 0 does not necessarily increase the maximum length of your string. 
#23




Having 9 consecutive 0's can be broken into two cases:
(1) Having 9 consecutive 0's within the first 1999 elements of the sequence or (2) Not having it in the first 1990 elements, a 1 for element 1991 and then 9 consecutive 0's at the end: So let be the probability that 9 consecutive 0's are found in the first elements of the sequence. Then the two cases give: So if you calculate and let then you can just use the recursion above to find right? Starting with the sequence looks like this: 
#24




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Last edited by Academic Actuary; 12292017 at 05:15 PM.. 
#26




I found other solutions and they are waaay more complicated. Seems like this problem is super hard.
http://mathforum.org/library/drmath/view/56637.html 
#28




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Tags 
binary, martingale, sequence, string 
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