Understanding Conditional Probability with Intersections of Events I am attempting to ensure I understand conditional probability, so I came up with the following problem which seems to be posing some issues with regards to my understanding.  Say we have a deck of 16 cards.  There are four red, four green, four blue, four yellow cards.  Every color has a version with a star, circle, square, and triangle, so in total we have sixteen unique cards.  John has some random four of these sixteen cards.
Say we know John has the red circle card.  What is the probability he has 2 circle cards total?
Attempt at solution:
We are given that John has the red circle card, so we can look at this as 
$Pr[A|B]$, where A is the event John has 2 circle cards and B is the given event, that he has a red circle card.
I expanded this to:
$$\frac{Pr[A\cap B]}{Pr[B]}$$
The top portion represents all of the combinations where John has 2 circle cards, and one of them must be the red circle card.  This should be divided by the possible combinations of four cards we can have. So
$$Pr[A\cap B] = \frac{\binom{1}{1}\binom{3}{1}\binom{12}{2}}{\binom{16}{4}}$$
We choose 1 of the 1 red circle cards, 1 of the 3 other circle cards, and 2 from the remaining cards to form our group of 4.  We divide this by the cases of all combinations of 4 cards we can have, 16 choose 4, which is 1820.  We divide this whole equation by $Pr[B]$, the probability of the given, that we have 1 red circle card, which is $1/16$ if I am not mistaken. However, when I evaluate, I reach a value larger than 1, which I know is impossible in a probability scenario.  At what point does my logic fail here?
 A: Event $B$ corresponds to having the red circle card among four cards randomly drawn without replacement from the $16$ cards.  This probability is not simply $1/16$:  it is $$\Pr[B] = \frac{\binom{1}{1}\binom{15}{3}}{\binom{16}{4}} = \frac{1}{4}.$$  This is because, in the deck with one such card (red circle), there is $\binom{1}{1}$ way to select one card that is the desired card, and $\binom{15}{3}$ ways to select the other three cards that are not the desired card.
The same logic applies to the numerator event $\Pr[A \cap B]$ which is the probability of selecting a red circle card and any other circle card.  This can be done in $\binom{1}{1}\binom{3}{1}\binom{12}{2}$ ways as you correctly calculated.  Therefore, the desired conditional probability is $$\Pr[A \mid B] = \frac{\binom{3}{1}\binom{12}{2}}{\binom{15}{3}} = \frac{198}{455}.$$
Notice that once written this way, it becomes clear that the conditional probability is itself hypergeometrically distributed, and that this makes sense:  once the red circle card is chosen, it is fixed, and the problem becomes simply a matter of the probability of selecting exactly one of the three remaining circle cards, and two of the 12 non-circle cards, out of a total of three cards chosen without replacement from the set of 15 cards that exclude the red circle card.
The takeaway here is that you actually enumerated the more difficult joint probability correctly but overlooked the easier marginal probability calculation because it was deceptively simple.  Take care to think through your calculation and not take shortcuts or "intuitive" arguments.
