Probability of drawing 2 aces that are not spades? I am trying to come up with a combinatorics formula for drawing 2 aces that are not spades. I know that I can use 13C1 • 4C2 to find the number of ways to draw two aces,but I Am not sure how to write it to not include spades.
 A: This is a trivial application of the hypergeometric distribution which is just a fancy name for counting the number of valid hands that we are interested in and dividing by the total number of hands.
Here, there are three cards that we are interested in possibly drawing, namely $A\heartsuit,~A\diamondsuit,$ and $A\clubsuit$.  We want to draw two of those and nothing else when drawing two cards.  There are $\binom{3}{2}=3$ ways that we can do this if order doesn't matter (able to be seen without the use of binomial coefficients by simply picking which of the three aces wasn't the one that was picked).
There are, as alluded to earlier, $\binom{52}{2}$ equally likely ways in which we pick two cards from the deck if order doesn't matter.
The probability is then:
$$\frac{\binom{3}{2}}{\binom{52}{2}}$$
Or, if you prefer so that you stick with the format of the hypergeometric distribution how it is usually written: $\dfrac{\binom{3}{2}\binom{49}{0}}{\binom{52}{2}}$, noting that the extra term of $\binom{49}{0}$ is unnecessary since it simply evaluates to $1$.

If you prefer, you can think about this where order matters instead and it won't change the final result.  You get $3\times 2$ ways you can pick two non-spade aces in sequence compared to $52\times 51$ ways you can pick two cards of any type in sequence, giving
$$\frac{3\times 2}{52\times 51}$$
Yet another way you can see this is by looking at the probability the first card is a non-spade ace which is $\frac{3}{52}$ and multiplying this by the probability that the second card is also a non-spade ace given that the first was too which is $\frac{2}{51}$ giving
$$\frac{3}{52}\times\frac{2}{51}$$
which of course equals the same as we got in the other ways of looking at the problem.
