Finding the expectation of a truncating event given a particular outcome? Approaching the following problem:

Gambles are independent, and each one results in the player being
  equally likely to win or lose 1 unit.  Let $W$ denote the net winnings
  of a gambler whose strategy  is to stop gambling immediately after his
  first win. Find $E[W]$

Realizing that the expectation is not just $E[ W] = \sum\limits_{x} i \left(\frac{ 1}{ 2}\right)$ I am unsure of how to approach the problem.
Letting $W_L$ be the value of $W$ accumulated by loses and $W_W$ be the value of $W$ accumulated by wins, I am inclined to believe we are looking at $E[ W_L] = \sum\limits_{i = 0}^\infty -i \left(\frac{ 1}{ 2}\right)^i$ and the $E[ W_W] = 1$.  
However, 1) I do not know if this is the correct approach and 2) should this indeed be the correct approach, I do not understand [conceptually ] how to merge these two summations? 
 A: Let $X_1,X_2,\dots$ be the winnings on the individual bets, and $N$ the stopping time representing the gambler's strategy. Since it is a fair game, we have $E(X)=(-1){1\over 2}+(1){1\over 2}=0$. The gambler's total winnings are $W=X_1+X_2+\cdots+ X_N$ and by Wald's equation we get  $\mathbb{E}(W)=\mathbb{E}(N)\mathbb{E}(X)=0.$
The moral of the story is that the average winnings of a fair game is zero, no matter what stopping strategy you adopt.
A: Let $W$ be the "winnings," and $X$ the number of trials until the first head. Then $W=-(X-1)+1=2-X$.
Now use the fact that $E(X)=\frac{1}{1/2}$ and the fact that $E(2-X)=2-E(X)$.
Here we used the fact that if $p$ is the probability of success on any trial, and $p\ne 0$, then $E(X)=\frac{1}{p}$.  (The random variable $X$ has geometric distribution.)
There are various ways to prove the required result. For example, when $p=1/2$, the probability that $X=n$ is equal to $\frac{1}{2^n}$. So
$$E(X)=1\cdot \frac{1}{2}+2\cdot \frac{1}{^2}+3\cdot \frac{1}{2^3}+\cdots.\tag{$1$}$$
Multiply by $2$. We get
$$2E(X)=1+2\cdot \frac{1}{2}+3\cdot \frac{1}{2^2}+4\cdot\frac{1}{2^3}+\cdots.\tag{$2$}.$$
Subtract $(1)$ from $(2)$. We get
$$E(X)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots.$$
The series on the left is an infinite geometric series with sum $2$.
Another way: We find $E(W)$ directly.  With probability $\frac{1}{2}$ we win immediately, ending up with $1$ dollar.  With probability $\frac{1}{2}$ we "win" $-1$ dollar on the first throw, and in effect the game starts all over again, and our expected net gain is $E(W)$. It follows that
$$E(W)=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot(-1)+\frac{1}{2}E(W)).$$
On the assumption that $E(W)$ is finite, we can now solve the above equation for $E(W)$, and find that $E(W)=0$.
A: Let $X$ denotes the number of trial of which winning occurs. Then we have:
$W=1+(-1)(X-1)=2-X$
Hence the expectation would be:
$E[W]\\=E[E[W|X]]\\=\sum\limits_{x} E[W|X=x]P\{X=x\}\\=\sum\limits_{x=1}^\infty \frac{(2-x)}{2^x}\\=0$
Note: The only thing that is little tricky here is to find $\sum\limits_{x=1}^\infty \frac{x}{2^x}$, however we have:
$\frac{1}{2}\sum\limits_{x=1}^\infty \frac{x}{2^x}\\=\sum\limits_{x=2}^\infty \frac{x-1}{2^x}\\=(\sum\limits_{x=1}^\infty \frac{x}{2^x}-\frac{1}{2})-\frac{1}{2}$
So that $\sum\limits_{x=1}^\infty \frac{x}{2^x}=2$.
