# 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.

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 $$Var(N) = 6 - 2^2 = 2$$.

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:

$$\mathbb{V}(N) = \frac{1-p}{p^2} = \frac{1-\tfrac{1}{2}}{(\tfrac{1}{2})^2} = \frac{\tfrac{1}{2}}{\tfrac{1}{4}} = 2.$$

• +1. My thoughts exactly, I immediately thought: "This is a geometric, let's plug the numbers!" – usεr11852 Oct 22 '18 at 23:09

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, $$E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$$.