text: # Probability of binary consecutive occurence with unequal probability
  
  #### cygi1989
  
  ##### New member
  I have been with this to some forums but it didn't help and I was advised to come to this one, so here is the question.
  
  My task is to compute given N - length binary series and P = p(0) and 1-P = p(1) expected number of consecutive occurence of 0 or 1 of k - length.
  For example 10011100 has one serie of 1 with length 1 and one serie with length 3,
  and also have two series of 0 of length 2.
  I know that when the probability of 0 and 1 is equal than
  Expected number of series of 0 or 1 of k-lenth in n-length binary numbers is = (n-k+3)/2^(k+2).
  But what the case when probability of 0 is for example (0.4) or any another?
  
  Sebastian
  
  Last edited by a moderator:
  
  #### chisigma
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  In order to clarify the question let's consider an example: if $p(0)=p(1)=\frac{1}{2}$ what is the probability to have a sequence of three ones in a string of four binary symbols?... in this case the possible strings are 1110 and 0111 on a total of 16 strings so that the requested probability is $\displaystyle P=\frac{2}{16}=\frac{1}{8}$...
  
  Is that correct?...
  
  Kind regards
  
  $\chi$ $\sigma$
  
  #### cygi1989
  
  ##### New member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Yes, it's correct.
  Regards to my formula it gives
  (4 - 3 + 3 ) / (2^ 5) = 4 / 32 = 1 / 8.
  My exact task is to find expected number of series of k - lenth for k = 1 .. 6, with P(1) = 800 / 2007 and N = 2007
  
  #### CaptainBlack
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Yes, it's correct.
  Regards to my formula it gives
  (4 - 3 + 3 ) / (2^ 5) = 4 / 32 = 1 / 8.
  My exact task is to find expected number of series of k - lenth for k = 1 .. 6, with P(1) = 800 / 2007 and N = 2007
  Does this need to be an exact closed form solution or will an Monte-Carlo estimate be OK?
  
  CB
  
  #### cygi1989
  
  ##### New member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Exact result will be better I guess because I have calculated estimation by myself (I run this test for 1000 esembles) and I need to compare my result with expected ones. Anyway, what result will come from Monte Carlo method and how to calculate it?
  
  #### CaptainBlack
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Exact result will be better I guess because I have calculated estimation by myself (I run this test for 1000 esembles) and I need to compare my result with expected ones. Anyway, what result will come from Monte Carlo method and how to calculate it?
  It sounds like you already have a Monte-Carlo estimate (which would be produced by generating random strings of length 2007 and counting run lengths for lengths 1..6 and dividing the totals by 2007).
  
  CB
  
  #### chisigma
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  I have been with this to some forums but it didn't help and I was advised to come to this one, so here is the question.
  
  My task is to compute given N - length binary series and P = p(0) and 1-P = p(1) expected number of consecutive occurence of 0 or 1 of k - length.
  For example 10011100 has one serie of 1 with length 1 and one serie with length 3,
  and also have two series of 0 of length 2.
  I know that when the probability of 0 and 1 is equal than
  Expected number of series of 0 or 1 of k-lenth in n-length binary numbers is = (n-k+3)/2^(k+2).
  But what the case when probability of 0 is for example (0.4) or any another?
  
  Sebastian
  Let p the probability of the symbol 1 [and 1-p the probability of the symbol 0...]. A sequence of k consecutive ones is detected if in a string of k+2 symbols the first and the last are 0 and all others are 1. The probability of such a recurrence is...
  
  $\displaystyle P_{k}= p^{k}\ (1-p)^{2}$ (1)
  
  ... and because in a sequence of n symbols there are n-k+3 possible subsequence on k+2 symbols, the expected number of k consecutive 1 is...
  
  $\displaystyle E_{n,k}= (n-k+3)\ P_{k}=(n-k+3)\ p^{k}\ (1-p)^{2}$ (2)
  
  For $p=\frac{1}{2}$ we find Your formula...
  
  Kind regards
  
  $\chi$ $\sigma$
  
  #### CaptainBlack
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Let p the probability of the symbol 1 [and 1-p the probability of the symbol 0...]. A sequence of k consecutive ones is detected if in a string of k+2 symbols the first and the last are 0 and all others are 1. The probability of such a recurrence is...
  
  $\displaystyle P_{k}= p^{k}\ (1-p)^{2}$ (1)
  
  ... and because in a sequence of n symbols there are n-k+3 possible subsequence on k+2 symbols, the expected number of k consecutive 1 is...
  And are they independent/uncorrelated?
  
  CB
  
  #### chisigma
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  And are they independent/uncorrelated?
  
  CB
  Is the 'they' related to the symbols, to the strings of k+2 symbols or to something else... please?...
  
  Kind regards
  
  $\chi$ $\sigma$
  
  #### CaptainBlack
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Is the 'they' related to the symbols, to the strings of k+2 symbols or to something else... please?...
  
  Kind regards
  
  $\chi$ $\sigma$
  The strings of k+2 symbols.
  
  CB
  
  #### chisigma
  
  ##### Well-known member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  The strings of k+2 symbols.
  
  CB
  Very well!... let's suppose to have a sequence of random binary symbols $a_{i}$ and, independently from i, is $P\{a_{i}=1\} = p \implies P\{a_{i}=0\}=1-p$. Now we consider a subsequence of k+2 consecutive symbols $\overrightarrow {a}_{i}= [a_{i},a_{i+1}, ... , a_{i+k+1}]$. Independently from i is...
  
  $\displaystyle P_{i,k}=P\{\overrightarrow {a}_{i}= [0,1, ... . 1, 0]\}= p^{k}\ (1-p)^2$ (1)
  
  ... so that, if You observe m subsequences of k+2 consecutive symbols, the expected value of times on which is $\overrightarrow {a}_{i}= [0,1, ... , 1, 0]$ is $m\ P_{i,k}$...
  
  Kind regards
  
  $\chi$ $\sigma$
  
  #### cygi1989
  
  ##### New member
  Re: Probability of binary consecutive occurence with uneqaul probability
  
  Let p the probability of the symbol 1 [and 1-p the probability of the symbol 0...]. A sequence of k consecutive ones is detected if in a string of k+2 symbols the first and the last are 0 and all others are 1. The probability of such a recurrence is...
  
  $\displaystyle P_{k}= p^{k}\ (1-p)^{2}$ (1)
  
  ... and because in a sequence of n symbols there are n-k+3 possible subsequence on k+2 symbols, the expected number of k consecutive 1 is...
  
  $\displaystyle E_{n,k}= (n-k+3)\ P_{k}=(n-k+3)\ p^{k}\ (1-p)^{2}$ (2)
  
  For $p=\frac{1}{2}$ we find Your formula...
  
  Kind regards
  
  $\chi$ $\sigma$
  Absolutely brilliant. Many thanks!