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
    $\begingroup$ +1. My thoughts exactly, I immediately thought: "This is a geometric, let's plug the numbers!" $\endgroup$ – usεr11852 says Reinstate Monic 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, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.


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