Flip a coin multiple times. When the process stops, we see one head. Question, how many tails in expectation?

Question

Intuition says one, but calculation using negative binomial distribution says two. got confused? Help please.

Here is my calculation.

Note that when the process stops, we observe $x$ heads. The process is equivalent to that once we observe the $x+1$ heads, the process stops immediately. So, the number of tails follows the negative binomial distribution with parameters ($x+1$,$1/2$) and its mean is $x+1$.

For the case, $x=1$, on average we observe $2$ tails when the process stops and $1$ head is observed.

Does it make sense? Thank you again!

Answer 1

The formula should be correct, if you are counting the right thing.

You want the expected count of successes (tails) before the first failure (head). $$X_{r,p}\sim\mathcal{NegBin}(r, p)\implies \mathsf E(X_{r,p})=\dfrac{pr}{1-p}$$

$$X_{1,1/2}=1$$

You may be using the formula for the expected count of trials until the first failure.   That counts the head as well as the tails.

Answer 2

It is easier to analyze as expected number $(N)$ of tosses for the first head to show up.
Assuming a fair coin $N=\sum_{n=1}^\infty n(\frac{1}{2})^n$. To carry out the calculation let $f(x)=x\sum_{n=1}^\infty nx^{n-1}=x\frac{d}{dx}\sum_{n=1}^\infty x^n=x\frac{d}{dx}\frac{x}{1-x}=\frac{x}{(1-x)^2}$
This gives us $N=f(\frac{1}{2})=2$ This means that the expected numbe of tails $=N-1=1$.