How many HH pairs appear in a 100-coin toss sequence?

Question

What is the expected number of 2-consecutive-head pairs in a sequence of 100 fair coin flips, allowing overlap (HHH equals two pairs)?

Answer 1

Quick estimate. Think of the sequence as wrapping around (makes the calculation easier). Then there are $100$ pairs. Each has a $25\%$ chance of being HH. So you expect $25$ such pairs.

Another way to look at it: each of the expected $50$ heads is equally likely to be followed by a head or a tail.

Answer 2

Consider the independent random variables $X_1,X_2,\dots, X_N$, where $N=100$ for short, taking the values $0,1$ with equal probability. We consider here an H to be one, a T to be zero. Then we need the mean (or expectation, expected value $\Bbb E$) of the variable $X_1X_2+X_2X_3+\dots+X_{N-1}X_N$, which is: $$ \begin{aligned} &\Bbb E[\ X_1X_2+X_2X_3+\dots+X_{N-1}X_N\ ]\\ &\qquad =\Bbb E[X_1X_2] + \Bbb E[X_2X_3]+\dots+\Bbb E[X_{N-1}X_N ]\\ &\qquad =\Bbb E[X_1]\;\Bbb E[X_2] + \Bbb E[X_2]\; \Bbb E[X_3]+\dots+\Bbb E[X_{N-1}]\; \Bbb E[X_N ]\\ &\qquad =\underbrace{\frac 12\cdot\frac 12 + \frac 12\cdot\frac 12+\dots+\frac 12\cdot\frac 12 }_{(N-1)\text{ times}}\\ &\qquad =\frac 14(N-1)\ . \end{aligned} $$

Answer 3

Let $Y_i=1$ if $X_i=X_{i+1}=H$ , $Y_i=0$ elsewhere. We have $E[Y_i]=P(Y_i=1)=\frac14$.

Further, the number of HH pairs is $T=\sum_{i=1}^{N-1} Y_i$ and

$$E[T]= \sum_{i=1}^{N-1} E[Y_i]=(N-1) \frac14$$