How to find the expected number of running heads of a specific length (say 'k' exactly) in 'n' tosses of a coin (fair/biased). For example, consider the output of a coin toss as follows "THHHTHTHHHTTTHHTHHT". The number of running head of length 1,2 and 3 are 1,2 and 2 respectively.


1 Answer 1


Index the tosses by $i=1,2,\ldots, n.$ Let $X_{i;k}$ be the indicator of the event "a run of exactly $k$ heads terminates at toss $i.$" From the formula of expectation (as the product of values by their probabilities), the expectation of any indicator is its chance of being $1:$

$$E[X_{i;k}] = \Pr(X_{i;k}=1).$$

When the coin has a chance $p$ of heads (independently at each toss) and therefore a chance $q=1-p$ of tails, the independence assumption shows the chance that $X_{i;k}=1$ must be the chance of $k$ heads preceded and followed by either (a) a tails or (b) the terminus of the sequence. Thus,

  1. When $i\lt k,$ $\Pr(X_{i;k}=1)=0.$

  2. When $i=k,$ $\Pr(X_{i;k}=1) = p^kq$ if $k\lt n$ and otherwise $\Pr(X_{i;k}=1)=p^k.$

  3. When $k \lt i \lt n,$ $\Pr(X_{i;k}=1)=p^kq^2.$

  4. When $i=n$ and $n\gt k,$ $\Pr(X_{i;k}=1)=p^k q.$

Let $N_{k;n}$ be the number of runs of exactly length $k$ in the sequence. Since

$$N_{k;n} = \sum_{i=1}^n X_{i;k},$$

taking expectations gives

$$E[N_{k;n}] = \sum_{i=1}^n E[X_{i;k}]$$


$$E[N_{k;k}] = p^k,$$

$$E[N_{k;k+1}] = 2p^kq,$$

and generally for $n\ge k+2,$

$$E[N_{k;n}] = 2p^kq + (n-k-1)p^kq^2.$$

  • $\begingroup$ Thank you very much @whuber. $\endgroup$ Commented Sep 6, 2020 at 12:02

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