Probability of Drawing Two Aces Conditioned on Drawing One Being Ace of Spades

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I am studying probability and came up upon something that I couldn't understand.

The problems is to calculate the probability of getting two aces, given that one of the two cards drawn is the ace of spades. I know that we need to calculate the intersection of the probability of the two events and divide that by the probability of getting the ace of spades. Now the answer that is provided is this:

$$\dfrac{2\times \frac{1\times 3}{52\times 51}}{\frac{1*\binom{51}{1}}{\binom{52}{2}}}=\frac{1}{17}$$

I understand the denominator part. Where we choose first choose the ace of spade, then choose any of the 51 cards and divide that by all possible outcomes (52C2).

However, what I do not understand is why is the numerator multiplied by two. I reckon that the probability of getting the ace of spade and another ace should be just 1/52 multiplied by 3/51

$$\frac{1}{52} \times \frac{3}{51}$$

Why do we multiply by two?

Thank you

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Maybe if I write your solution like this, it will be more clear:

$$\cfrac{\binom{2}{1} \cdot \cfrac{1}{52} \cdot \cfrac{3}{51}}{\cfrac{\binom{1}{1}\binom{51}{1}}{\binom{52}{2}}} = \cfrac{1}{17}$$

There are two ways to draw one ace of spades when drawing two cards: either you draw the ace of spade first, or you draw it second. Thus, $\binom{2}{1}$.