# Expected number of running heads in coin toss

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.

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

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

yielding

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

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

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

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

• Thank you very much @whuber. Sep 6 '20 at 12:02