Expectation value : trials with and without replacement 
Two cards are drawn at random from a deck of cards. Let X be the
  number of aces obtained. Then the value of E(X) is

Direct Method
$$
P(X=0)=P(\text{no ace})=\frac{{}^{48}C_2}{{}^{52}C_2}=\frac{48*47}{52*51}\\
P(X=1)=P(1 \text{ ace and }1\text{ non-ace})=\frac{{}^{4}C_1.{}^{48}C_1}{{}^{52}C_2}=\frac{4*48*2}{52*51}\\
P(X=2)=P(2 \text{ non-ace})=\frac{{}^{48}C_2}{{}^{52}C_2}=\frac{4*3}{52*51}
$$
$$
E(X)=\frac{384+24}{52*51}=\frac{408}{2652}=\frac{2}{13}
$$
Doubt
In some references it is solved by considering it a Bernoulli trial, Pls check Method.
If the experiment was with replacement, without doubt I'd say it a Bernoulli trial.
$$n=2\\p=\frac{4}{52}=\frac{1}{13}\\
E(X)=np=2*\frac{1}{13}=\frac{2}{13}$$
Why do we have the expectation value matching the one with replacement (Bernoulli trial) ?
 A: As there is some confusion in the comments, I'll post some computations.
First of all, to specify the process here you need to say whether or not you replace cards after each draw.  That has no effect on the expectation, but it will change the case by case probabilities.
In either case (with or without replacement) the probability that a single draw is an Ace is $\frac 4{52}$ hence, by Linearity of Expectation $$E=\frac 8{52}=0.153846154$$
With replacement:  Here we have a straight Binomial process.  The probability of drawing exactly $i$ Aces is $$P_i=\binom 2i\times \left(\frac 1{52}\right)^i\times \left(\frac {51}{52}\right)^{2-i}$$ whence
The case by case probabilities are $$P_0=0.852071006\quad P_1=0.142011834\quad P_2=
0.00591716$$ and we can then confirm the expectation.
Without replacement:  Now we get $$P_0=\frac {48}{52}\times \frac {47}{51}=0.850678733\quad P_1=2\times \frac 4{52}\times \frac {48}{51}=0.14479638\quad P_2=\frac 4{52}\times \frac {3}{51}=0.004524887$$ and, again, it is easy to verify the expectation.
A: "First card is an Ace" and "Second card is an Ace" are indeed Bernoulli events each with probability $\frac{4}{52}$, so the expected number of Aces as the first card is $\frac{4}{52}$ and the expected number of Aces as the second card is also $\frac{4}{52}$ 
They are not independent so you do not have a binomial distribution for the two cards together; you in fact have a hypergeometric distribution
But you do not need independence or a binomial distribution to find the expected number of Aces in the first and second cards. Linearity of expectation is enough to give you $E[X]= \frac{4}{52}+\frac{4}{52} = \frac{2}{13}$   
A: Here is a great article on the topic of Linearity of Expectation. To understand your specific question in the case without replacement (and why the expectations are still the same for all 3 cases), let X be the random variable that denotes the number of aces drawn after 2 draws without replacement. We have:
$$ E(X) = E(A) + E(B) $$
where A is the random variable s.t. A = 1 when the 1st card is an ace, and 0 otherwise; B = 1 when the 2nd card is an ace, and 0 otherwise.
E(A) is a simple case:
$$ E(A) = \sum_A P(A) * A $$
$$ = 0 * P(A=0) + 1 * P(A=1) $$
$$ = P(A=1) $$
$$ = 4/52 $$
E(B) will be the same as E(A), even though A and B are not independent:
$$ E(B) = \sum_B P(B) * B $$
$$ = 0 * P(B=0) + 1 * P(B=1) $$
$$ = P(B=1) $$
$$ = P(B=1 \cap A=1) + P(B=1 \cap A=0) $$
$$ = P(B=1 | A=1) * P(A=1) + P(B=1 | A=0) * P(A=0) $$
$$ = \frac{3}{51} * \frac{4}{52} + \frac{4}{51} * \frac{48}{52} $$
$$ = \frac{4}{52} * (\frac{3}{51} * \frac{48}{51}) $$
$$ = \frac{4}{52} * 1 = \frac{4}{52} $$
You can read the article to see the proof of why Linearity of Expectation applies to dependent variables.
