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What is the expectation of two consecutive heads in first time? I tried something like this: let's say that $\xi$ is a random variable that says the first time when we have two consecutive heads. For example, $\xi$ = 2 if we have two consecutive heads in two first attempts, $\xi$ = 3 if we have two consecutive heads from the second and third atempts etc. if $p$ - probability of head ($\frac{1}{2}$), $q$ - probability of tail ($\frac{1}{2}$), $P(\xi = 2) = (\frac{1}{2})^2$, $P(\xi = 3) = (\frac{1}{2})^3$, $P(\xi = 4) = 2*(\frac{1}{2})^4 (TTHH, HTHH)$, $P(\xi = 5) = 3*(\frac{1}{2})^5 (TTTHH, HTTHH, THTHH)$ etc. Then we have $E(\xi) = 2*(\frac{1}{2})^2 + \sum_{k=3}^\infty k * (k-2)*(\frac{1}{2})^k$. But the sum of this condition is 3 although it should be 6. What is wrong?

prostak
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    You seem to be extrapolating from a very small sample. Already, with $\zeta=6$, there are $TTTTHH, HTTTHH, THTTHH, TTHTHH, HTHTHH$, so $5$ cases...where your linear extrapolation would yield $4$. – lulu Jul 22 '23 at 00:08
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    here is a duplicate for your underlying question. – lulu Jul 22 '23 at 00:10
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    And here is a way to count the binary strings with no $11$, which is basically what you need to do the problem your way. (Technically, you need to count those strings that do not end in $1$, but it should be easy for you to prove that this is the same as the general number shifted in length by one). – lulu Jul 22 '23 at 00:13
  • @lulu Thank you – prostak Jul 22 '23 at 04:27

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