Why probability of a diamond followed by an ace on drawing $2$ cards from a shuffled standard deck is not $1/51$? Two cards are dealt at random from a standard deck of $52$ cards. What is the probability that the first card is a $\diamondsuit$ and the second card is an ace?
I thought I got this question right but the system said it was wrong. This is how I did it.
So first I know that there are $13$ diamonds in a deck of cards so the probability of getting a $\diamondsuit$ is $13/52$. And there are $4$ aces so the probability of the second card being an ace is $4/51$ and then you multiply them together to get $1/51$.
But it is wrong can someone help me?
 A: HINTS: The existence of the diamond ace complicates this, as Buraian mentioned at in the comments. Consider two cases:
Case 1: You draw a diamond other than the ace for your first card. The probability of doing this is $12/52$, and the probability of then getting a diamond is $4/51$, like you said. Multiply them to get the probability of both.
Case 2: You draw the ace of diamonds for your first card (with probability $1/52$). Then, the probability of drawing an ace is not $4/51$, because there aren't 4 aces left in the deck...
The important point of decomposing the cases like this is that they're disjoint, which is useful. Can you take it from here?
A: One may specify a card from the deck by picking a card and noting the suit; and then, without replacement, drawing a second card and noting the rank. [Example: $\clubsuit 7$ followed by $\heartsuit 3$ specifies the card $\clubsuit 3$.] By symmetry, any card is equally likely to be specified in this manner, hence the desired probability for your specific card (i.e., your specific suit-rank combination) is $\frac{1}{52}$.
A: Ok I readjusted my solutions a bit.
Solution 1:
We have two cases because if the first card is a $\diamondsuit$, it could be an ace or not be an ace.
There is a $\dfrac{1}{52}$ chance that the ace of $\diamondsuit$ is drawn first, and a $\dfrac{3}{51} = \dfrac{1}{17}$ chance that the second card drawn is one of the three remaining aces, which gives a probability of $\dfrac{1}{52}\cdot \dfrac{1}{17} = \dfrac{1}{884}$ chance that this occurs.
There is a $\dfrac{12}{52} = \dfrac{3}{13}$ chance that a $\diamondsuit$ other than the ace is drawn first, and a $\dfrac{4}{51}$ chance that an ace is drawn second, giving a $\dfrac{3}{13}\cdot \dfrac{4}{51} = \dfrac{4}{221}$ chance that this occurs.
So the probability that one of these two cases happens is $\dfrac{1}{884} + \dfrac{4}{221} = \boxed{\dfrac{1}{52}}$.
Notice that we can avoid some of the large denominators above by organizing this computation as follows:$$\dfrac{1}{52}\cdot\dfrac{3}{51}+\dfrac{12}{52}\cdot\dfrac{4}{51} = \dfrac{1\cdot 3+12\cdot 4}{52\cdot 51} = \dfrac{51}{52\cdot 51}=\boxed{\dfrac{1}{52}}.$$
Solution 2:
We can solve this problem by using symmetry. Make a new "card" by combining the suit of the first card with the rank of the second card; for example, if the first two cards are $\spadesuit$2 and $\heartsuit$Q, then the new "card" would be $\spadesuit$Q. Then this new "card" is equally likely to be any of $52$ possibilities, one of which is the desired $\diamondsuit$A. So, the probability that the first card is a $\diamondsuit$ and the second card is an ace is $\boxed{\frac1{52}}$.
