Consider that we run consecutive identical independent random experiments where we pick a number from the set $\{0,1,2\}$. The probability of having $0$ is $p_0$ and the probability of having $1$ is $p_1$. What is the expected number of trials in order to have either $n$ consecutive $0$'s OR $n$ consecutive $1$'s?
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hi, I think the essence is captured here – P. J. Jan 19 '21 at 11:48
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@P.J. I understand, but the situation is somewhat more subtle. For example, if you have n-1 zeros and a 1 appears then you lose the current sequence but you have the first element of a possible sequence of n 1's. So, the question really lies on how to combine these two possible outcomes. – RTJ Jan 19 '21 at 11:54
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right! Maybe we can try something like this - $$E[N_k] = E[N_k^0 +N_k^1]$$ where $N_k$ is the number of trials for either k consecutive 0's ($N_k^0$) or k consecutive 1's ($N_k^1$). Linearity of conditional expectation should help things greatly. – P. J. Jan 19 '21 at 12:01