
#11




Quote:
Also, my first question is also a hint/key to answering the original problem. But the main point is don't try to tackle the n=2000 head on. Make a smaller and more tractable problem first; get a solid handle on it and look to extract generalizations to see how to solve the larger (and more generic) case.
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#12




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#13




Quote:
with n positions: Code:
you have 1 string of n 0's you have 2 strings of n1 0's you have 3 strings of n2 0's . . . you have k strings of n+1k 0's . . .
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I find your lack of faith disturbing Why should I worry about dying? It’s not going to happen in my lifetime! Freedom of speech is not a license to discourtesy #BLACKMATTERLIVES 
#14




Quote:
Let F(1), F(2), ... denote the Fibonacci numbers. As you know, F(1)=F(2)=1 and F(i)=F(i1)+F(i2). Then the desired probability turns out to be 1[F(12)/2^10] ~= 86%. Now assume you want 3 consecutive zeros in 10 flips. The right sequence to use is the socalled Tribonacci numbers T(i) satisfying T(1)=0, T(2)=T(3)=1, T(i)=T(i1)+T(i2)+T(i3), and the correct answer is 1[T(13)/2^10] ~= 51%. For 9 consecutive zeros in 2000 flips, you will use the Enneanacci numbers E(i) (https://oeis.org/A104144) and compute 1[E(2009)/2^2000]. Edit: I computed E(2009) as 1.60988E+601 in Excel. Answer is 86%. Last edited by stefanos; 12292017 at 01:23 PM.. Reason: add final answer 
#16




Quote:
00100 00101 00111 00101 10010 10011 11001 01001 11100 there are not 4 with a longest string of 2 zeros 
#17




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I ran the simulation and that does look right. The fact that this works is not something that is in any way obvious to me. Last edited by Academic Actuary; 12292017 at 02:18 PM.. 
#18




my friend who bets red/black roulette tells me 0 chance. the wheel learns and doesn't have sequences that long.
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#19




Quote:
0101 `10 `1 1010 `1 `101 `10 `1 For n=1, we have F(3)=2 outcomes, for n=2 we have F(4)=3 outcomes, for n=3 we have F(5)=5 outcomes, for n=4 we have F(6)=8 outcomes, etc. Last edited by stefanos; 12292017 at 07:40 PM.. 
#20




My intuition is telling me that there is some trick where you look at this problem in the context of group theory and that there could be some subgroup where the elements are easy to count and line up nicely with the target or its complement.
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Tags 
binary, martingale, sequence, string 
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