Deck of cards probability I was doing a question off my exam review but I'm not sure if I'm getting this correctly.

We take a uniformly random permutation of a standard deck of 52 cards, so
  that each permutation has a probability of 1=52!.
   A = "the top card is an Ace",
   B = "the bottom card is the Ace of spades",
   C = "the bottom card is the Queen of spades".
  Determine
  Pr(A | B), and Pr(A | C)

I'm thinking Pr(A|B)=Pr(A and B)/Pr(B); 
where Pr(A and B) = (1/52!)(1/51!)
and Pr(B)=1/52!
Same with Pr(A | C) or am I wrong?
 A: The formula you are using for conditional probability is correct, but the probability calculations are not. And although the formula can be used to solve the problem, it is not the simplest way to do it. We give two solutions, one that uses the formula and one that does not. 
Let $A,B,C$ be as in your post.
To find $\Pr(A|B)$ using the formula, we need $\Pr(A\cap B)$ and $\Pr(B)$. 
The probability of $B$ is $\frac{1}{52}$. This is obvious (all cards are equally likely to be at the bottom). But we could also say that there are $51!$ permutations in which the Ace of spades is at the bottom, so $\Pr(B)=\frac{51!}{52!}=\frac{1}{52}$.
For $\Pr(A\cap B)$, let us count the permutations that satisfy $A\cap B$. The last card is determined. The first card can be any of $3$, and the rest can be permuted in $50!$ ways. Thus $\Pr(A\cap B)=\frac{3\cdot 50!}{52!}=\frac{3}{52\cdot 51}$. Now we can finish the calculation.
But the problem can be solved in a much simpler way. Once we have put the Ace of spades at the bottom, there are $51$ cards left. Of these, $3$ will satisfy the condition "I am a spade." Thus $\Pr(A|B)=\frac{3}{51}$. 
The same argument gives $\frac{4}{51}$ for $\Pr(A|C)$. 
A: $$P(A|B)=\dfrac{P(A\,and\ B)}{P(B)}$$
$$P(A|B)=\dfrac{(3/51!)(1/52!)}{1/52!}$$
$$P(A|B)=\dfrac{3}{51!}$$
Your mistake is that you have not taken into consideration that there are four aces in a deck. Read the question more carefully. I have done part A to show what I mean.
