Expected value of a coin toss You flip a coin. If you get heads you win \$2 if you get tails you lose \$1. What is the expected value if you flip the coin 1000 times? 
I know that the expected value of flipping the coin once is $\frac{1}{2}(2) - \frac{1}{2}(1) =0.50$
Would the expected value be 500? 
 A: Let $X$ be the random number of heads obtained in $n = 1000$ flips of a fair $(p = 1/2)$ coin.  Then $$X \sim \operatorname{Binomial}(n = 1000, p = 1/2),$$ and the expected value of the number of heads is $$\operatorname{E}[X] = np.$$  Now, the net winnings (or losses) is equal to 2 dollars for each head minus 1 dollar for each tail observed.  So if we observed $X$ heads, then presumably we also observed $1000-X$ tails, and the net winnings/losses is therefore $$W = 2X + (-1)(1000-X) = 3X - 1000.$$  Because $X$ is a random variable, $W$ is also a random variable.  The expectation of $W$ is simply $$\operatorname{E}[W] = \operatorname{E}[3X - 1000] = 3 \operatorname{E}[X] - 1000 = 3 np - 1000 = 3(1000)(1/2)-1000 = 500.$$
Notice that this approach lets us calculate the expected winnings for any general payoff structure where we win $s$ dollars for each head and lose $t$ dollars for each tail.  Then $$W = sX - t(n-x) = (s+t)X - nt,$$ and $$\operatorname{E}[W] = (s+t)(np) - nt = n(sp - t(1-p)).$$  What this tells us is that the expected winnings/losses for a fixed $s$ and $t$ will in general become more extreme as the number of trials increases (if $sp \ne t(1-p)$).
We can also use the above to calculate the variance; e.g., $$\operatorname{Var}[W] = (s+t)^2 \operatorname{Var}[X] = (s+t)^2 np(1-p).$$
