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Old 12-29-2017, 09:10 AM
Tsin Tsin is offline
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Default What is the probability...

there is a string of 2000 random binary numbers (0 and 1) each occurring with the same probability of 50%. What is the probability that in this binary string, there happen to be 1 or more sequence of 9 zeros in a row?

edited 10 to 9

Last edited by Tsin; 12-29-2017 at 09:20 AM..
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Old 12-29-2017, 09:14 AM
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Quote:
Originally Posted by Tsin View Post
there is a string of 2000 random binary numbers (0 and 1) each occurring with the same probability of 50%. What is the probability that in this binary string, there happen to be 1 or more sequence of 10 zeros in a row?

my intuition tells me its 1/2 but I am not sure how to prove it.
There either is such a sequence or there isn't. Two possibilities, so a 50/50 chance.

Right?
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Old 12-29-2017, 09:19 AM
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No, I was comparing it to martingale strategy which obviously cant work so the chance of doubling capital must equal the chance of losing it all which is 9 zeros in a row not 10 as I though.
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Old 12-29-2017, 09:39 AM
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http://sites.math.rutgers.edu/~prd41...ive_Heads.html

Try using this.
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Old 12-29-2017, 09:42 AM
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Solution is easy to find using numeric methods. I want to know how to prove that kind of problems analytically because it looks like it can be solved on paper.

Last edited by Tsin; 12-29-2017 at 09:48 AM..
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Old 12-29-2017, 10:01 AM
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You have to create a transition probability matrix. In your case because you are interested in a string of 9 zeros it would be 10 by 10. Your initial state vector is [1,0,......,0] as you are in state 0 with probability 1.

Any row of the matrix takes you to the next higher state with probability .5 and to 0 with probability .5 except the 10th row were 9 is an absorbing state. Raise the matrix to the power 2000 and multiply by the initial state vector. The last entry would be the probability of a string of at least 9 zeros.
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Old 12-29-2017, 10:16 AM
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What's the probability that that such a sequence does not exist?


Note: you might start by working out the situation with a much smaller number than 2000; say, 10. Get your answer there and then extend the number to 15 (instead of 2000). If the solution doesn't hit you here, then extend it to 20 and see what happens.
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Old 12-29-2017, 10:31 AM
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Quote:
Originally Posted by Vorian Atreides View Post
What's the probability that that such a sequence does not exist?


Note: you might start by working out the situation with a much smaller number than 2000; say, 10. Get your answer there and then extend the number to 15 (instead of 2000). If the solution doesn't hit you here, then extend it to 20 and see what happens.
10 and 15 flips are easy because you can have only one consecutive string of nine or more zeros. With 2000 flips it isn't so easy unless I am missing something.
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Old 12-29-2017, 10:32 AM
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If we assume equal number of 0s and 1s in the sequence will it affect the answer? They happen with the same probability so the average would be 1000 0s and 1000 1s after n trials when n->oo. It would greatly simplify this problem.

Last edited by Tsin; 12-29-2017 at 10:55 AM..
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Old 12-29-2017, 10:56 AM
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If you assume an equal number of 1's and 0's it is a different problem. You would have 2000C1000 instead of 2^2000 sequences. You want the probability of a sequence of at least 9 zeros. It could begin in position one through position 1992. Because they are not independent you would have to use inclusion exclusion. Don't think it would be computationally feasible.

Last edited by Academic Actuary; 12-29-2017 at 01:03 PM..
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