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?

For me attaching some numbers to the problem makes it more intuitive.
In all, there are $52*51-48*47=396$ possible ways in which you can draw at least 1 ace from the deck. There are $1*51+51*1=102$ ways in which you can draw the Ace of Spades from the deck. So it's about $\frac14$ as likely to draw specifically the Ace of Spades. The deviation from exactly $\frac14$ comes from the 12 ways in which is possible to draw 2 aces. ($102*4-12=396$) 
Now, out of the 396 ways you can draw at least one ace from the deck, there are only 12 ways to draw 2 aces. However, out of the 102 ways in which you can specifically draw the Ace of Spaces, there are 6 total ways in which you can draw two aces. 
So drawing a specific Ace from the deck is only about a fourth as likely, but drawing a specific ace includes half the ways in which you can draw two aces. This meshes perfect with your calculated probabilities, since $\frac1{33}$ is almost half of $\frac1{17}$.
