Probability of drawing 4 aces when drawing 5 cards from a regular deck of cards. I cannot seem to wrap my head around how to calculate the probability of drawing four ace when drawing five cards from a deck of cards. My intuition tells me the math below but its wrong for some reason...
$
\frac{\frac{4}{52}\frac{3}{51}\frac{2}{50}\frac{1}{49}C^{48}_{1}5!}{C^{52}_{5}}.$
$\frac{4}{52}\frac{3}{51}\frac{2}{50}\frac{1}{49}$ is the probability of drawing the four aces.
$C^{48}_{1}$ is all the different combinations we can draw all other cards
$5!$ are all the combinations that this can happen. For example, we can draw any card but an ace first, then four aces. We can draw an ace then the other three aces and some card, and so on. Thus we get 5 factorial.
$C^{52}_{5}$ is all of the combinations one can draw five cards from a deck of cards.
What is wrong with my reasoning and equation above?
It seems from other examples online that if we draw one type of ace (let's say heart) it does not matter that there are three other types of ace nor that we could draw a card that is not an ace at a certain other index (second, third fourth card, etc). How come?
From other posts and my textbook it seems that we can just assume we drew four ace and then we calculate the remaining cards. That yields the equation: $\frac{1*C^{48}_{1}}{C^{52}_{5}}$ where the $1$ is the fact that we are going to draw four ace so we can assume we already have. I am confused about where the different combinations of aces go and the different combinations in which we draw them together with the "extra" non ace card.
I have specified the apparent solution in the above paragraph. It has been given on multiple occasions throughout stackexchange and other parts of the internet. What these answers do not do is explain WHY.
 A: The problem can be solved in two ways.  We can either take the order of selection into account or not take it into account.
Taking the order of selection into account:  There are $52 - 4 = 48$ ways to select the card which is not an ace.  There are $5!$ ways to arrange the four aces and that card.  There are $P(52, 5) = \binom{52}{5}5!$ ways to select five cards in order.  Hence, the probability of selecting four aces when five cards are drawn is
$$\Pr(\text{all four aces are selected}) = \frac{48 \cdot 5!}{P(52, 5)} = \frac{48 \cdot 5!}{\dbinom{52}{5} \cdot 5!} = \frac{48}{\dbinom{52}{5}}$$
Not taking the order of selection into account:  We must select all four aces and one of the other $48$ cards in the deck while selecting five of the $52$ cards in the deck.  Hence, the probability of selecting four aces is
$$\Pr(\text{all four aces are selected}) = \frac{\dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}} = \frac{48}{\dbinom{52}{5}}$$
What is wrong with your calculation?
If you take the order of selection into account in the numerator, you must also take it into account in the denominator.  Also, in your numerator, you multiplied the probability of selecting four aces by the number of ways of selecting a non-ace.  You need to multiply the number of ways of taking all four aces by the number of ways of selecting a non-ace.
Addendum:  Here is another way of handling the ordered selection.  There are $48$ ways to select the non-ace.  There are four ways to select the first ace, three ways to select the second ace, two ways to select the third ace, and one way to select the fourth ace.  We will consider cases, depending on the position of the non-ace.
\begin{align*}
\Pr(\text{all four aces are selected}) & = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}  \cdot \frac{48}{48} + \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49}  \cdot \frac{1}{48}\\
& \quad + \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{2}{49}  \cdot \frac{1}{48} + \frac{4}{52} \cdot \frac{48}{51} \cdot \frac{3}{50} \cdot \frac{2}{49}  \cdot \frac{1}{48}\\
& \qquad + \frac{48}{52} \cdot \frac{4}{51} \cdot \frac{3}{50} \cdot \frac{2}{49}  \cdot \frac{1}{48}\\
& = \frac{5 \cdot 48 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{52 \cdot 53 \cdot 50 \cdot 49 \cdot 48}\\[2 mm]
& = \frac{48 \cdot 5!}{\frac{52!}{47!}}\\[2 mm]
& = \frac{48 \cdot 5!}{\frac{52}{47!5!} \cdot 5!}\\[2 mm]
& = \frac{48 \cdot 5!}{\binom{52}{5}5!}\\[2 mm]
& = \frac{48}{\binom{52}{5}}
\end{align*}
which agrees with the answer obtained above by first selecting the five cards and then arranging them.
A: In your attempt, you are mixing up probabilities and # of ways
If you want to use probability the way you tried, the ans would just be
$Pr = \dfrac4{52}\dfrac3{51}\dfrac2{50}\dfrac1{49}\dfrac{48}{48}\times5\quad$
[The $5$ multiplier for the $5$ places for non-ace]
The book's answer counts ways using the hypergeometric distribution
It would have been clearer if written as $\dfrac{\binom44\binom{48}1}{\binom{52}5}$
Please remember, numerators and denominators must match in type
A: If the aces and any one of the card are in a fixed order for example heart, diamond, club, spade, any card.
$$P(\text{fixed order})=\frac{1}{52}\times \frac{1}{51}\times \frac{1}{50}\times \frac{1}{49}\times \frac{48}{48}$$
But if they can be in any order then
$$P(\text{in any order})=\frac{1}{52}\times \frac{1}{51}\times \frac{1}{50}\times \frac{1}{49}\times \frac{48}{48}\times 5!$$
A: Let's look at the problem in the following way:
Assume that aces are cards of type $A$ and others are of type $B$. So, you have 4 cards of type $A$ and 48 cards of type $B$. You want to choose 4 cards of type $A$ and 1 card of type $B$. Hence the answer is
$$\dfrac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$$
