This is an example question from my textbook that I do not understand.
The solution states, if we define an indicator function $I_{i}$ that equals 1 when the $i^{th}$ card is an ace, and zero otherwise.
Clearly X=$\Sigma_{i=1}^{5}I_{i}$.
They state that, since we are sampling without replacement, $I_{i}$ are all identically distributed. Therefore the solution they give is $E(X)=5E(I_{1})=5 \cdot \frac{4}{52}$ (text book solution).
However, if they are sampling without replacement, wouldn't the individual probabilities change?
I.e. $I_{1}=P(1^{st} card \ is \ ace)=\frac{4}{52}$
$I_{2}=P(2^{nd} card \ is \ ace)=\frac{3}{51} $
etc.
The solution I would have come up with is
$E(X)=E(I_{1})+...+E(I_{5})=\frac{4}{52}+\frac{3}{51}+\frac{2}{50}+\frac{1}{49}+0.$
Why is this not correct?
The way you wrote $I_2$, it's the probability the second card is an ace assuming the first card is an ace (i.e., assuming there are $3$ aces and $51$ cards left in the deck). The probability the first card is an ace is only $\frac 4 {52}$.
You must add the probability that the second card is an ace assuming the first card is not an ace. The probability the first card is not an ace is $\frac {48} {52}$, and the probability the second card is an ace in that scenario is $\frac {4}{51}$.
So looking at it this way comes to the result in the textbook, since $$\frac 4{52} \frac 3{51}+\frac {48}{52} \frac 4 {51}= \frac 4{52}\left(\frac3{51}+\frac{48}{51}\right)=\frac4{52}$$ is the probability the second (and in fact each) card is an ace.