Suppose each coin toss is independent, what is the expected number of coin tosses until a run of "k" successive heads occur? Tried finding a recursive expression to solve the problem but got completely lost.
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You might want to look here. – rogerl Jul 27 '14 at 17:12
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and https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads – PatrickT Feb 11 '19 at 05:22
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Let $R_k$ be the number of tosses until the first run of $k$ heads and $E_k := E(R_k)$. The event that the first run of $k$ heads appears is the disjoint union of the event that the first run of $k-1$ heads appears followed by a head and the event that the first run of $k-1$ heads appears followed by a tail followed eventually by a run of $k$ heads.
If the probability of heads is $p$, then using the properties of conditional expectation
$$E_k = p(E_{k-1} + 1)+ (1-p)(E_{k-1} + 1 + E_k)$$
and
$$E_k = \frac1{p}(E_{k-1} +1).$$
Since $E_1 = 1/p$, we can solve recursively to obtain
$$E_k = \sum_{j=1}^{k}p^{-j}= \frac{(\frac1{p})^k-1}{1-p}.$$
RRL
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