Probability Question about deck of cards maybe you can help me out.
The following holds: There is a deck of 52 cards. 13 clubs, 13 spades, 13 hearts, 13 diamonds). Normal deck of cards. Nothing special.
And now the following happens. I draw a card, and put it away, without looking at it. I then draw 2 more cards and look at them. Each of those 2 cards were spades. And now I want to know, given this information, what the probability is, that the first card I didn't look at, was spades as well.
I don't even know, how to start, notation wise. Like P(first drawn card = spades| 2nd and 3rd drawn card = spades)? And normally, should it not just be P(first card = spades) =13/52? But I feel, that would be wrong here.
I would try to calc 13/52 x 12/51 x 11/50 = 0.0129. So that would be 1.3% But that doesn't consider the information properly, I think.
 A: There are $50$ equiprobable candidates for the first card and $11$ of these candidates are spades. 
So the probability that the first card is a spade equals: $$\frac{11}{50}$$
A: How about writing $S_i$ for the event that the $i$th card drawn is a spade? Now you need Bayes' formula:
$$\Pr(S_1\mid S_2,S_3)=\frac{\Pr(S_1,S_2,S_3)}{\Pr(S_2,S_3)}.$$
A: Let us look at the desired conditional probability from the point of view of the classical definition of probability. 
We need to find the proportion of triples of cards, in which the first card is a spade, among all the triples of cards, where the second and third are spades. It is exactly the same as the proportion of fully spades triples among triples with a second and third are spades. And there is the same number of triples of cards, where the second and third are spades and where the first and the second are spades. 
Therefore this proportion is exactly the same as the proportion of fully spades triples among all triples, where the first and second are spades. And it coincides with the probability of pulling out the third spade if the first two spades are already removed. So it is $\dfrac{13-2}{52-2}$. 
