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...
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
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...