Why is the expected value, $E(X)$, after $5$ games $0$? In a game the probability that you win is $1/2$. If you win the game, you get $1\$$ and if you loose the game you loose $1\$$. Let $X$ denote the total amount of money after $5$ games. The expected value $E(X)=0$. Why is that?
 A: The chance you win is $50$ percent and the chance you lose is $50$ percent and both incur a win and lose of a dollar respectively. This implies that the expected value of the amount of money you will make after each game, $E(X_1)$, is $0$. The way this is calculated is you take the fractional probability you win and multiply it by the gain you receive and add that to the product of the fractional probability you lose and the money you give up when you lose. Mathematically this means:
$E(X_1)= \frac{1}{2}(1)+\frac{1}{2}(-1) = 0$
Now as mentioned in the comment by Aravind, expected value is linear. This means that the expected value of the money you make from one game can be multiplied by 5 to get the expected value of the money you make from five games. Thus:
$E(X)= E(X_1)*5 = 0*5 = 0$
To clarify the issue regarding your comment about the arithmetic mean argument. The expected value of playing three games for example is the arithmetic mean of every possible iteration of playing three games (each weighted equally in this case because the probably that each iteration happens is the same). So it is the mean of this quantity:
Situation 1: w, w, w = 1+1+1 = 3
Situation 2: w, w, l = 1+1-1 = 1
Situation 3: w, l, w = 1-1+1 = 1
Situation 4: w, l, l = 1-1-1 = -1
Situation 5: l, w, w = -1+1+1 = 1
Situation 6: l, w, l = -1+1-1 = -1
Situation 7: l, l, w = -1-1+1 = -1
Situation 8: l, l, l = -1-1-1 = -3
Note: w = win, l = loss, and a sequence w, w, l means win game 1, win game 2, lose game 3.
The arithmetic mean of this is: $\frac{3+1+1-1+1-1-1-3}{8} = \frac{0}{8} = 0$
A: There is an easier way:
Your random variable (each game) pmf is this
$$ p_Y(y) =
\begin{cases}
\frac{1}{2},  & \text{if $y=-1$ } \\
\frac{1}{2},  & \text{if $y=1$ }
\end{cases}$$
The question (trivial) is to calculate $\mathbb{E}[X=5Y]=0$
A different and more useful question could be: What's the probability that $X=k$?
SOLUTION
Observing that $\frac{Y+1}{2}\sim Bin(1;\frac{1}{2})$,  that is a Bernulli distribution with $p=\frac{1}{2}$, it is known that the sum of this bernulli is a binomial. In our case we have a binomial $Bin(5;\frac{1}{2})$ so, in other words,
$\mathbb{P}[X=k]=\binom{5}{\frac{k+5}{2}}(\frac{1}{2})^5$
$k =-5;-3;-1;1;3;5$
