# Expected number of coin tosses for consecutive heads or tails

A fun elementary exercise involving expectation is to compute the number of times we expect to toss a coin before seeing consecutive heads or tails.

Let $p_n$ denote the probability that it takes us $n \geq 2$ tosses to witness the first occurrence of consecutive heads or tails. Of all the $2^n$ possible sequences of flips, there are only $2$ that satisfy the condition that the last two flips are identical and the first $n-1$ flips are strictly alternating. For example, if it took us 3 tosses then we either observed the sequences HTT or THH. Thus we have

$\begin{equation*} p_n = \frac{2}{2^{n}} = 2^{1-n}. \end{equation*}$

These probabilities $p_n$ define a discrete probability distribution over the natural numbers. The expectation (or mean value) of this distribution is precisely the number of coin tosses needed to observe consecutive heads or tails. Letting $E$ denote this expectation, we have by definition

\begin{align} E &= \sum_{n=2}^{\infty} n \cdot p_n \\ &= \sum_{n=2}^{\infty} \frac{n}{2^{n-1}} \label{eq:def}. \end{align}

The right-hand expression \eqref{eq:def} can now be split into a series whose terms are themselves geometric series via

\begin{align*} \eqref{eq:def} &= \left(\sum_{n=2}^{\infty} \frac{1}{2^{n-1}} \right) + \sum_{n=2}^{\infty} \left( \sum_{j=n}^{\infty} \frac{1}{2^{j-1}} \right) \\ &= \sum_{n=1}^{\infty} \frac{1}{2^{n}} + \sum_{n=1}^{\infty} \sum_{j=n}^{\infty} \frac{1}{2^{j}} \\ & = 1 + \sum_{n=1}^{\infty} \frac{1}{2^{j-1}} \\ &= 3. \\ \end{align*}

We have thus proven that the expected number of coin tosses is in fact only 3.