Why doesn't my combinatorial solution to simple probability problem work? I need to find the probability of NOT getting an ace card on two draws from a deck of 52 cards.
My first thought (which I really think is correct) was to get the probability from taking $\frac{48}{52}\frac{47}{51}=\frac{\binom{48}{2}}{\binom{52}{2}}\approx0.85$. Isn't this correct?
Then I though about it in another way. There are $\binom{52}{2}$ ways too choose two cards from 52. To not get an ace, you can to choose two out of 12 values (where the ace is excluded) in $\binom{12}{2}$ ways, and then you can choose $\binom{4}{1}=4$ different cards from each of the chosen values. Then I'm thinking you could calculate the probability by taking
$$\frac{\binom{12}{2}\binom{4}{1}^{2}}{\binom{52}{2}}\approx0.80$$
This doesn't give the same as the first method I used. I'm obviously missing something, but can't really figure out what is wrong. I would appreciate some guidance on how I should think differently doing the later method! (I suppose the probability is approx. 0.85 as I got in the first place.)

In the second method you don't take the case that both cards have the same value into account. That would be $12 \cdot \binom42$ many possibilities. When you add those, you get the same result.
