Expected number of tosses to get 3 consecutive Heads I have a fair coin. What is the expected number of tosses to get three Heads in a row? 
I have looked at similar past questions such as Expected Number of Coin Tosses to Get Five Consecutive Heads but I find the proof there is at the intuitive, not at the rigorous level there: the use of the "recursive" element is not justified. The Expectation $\mathbb E[X]$ is a number, not a random variable, as it is treated there. Please make this clear.
 A: Consider the outomes $T,HT,HHT,HHH$. All but the last put you back to the start. So if $x$ is the expected number of tosses to get HHH we have $$x=\frac{1}{2}(x+1)+\frac{1}{4}(x+2)+\frac{1}{8}(x+3)+\frac{1}{8}3\ \ (*)$$ That is easy to solve for $x$ giving $$x=14$$
---------- Added 26 June 2016 ----------
Now let us consider this solution more carefully. Note first that the events $T,HT,HHT,HHH$ are disjoint and exhaustive. They have probabilities $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{8}$ respectively.
Let $3H$ be the random variable "a number of tosses until the first run of three $H$ is achieved". Now $(*)$ states: $$E(3H)=E(3H|T)p(T)+E(3H|HT)p(HT)+E(3H|HHT)p(HHT)+E(3H|HHH)p(HHH)$$ This is sometimes known as "computing expectations by conditioning" (see, for example, Sheldon Ross, Introduction to Probability Models 3.4 p100). It is often written more concisely as $$E(E(X|Y))=E(X)$$ where the outer expectation on the LHS is $E_Y(\cdot)$.
---Added 5-2-2022---
The expected number of tosses to obtain three consecutive heads given that the first toss is a tail equals one plus the expected number of tosses to obtain three consecutive heads (starting from that point). $$\mathsf E(3H\mid T) = 1+\mathsf E(3H)$$
This is the $(1+x)$ factor in the above. So too are the other terms evaluated.

It follows directly from the definitions of conditional probability and expectation. The discrete case is particularly straightforward.
It just comes down to what is sometimes called the "partition theorem": if $B_n$ is a partition of the sample space, then $E(X)=\sum_nE(X|B_n)p(B_n)$. Note that we have $$E(X|Y=y)=\sum_xxp(X=x|Y=y)=\sum_xx\frac{p(X=x\cap Y=y)}{p(Y=y)}$$ where the last equality is just the definition $$p(A|B)=\frac{p(A\cap B)}{p(B)}$$
Having written all that, I see that Wikipedia calls it the "Law of total expectation" and has an excellent article on it.
A: Although the question has already been answered, I would like to offer a very similar solution but a different approach mindset to it.

It is a crude image but it essentially explains the answer above very beautifully.
At the beginning, we have no coins tossed so we have no consecutive heads. Next, we toss an $H$ or $T$ with probability $\frac{1}{2}$. Thus, we go the next state $2$ with probability $\frac{1}{2}$, similarly with $3$ and $4$.
Let $g(x)$ be the expected time until we reach state $4$, $HHH$, from state $x\in\{1,2,3,4\}$. Obviously $g(4)=0$ since we are already at state $4$!
\begin{align}
g(1)&=\frac{1}{2}(g(2)+g(1))+1\\
g(2)&=\frac{1}{2}(g(3)+g(1))+1\\
g(3)&=\frac{1}{2}(g(4)+g(1))+1\\
\end{align}
Since whenever we move from one state to another we take or "waste" one step. However, we only take a step in the correct direction with probability $\frac{1}{2}$. Otherwise, we have to go back to state $1$ and need to find $g(1)$.
Thus, by substitution (or recursion),
$$g(1)=1+\frac{1}{2}g(1)+\frac{1}{2}\left(\frac{1}{2}g(3)+\frac{1}{2}g(1)+1\right)=1+\frac{1}{2}g(1)+\frac{1}{4}g(1)+\frac{1}{2}+\frac{1}{4}\left(\frac{1}{2}g(1)+1\right)=1+\frac{1}{2}+\frac{1}{4}+g(1)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)$$

$$g(1)=14$$

