Probability of being dealt four-of-a-kind in a set of $5$ cards? 
You are dealt a hand of five cards from a standard deck of playing cards. Find the probability of being dealt a hand consisting of four-of-a-kind. 

If possible, please provide a hint first before the answer.
One of the first things that came to me was $\dfrac{52}{52} \cdot \dfrac{51}{51} \cdot \dfrac{3}{50} \cdot \dfrac{2}{49} \cdot \dfrac{1}{48}$ but this was of course wrong.
Then, I realized that the third card could be the same as the first OR second card, so I tried $\dfrac{52}{52} \cdot \dfrac{51}{51} \cdot \dfrac{7}{50} \cdot \dfrac{2}{49} \cdot \dfrac{1}{48}$, which is also wrong. 
Then I realized that we probably need to add up the probabilities of situations where the first card is the one that doesn't match, or the second is the one that doesn't match, etc. I think this method leads to the solution, but I don't think we're intended to solve it this way. I think the solution looks something like $\dfrac{x}{ {{52}\choose{5}}}$, but I'm not sure what should be in the numerator.
 A: ${\underline{\text{Useful hint for future:}}}$
In drawing w/o replacement (hypergeometric distribution) 
when order doesn't matter, using combinations gives the answer directly:
$$\text{e.g. for your question}\;\; \frac{\binom{13}1\binom44\binom{48}1}{\binom{52}{5}}$$
whereas if order does matter, multiplying probabilities gives the answer directly,
$$\text{e.g. if the "single" must be second},\;\; \frac{52}{52}\cdot\frac{48}{51}\cdot\frac3{50}\cdot\frac2{49}\cdot\frac1{48}$$
and if you choose instead to use the other process, (which you can)
you must use a multiplication/division factor to compensate.

That is why, since order doesn't matter for your question,
if you use the multiplication of probabilities approach, you must multiply by $5$
A: First, because there are 13 sets of four-of-a-kind, number of ways of selecting one of the 13 ranks is $\begin{pmatrix}13\\1\end{pmatrix}$Then, once you have chosen the rank, the number of ways to draw the other three cards of that rank is $\begin{pmatrix}4\\4\end{pmatrix}$Finally, the number of ways to draw the last card. which is to choose from 48 other cards: $\begin{pmatrix}48\\1\end{pmatrix}$
The number of possible 5 card hands dealt from a 52 card deck is $\begin{pmatrix}52\\5\end{pmatrix}$
So the final answer is:$$\frac{\begin{pmatrix}13\\1\end{pmatrix}\begin{pmatrix}4\\4\end{pmatrix}\begin{pmatrix}48\\1\end{pmatrix}}{\begin{pmatrix}52\\5\end{pmatrix}}$$
A: You can compute the number of ordered draws or the number of unordered combinations of five cards.  The probability will come out the same as long as the denominator is computed the same way.  Your last sentence is correct for unordered combinations, so $x$ should be the number of unordered hands that contain four of a kind.  How many ways to select the rank you have four of?  How many ways to select the odd card, given the rank of four is already chosen?  
Your earlier approach is trying to count ordered ways to draw the hand, which is why the denominator comes out $52 \cdot 51 \dots 48$  In that case, you can compute the ordered number of ways to draw the four of a kind first, then multiply by the $5$ positions the odd card can be in.
A: Assuming all ${52 \choose 5}$ five card poker-hands are equally likely, the probability of being dealt 4 cards of the same rank is the ratio $x / {52 \choose 5}$, where $x$ is the number of five card poker-hands that contain 4 cards of the same rank.  For instance, a poker hand that contributes to $x$ is $\{2,2,2,2,Q\}$.  The number of ways to choose the rank is ${13 \choose 1}=13$, and the number of ways to choose the odd card is the number of ways to choose a card from the remaining 48 cards, which is ${48 \choose 1}=48$.  So $x=13 \cdot 48$. 
A: The simplest explanation might be the following:
there are ${52}\choose{4}$ possible combinations of 4 cards in a deck of 52. Then, with 5 cards, you can have 13 * 5 possible four of a kind. Divide the latter by the former.  
