Drawing 2 cards from a shuffled deck, what is the probability that the first is a spade and the second is an ace?

Question

Drawing 2 cards from a shuffled deck, what is the probability that the first is a spade and the second is an ace?

I was thinking that we'd do $\frac{13}{52} \cdot (\frac{4}{51} + \frac{3}{51})$, since the first card could have been an ace. Is this correct?

Answer 1

An alternate approach is that the first card is a spade with probability $\frac{13}{52}.$ Of those case, $\frac{1}{13}$ was the spade ace, and $\frac{12}{13}$ were not the ace. So the probability is:

$$\frac{13}{52}\left(\frac{1}{13}\cdot\frac{3}{51}+\frac{12}{13}\cdot\frac{4}{51}\right)$$

If $X$ is the even that the first card is a spade ace and $Y$ is the event that the first card is another spade we get, and $Z$ is the event that the second card is an ace, this formula is:

$$P((X\lor Y)\land Z)=P(X\lor Y)\left(P(X\mid X\lor Y)P(Z\mid X)+P(Y\mid X\lor Y)P(Z\mid Y)\right)$$

This is true in general if $X$ and $Y$ are disjoint events.

Answer 2

So the probability that the first card is a non-ace spade is $12/52$, and the probability that we pull an ace after that is $4/51$. On the other hand, the probability that we draw the ace of spades is $1/52$, and the chance of pulling another ace is $3/51$. All together we have

$$\frac{12}{52}\cdot\frac{4}{51}+\frac{1}{52}\cdot\frac{3}{51}$$

Answer 3

Another way to see this is to note that the events (1st card is Spade) and (2nd card is Ace) are independent because knowledge of the suit of the first card can't affect the probability of an Ace on the second card. Then

$P$(1st Spade and 2nd Ace) = $P$(1st Spade)$P$(2nd Ace) = (1/4)(1/13) = 1/52

which of course is the same as the solutions given above.