I said $P(E|F^cH) = (48C19)(3C1)(1C1)(31!)/(52!),$ i.e the first 19 cards have to not be any aces $(48C19)$, the 20th card has to be an ace but not the ace of spades,$(3C1)$, the 21st card is ace of spades (1C1) and the rest can be any of the remaining cards.
As your denominator of $52!$ makes clear, you’re counting permutations. There are $\binom{48}{19}$ ways to choose the first $19$ cards, but they can then be arranged in $19!$ different orders, so the first factor should be $19!\cdot\binom{48}{19}=\frac{48!}{29!}$: there are
$$\frac{48!}{29!}\cdot3\cdot 31!=3\cdot30\cdot31\cdot48!$$
permutations that result in the desired outcome $EH$, and the probability is therefore
$$\frac{3\cdot30\cdot31}{49\cdot50\cdot51\cdot52}=\frac{93}{5\cdot17\cdot49\cdot52}=\frac{93}{216580}\approx0.0004294\;.$$
Similarly, $P(F^c|H) = (48C19)(3C1)(32!)/(52!)$ Or would this simply be 3/4? My reasoning being you know the 20th card is an ace. So the probability of it being the ace of spades is 1/4 => not ace of spades with prob 3/4?
Yes, it’s simply $\frac34$, for the reason that you give.
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 $\Pr(A|C)$.
Best Answer
The conditional probability is a quotient,
P(AS at bottom and AH not at top)/P(AH not at top).
(Here AS and AH are the ace of spades and ace of hearts respectively.)
The numerator here is equal to
P(AS at bottom) - P(AS at bottom and AH at top)
and you can rewrite the second term as
P(AS at bottom) - P(AS at bottom) P(AH at top | AS at bottom)
and now put in numbers to get
$1/52 - (1/52)(1/51) = (50/51)(1/52)$
The denominator of the original quotient is P(AH not at top) or $51/52$, so the answer is
$$ {(50/51)(1/52) \over (51/52)} = {50 \over 51^2} $$
which agrees with your brother's answer.