I draw a card from a deck, and face it down. I draw another two cards of spades. What is the probability my first card is also a spade?

Question

Do I consider the probability before drawing both cards or after?

Question more clearly: A single card is removed at random from a deck of $52$ cards. From the remainder we draw $2$ cards at random and find that they are both spades. What is the probability that the first card removed was also a spade?

Answer 1

You are calculating the probability of the first card being a spade, therefor the next two additional draws mean nothing. The chance of drawing a spade from a deck of 52 would be 25%. It's as simple as that.

I hope this helps.

Answer 2

The trick here is to realize that the situation described is equivalent to removing two spades from the deck and then drawing a card at random from what remains. If you can calculate the probability of a spade occurring as the third card in that simpler setting, you're good to go.

Many probability problems can be similarly simplified by realizing that the specific order in which certain things are said to happen is irrelevant. That's not to say the order in which things happen is never relevant, just that the meanings of the letter strings (aka words) "first," "second," "third," etc., are sometimes interchangeable.

Answer 3

To emphasize, @Barry's answer is correct and is the easiest way to think of the answer.

Since that confuses people for whatever reason, a way to convince people of this is instead to approach directly via definitions.

Recall that $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}$ by definition. That is to say, the probability of an event $A$ occurring given that an event $B$ also occurs (whether that is past, present, or future... irrelevant) is the ratio of the probability of both $A$ and $B$ occurring over the probability that $B$ occurs regardless.

Here, letting $A$ be the event that the first card is a spade, $B$ the event that both the second and third cards are spades, we have:

$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}=\dfrac{\frac{13\cdot 12\cdot 11}{52\cdot 51\cdot 50}}{~~~~~\frac{13\cdot 12}{52\cdot 51}~~~~~} = \dfrac{11}{50}$$

In case $\Pr(B)$ confuses you, see this related question and/or again approach directly via definition. If you insist on doing this the long way, then recognize $\Pr(B) = \Pr(B\mid A)\Pr(A)+\Pr(B\mid A^c)\Pr(A^c)$ by law of total probability and see that it simplifies to what I claimed above.