TITLE: 5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades?
QUESTION [13 upvotes]: We have $\binom{52}{5}$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $\binom{4}{4}$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $\binom{48}{1}$ ways.
So is the answer:
$$\dfrac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}=\dfrac{1}{54145}?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)\times (3/51) \times (2/50) \times (1/49) \times (1/48)$ which is a completely different here.
Which is the correct approach?

REPLY [7 votes]: The first way:
There is only one set of cards, that fits the specifications (all 4 aces and the king of spades) and $\binom{52}{5}$ different sets of cards that can all be drawn with equal probability. The chance to draw your set is
$$
\frac{1}{\binom{52}{5}}
$$
The second way:
The chance of drawing a specific card out of $n$ is $\frac{1}{n}$. So a specific sequence or order of 5 cards has the chance $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}\cdot\frac{1}{48}$ to be drawn. There are $5!$ different sequences or orderings of 5 cards, that countain the 5 specified cards (s. permutations). So the chance to draw the set is
$$
\frac{5!}{52\cdot51\cdot50\cdot49\cdot48} = \frac{5!\cdot 47!}{52!}
= \frac{1}{\frac{52!}{5!\cdot 47!}} = \frac{1}{\binom{52}{5}}
$$
For the last equality see Binomial coefficient.