Expected number of aces in a poker hand So I'm having difficulty understanding a certain idea. Say I have a standard $52$ card deck (fairly shuffled, etc.) and I pull $5$ cards from it. We ask, what is the expected value of the number of aces in this hand.
Obviously we could indicator functions, so we can let $I_i$ denote the indicator function for the $i$th card and let $X = I_1 + \cdots + I_5$ denote the value we are trying to find. In essence we're trying to find $E(X)$.
(Indicator function basically means $I_i = 1$ if the $i$th card is an ace and $I_i = 0$ if the $ith$ card is not an ace. So $X$ would count the number of aces.)
I know that expected value doesn't care about dependence, so in particular $E(X) = E(I_1) + \cdots E(I_5)$. And since each $I_i$ is an indicator function, letting $p_i$ be the probability that the $i$th card pulled is an ace, we get $E(X) = p_1 + \cdots + p_5$.
At this point, every solution I've seen says, "The chance of any particular card being an ace is $\frac{4}{52}$ and so you get $5 \cdot \frac{4}{52} = \frac{5}{13}$", but this is precisely where I have a problem. I don't understand how the chance of any card being chosen being an ace has probability $\frac{4}{52}$. I totally understand that the 1st card has probably $\frac{4}{52}$ and so $p_1 = \frac{4}{52}$, but after that, the 2nd card no longer has probability $\frac{4}{52}$ since it's dependent on the first card. If the first card is an ace, the chances becomes $\frac{3}{51}$, if the first card isn't an ace, the chances become $\frac{4}{51}$. In either case, we get something different than $\frac{4}{52}$.
The standard definition in this case makes perfect sense:
$$E(X) = \sum_{x = 0}^4 x P(X = x) = \sum_{x = 0}^4 x \cdot \frac{\binom{4}{x} \binom{48}{5-x}}{\binom{52}{5}}$$
This is the birthday problem and obviously my brain is trying to work in this way, which is why it's having difficulty seeing the other way.
Can anyone help explain why there magically is no dependency when we look at it from the perspective of an indicator function, i.e. when $E(X) = \sum p_i$?
 A: Answer on request of the OP.


"but after that the 2nd card no longer has probability $\frac4{52}$ to be an ace..."

Then can you give me any reason why the second card should have less or more chance to be an ace than the first?... It seems that you are not looking at the plain probability that the second card is an ace but at conditional probabilities (which is unnecessary and confusing here).
If you still insist on that approach have a look at the following justification.
Let $A$ be the event that the first card is an ace and let $B$ be the event that the second card is an ace. Then: $$P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)=\frac3{51}\frac4{52}+\frac4{51}\frac{48}{52}=\frac4{52}$$
It must be said though that this is a somewhat artificial/unnatural effort to convince you. Much better is it if your intuition on this all can be enlightened.
If someone is asked to take one by one two cards of a deck and questions are asked about the second card then often (and in good caution) his mind wonders back to the first card that was drawn already and to the fact that both cards were drawn in somewhat different circumstances. If however the cards are put on $52$ distinct spots and questions are asked about the card that happens to be put on spot number $2$ (or any other spot) then this temptation is not present.
So in situations as sketched in your question just wonder: "what is the probability that the second card is an ace?" This without spending any thoughts on the character of the first card (or any other card). Then the answer is clearly $\frac4{52}$. And of course same is true for the other $4$ cards.

I sincerely hope that your intuition is enriched by this.
A: The expected number =$\sum\limits_{k=0}^4kP_k$ where $P_k=S_k\frac{Q_kR_k}{D}= \frac{\binom{4}{k}\binom{48}{5-k}}{\binom{52}{5}}$.
Term explanation for Hand that has $k$ aces. $S_k=\binom{5}{k}$ describes permutations of positions of $k$ aces in $5$ card hand.  $Q_k=\binom{4}{k}k!$ describes combinations of $k$ out of $4$ aces in hand.  $R_k=\binom{48}{5-k}(5-k)!$ describes possible non-aces in hand.  $D=\binom{52}{5}5!$ describes cards available before each card is dealt.
