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You throw $N$ times a fair coin. What is the expected number of subsequences of $k$ consecutive heads?

For instance, with $N=6$ and $k=2$, for the outcome HTHHHH, we would get $3$ such subsequences.

I know the solutions to two related problems, but did not find this one being asked:

  • what is the expected number $N$ needed to get $k$ consecutive heads when throwing coins (see here)

  • what is the probability of having at least $k$ consecutive heads for $N$ throws (see here)

charmd
  • 6,212

3 Answers3

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For $j = 1,\ldots,n$, let $X_j = 0$ if the $j$-th coin flip is tails and $X_j = 1$ if the $j$-th coin flip is heads. The number of subsequences of $k$ consecutive heads is $$N = \sum_{j = 0}^{n-k}X_{j+1}X_{j+2}\cdots X_{j+k}.$$ Do you see why this is true? Now, using linearity of expectation followed by the fact that $X_1,\ldots,X_n$ are independent, we have \begin{align*} E[N] &= E\left[\sum_{j = 0}^{n-k}X_{j+1}X_{j+2}\cdots X_{j+k}\right] \\ &= \sum_{j = 0}^{n-k}E[X_{j+1}X_{j+2}\cdots X_{j+k}] \\ &= \sum_{j = 0}^{n-k}E[X_{j+1}]E[X_{j+2}] \cdots E[X_{j+k}]. \end{align*}

Can you finish the problem from here?

JimmyK4542
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The number of possible sub-sequences is (n-k+1). Multiply these by the probability for each, p = (.5)^k. So, for n=6, k=2, E [k consecutive heads] = (6-2+1)(.25) = 1.25.

Bafs
  • 102
  • I agree that "the number of possible sub-sequences [can be counted as] $n-k+1$", but I am not sure what argument you are thinking of when writing this – charmd Nov 28 '20 at 06:48
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All head streaks either end $\ldots HT$ or are at the far end ending $\ldots H$ and all such cases are head streaks

The expected number of times $TH$ appears is $\frac{n-1}{4}$ and the probability the final throw is an $H$ is $\frac12$

So add these together to get $\dfrac{n+1}{4}$ expected head streaks

Henry
  • 169,616