Variance of Coin Flips Until H If we flip a fair coin until we get heads, what is the variance of the number of flips to do this?
My attempt is:
$$E(flips):=Y=1\times P(H)+(1+Y)\times P(T)$$
$$\Rightarrow Y=\frac{1}{2}+\frac{1+Y}{2}$$
$$\Rightarrow Y=2$$
I know this is correct. Now we attempt to compute:
$$E(flips^2):=X=1^2\times P(H)+(\sqrt{X}+1)^2\times P(T)$$
$$\Rightarrow X=\frac{1}{2}+\frac{(\sqrt{X}+1)^2}{2}$$
$$\Rightarrow X=2(2+\sqrt 3)$$
I know the right answer is $E(flips^2)=6$, but I'm not sure how to solve it using this recursive strategy, as the above seemed most natural to me.*
*ie take the expected value $X$, square root it to get the number of flips (not squared), add one, then square it again.
 A: You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails.  We will compute $E(N^2)$ using the Law of Total Expectation: 
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$.  Let's find its distribution.  Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$.  Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$.  Thus
\begin{align} E(N^2) & = E(E(N^2 | H_1)) \\
& =  1^2 \cdot \frac{1}{2} + (E(N^2) + 2E(N) + 1)\cdot 1/2 \\
& =  \frac{1}{2} + (E(N^2) + 2(2) + 1)\cdot 1/2. \\
\end{align}
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
A: The excellent answer by Gordon Honerkamp-Smith already answers your question, but I will give you an alternative derivation that goes directly to the distributional form.  Let $N$ be the number of flips until the first head appears.  This random variable has a geometric distribution:
$$N \sim \text{Geom}(p = \tfrac{1}{2}).$$
Using the known form for the variance of this distribution, you get:
$$\mathbb{V}(N) = \frac{1-p}{p^2} = \frac{1-\tfrac{1}{2}}{(\tfrac{1}{2})^2} = \frac{\tfrac{1}{2}}{\tfrac{1}{4}} = 2.$$
A: Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=\frac{1-p}{p^2}$ and $E(flips)=\frac{1}{p}$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
