Choose 3 cards from a deck, if last two are spades what is the chance of first card being a spade? I'm stuck on this problem, and need some explanation if possible :

From 52 cards we take 1, after that we take 2 more, both of which are spades. What are the odds that first card is also a spade?

In particular, I'm not sure how to set up my Hypotheses/Evidence. I have a feeling that there might be some case branching or something but I'm unsure on how to set up the whole problem.
 A: Hint: This question could be asked in a very strange way that might actually be easier to understand: Knowing that the NEXT TWO cards MUST be spades, what are the odds that this first card is a spade?
I should add that JMoravitz's comment is really excellent.
A: The probabilities of drawing the three cards are the same in any order, so imagine drawing the two known spades first.  What is the chance the third card is a spade?
A: The question is to ask the conditional prob
$$
P(\text{1st card is spade|2nd and 3rd cards are spades})
$$
Let's break it down like
$$
\frac{P(\text{1st card is spade AND 2nd and 3rd cards are spades})}{P(\text{2nd and 3rd cards are spades})} = \frac{P(\text{1st, 2nd and 3rd cards are spades})}{P(\text{2nd and 3rd cards are spades})}
$$
$$
P(\text{1st, 2nd and 3rd cards are spades}) = \frac{13}{52}\frac{12}{51}\frac{11}{50}
$$
$$
P(\text{2nd and 3rd cards are spades}) = P(\text{1st, 2nd and 3rd cards are spades}) + P(\text{1st is NOT spade, 2nd and 3rd cards are spades}) = \frac{13}{52}\frac{12}{51}\frac{11}{50} + \frac{39}{52}\frac{13}{51}\frac{12}{50}
$$
Plug in all numbers back to the question I got
$$
\frac{11}{50}
$$
