Setting up questions relating to expected number I'm having difficulty setting up questions relating to finding the expected number of something. So, for example, this question "Eight cards are drawn with replacement from a standard 52 card deck. Find the expected number of different ranks (Ace, 2, 3, ..., King) that occur among the eight cards."
So the set up looks something like
0
(12/13)^8          
1
1-(12/13)^8
with x = # diff ranks
1, at least one of rank i in the 8
0, no rank i in 8
What I don't get is why is (12/13) raised to the 8th power. I can see why it is 8, because we have 8 cards, but why is it necessary to raise it to the power. Using the stuff from above, E(x) = 13 (1-(12/13)^8). I understand why we're using the probability from 1, but why do we need to multiply it by 13?
 A: The essential point here is that the expectation of the number of ranks is simply the sum of the probabilities that each rank is represented.  Since this is the same for each rank, we can simply compute it for one rank and multiply by $13$.  
The probability that we get at least one Ace is $1$ minus the probability that we get no Aces, which is how we get $$1-\left(\frac{12}{13}\right)^8$$  When we do the problem with replacement, as above, we have a probability of $\frac{12}{13}$ of not getting an Ace on each draw.  The draws are independent, so to get the probability that we don't get an Ace on any of the eight draws, we just multiply the eight probabilities together.
I guess it's the first part you're having trouble with, from what you said.  For $i=1,2,3\dots,13$ let $$I_i=\cases{1, &if we draw a card of rank $i$\\0,&otherwise}$$  Then we want to compute the expectation of$$\sum_{i=1}^{13}{I_i}$$  Now if $p_i$ is the probability of getting at least one card of rank $i$, I think you can see that the expectation is just $$\sum_{i=1}^{13}{p_i}$$ which simplifies to $13p$ in this case, as discussed earlier. 
If the problem were "without replacement" all that changes is computing the probability of not getting an Ace.  We still would multiply it by $13$ to get the final answer.  Here we would say that there are $\binom{52}{8}$ ways to draw $8$ cards, and there are $\binom{48}{8}$ ways to draw eight cards that aren't Aces, so the the probability of drawing no Aces is $$
\frac{\binom{48}{6}}{\binom{52}{8}}=\frac{48\cdot47\cdot46\cdot45}{52\cdot51\cdot50\cdot49}$$
