What is the expected number of coin tosses needed to obtain a head? Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.
I understood that the formula for the expected value is:
$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$
The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.
So, if I get a head on the first try, then  $ p_1 = \frac{1}{2} , x_1 = 1 $
If I get a head on the second try, then $ p_2 = \frac{1}{4} , x_2 = 2 $
And then, I woudl have that:
$$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$
So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.
My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?

To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with
$$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},
$$
hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$
hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete.
