Playing card game probability problem! So the problem is stated below:

A card game of 52 cards are dealt between 4 people (including me), each receiving 13 cards. 4 of the cards are Aces; one is called Ace of Spades and the other Ace of Clubs.
I wish to work out the probability that I receive neither the Ace of Spades nor the Ace of Clubs.

Progress
let S=receiving one Ace of Spades in hand,
$P(S)=1/4$
let C= Ace of Clubs in hand
$P(C)= (1/51)/(1/12)=4/17$
$=> P(S)*P(C)=1/17$ which gives the probability of getting both an Ace of Spades and an Ace of Clubs.
If I let A=getting all 4 Aces in my hand
$P(A)=[(52-4)!13!]/[52!(13-4)!]$
How do I carry on from there to find the probability that I receive neither the Aces of Spades and Clubs? And, what will be the probability that I receive at least one?
Thank you so much! :)
 A: A simple picture will give a simple answer:
Visualize 52 ordered slots into which to put the 52 distinct cards;  the first thirteen slots will hold your cards.
Put the Ace of Spades in a slot first,  For you to not get it, it must land in one of $39$ slots that you don't own, out of the $52$ possible.  Next, put out the Ace of Clubs.  To miss it, you must put it in one of the $38$ remaining slots you don't own, out of the $51$ remaining slots.  So the answer is: $$P_{neither}=\frac{39}{52}\times \frac{38}{51}$$
As to the probability of getting at least one:  Either you get none, or you get some.  The two probabilities will add to $1$.
A: The result of P(A) is also right.
The probability to receive neither the Aces of Spades and Clubs is
$P(A^c \cap S^c)= \frac{50}{52}\cdot \frac{51}{49}\cdot\frac{50}{48}\cdot \ldots \cdot \frac{49}{41}\cdot \frac{48}{40}=\frac{{2 \choose 0} \cdot {50 \choose 13}}{{52 \choose 13}} $
Just draw one card after the other. After each drawing, the amount of all cards (denominator) and the amount of cards without aces of spades and clubs (denominator), are reduced by one card.
The probability that you receive at least one is the converse probability of $P(A^c \cap S^c)$
