On the intuition behind a conditional probability problem. This is a very similar question  to this one. But notice the subtle difference that the event that I define $B$ is that I am dealt at least an ace.
Suppose I get dealt 2 random cards from a standard deck of poker (52 cards).
Let the event $A$ be that both cards are aces, let $B$ be the probability that I am dealt at least an ace and let $C$ be the probability that I am dealt the ace of spades.
We have
$P(A| C) = \frac{3}{51}= \frac{1}{17} $  and $$P(A|B) = \frac{P(A,B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1 - \frac{\binom{48}{2}}{\binom{52}{2}}} = \frac{1}{33}$$ 
I do not intuitively uderstand  how $P(A|B) < P(A|C)$. Surely the probability of having been dealt 2 aces given that I am dealt at least an ace should be higher than the probability of having been dealt 2 aces given that I am dealt the ace of spades?
Could I get some help in understanding where my intuition is failing? Is there an other way to approach this problem?

I like pictures.
$A$  is the event that both cards are aces, B is the event that at least one card is an ace and $C$ is the event that one of the cards is the ace of spades.  Since, $ C \subset B$ and $A \subset B$, the size of these events can be visualized as follows

$$P(A|B) = { P(A)\over P(B) }$$ and $$P(A|C) = { P(A \cap C) \over P(C) }.$$
