What is the conditional probability that the second card is a Spade given that the second-to-last card is a Spade?

Question

What is the conditional probability that the second card is a Spade given that the second-to-last card is a Spade? Cards are dealt without replacement.

I know conditional probability is $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ My question is how do we find $P(B)$ and $P(A \cap B)$. Finding the P(B) means what is the 51st card place down is spades. Would $$P(B) = \frac{\binom{51}{13}}{\binom{52}{13}} \cdot \frac{13}{51}$$ and $$P(A \cap B) = \left(\frac{13}{52} + \frac{12}{51} + \frac{1}{2}\right)?$$ Any hint or advice will help. Thanks.

Answer 1

I spread out a shuffled deck of cards, face down, before you.

I point to any one among these fifty-two and ask you: What is the probability that that card is one from the thirteen spades?   Does your answer depend on to which is the card that I point?

Well, it so happens that I have pointed to the second from the bottom of the deck, and when I turn it over it is revealed to be a spade.

Now I point to any other card, among the fifty-one remaining in the deck, and now ask: What is the probability that that card is one among the twelve remaining spades?

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Answer 2

Let $A=$ {2nd card is a spade} and $B=$ {penultimate card is a spade}.

All the condition tells you is that for each of the other 51 positions, you have one fewer spade that it could be. So $P(A)=P(B)=\frac{13}{52}=\frac{1}{4}$, but $P(A|B)=P(B|A)=\frac{12}{51}=\frac{4}{17}$.

From there, it's easy to get $P(A \cap B)$ using the formula $P(A \cap B) = P(A|B) \cdot P(B)$.