1

David repeatedly flips a fair coin. Find the expected value of the total number of heads he will flip before flipping two consecutive tails.

(A) $2$ (B) $5/2$ (C) $3$ (D) $4$ (E) $9/2$

I'm totally confused on this problem. Here's the two posts that go with it:

The first is by a guy named xwang1 who posted an incorrect solution, and that's what my question is about:

Let $E_x$ ($x=0$ or $1$)be the expected number of heads we need to flip until we have $2$ consecutive tails. So then we have $E_0=1+\frac{1}{2}E_0+\frac{1}{2}E_1$ and $E_1=1+\frac{1}{2}E_0$. Solving we get $E_0=\boxed{6}$ and $E_1=4$.

I know that it's actually 3 and 2 because the next post corrects it, but what does "Let $E_x$ ($x=0$ or $1$)be the expected number of heads we need to flip until we have $2$ consecutive tails" mean and how does it go to the next step?

My main question is what does $E_x$ mean.

asdf334
  • 376

1 Answers1

1

$x$ is the number of tails after the most recent head. At the start, there have been no recent tails so the expected number is $E_0$.

Empy2
  • 50,853