what is the conditional probability that the card following it is the ace of spades? i have seen a couple of different answers for this question, but i still cant understand why.
Suppose that an ordinary deck of 52 cards is shufﬂed and the cards are then turned over one at a time until the ﬁrst ace appears. Given that the ﬁrst ace is the 20th card to appear, what is the conditional probability that the card following it is the ace of spades?
this is what i was trying.
\begin{equation*}
\begin{aligned}
P(\text{following is the ace of spades} | \text{first ace is the 20th card}) 
& = \dfrac{P(\text{following is the ace of spades} \cap \text{first ace is the 20th card})}{P(\text{first ace is the 20th card})} \\
& = \dfrac{\dfrac{4C1}{32C1}\times\dfrac{3C1}{31C1}}{\dfrac{4C1}{32C1}} \\
& =\dfrac{\dfrac{4!/3!}{32!/31!}\times\dfrac{3!/2!}{31!/30!}}{\dfrac{4!/3!}{32!/31!}}
\end{aligned}
\end{equation*}
 A: There's a $\frac 14$ probability that the first ace is $A\spadesuit$, in which case the answer would be $0$.  Otherwise, there are $32$ remaining cards of which one is $A\spadesuit$ so $\frac 1{32}$ in that case.  Hence $$\frac 14\times0+\frac 34\times \frac 1{32}=\frac 3{128}$$
A: If you must do these as conditional events then let's define some events: 


*

*$A$ is the event that the first Ace appears as the $20$th card

*$B$ is the event that the first Ace is the Ace of Spades

*$C$ is the event that the card following the first Ace is the Ace of Spades


Clearly $B \cap C = \emptyset$ and $P(B \cap C) = P(B \mid C)=P(C \mid B)=0$ so $P(C \mid A \cap B)=0$  
By symmetry $P(B)=\frac14$ and $P(B \mid A)=\frac14$ with $A$ and $B$ independent and $P( B^c\mid A)=P(B^c) = \frac34$ 
As lulu says, you can easily calculate $P(C \mid A \cap B^c) = \dfrac{1}{52-20} = \dfrac{1}{32}$ 
So $P(C \mid A) =  \dfrac{P(C \cap A)}{P(A)} = \dfrac{P(C \cap A \cap B)+P(C \cap A \cap B^c)}{P(A)} = \dfrac{P(C \mid A \cap B)P(B \mid A)P(A)+P(C \mid A \cap B^c)P(B^c \mid A)P(A)}{P(A)} $ $= 0\times \dfrac14 + \dfrac1{32} \times \dfrac34 =\dfrac3{128} $
A: Given that the ﬁrst ace is the 20th card to appear:
This suggests no Ace was chosen in the first 19 card flips.  No other information was given, so every other card is subject to random choice in the first 19 card flips.  On the 20th flip we are told an Ace is flipped.  This gives the equation for P(A)
$P(A) = \binom{48}{19}  \binom{4}{1}$
On the 21st flip we are told an Ace of Spades is flipped. This means there are only 3 Aces to be randomly selected on the 20th flip.  On the 21st flip there are 32 cards to chose from and we need the probability that the Ace of Spades is chosen.
$P(BA) = \binom{48}{19}  \binom{3}{1}  \dfrac1{32}$
So the final equation is
$P(B \mid A) = \dfrac{\binom{48}{19}  \binom{3}{1}  \dfrac1{32}}{\binom{48}{19}  \binom{4}{1}}$
