Probability of four-of-a-kind flawed logic I know my logic is completely wrong and I want to know why. I'm faced with the question of counting the probability that a hand of ($5$) cards in poker contains
four cards of the same kind (four of a kind) in a standard $52$ card deck
$$_ _ _ _ _$$
Now:
First card:
Can be any card so probability is 1
Second card:
The second card has to be the same rank as first one, and since there are only 4 cards of that rank in a deck and one has already been used in the first card then there is a probability of $\frac{3}{51}$ of picking a second card same as the first
Third card:
Similarly the third card has probability $\frac{2}{50}$ of being chosen since there are only two cards of the rank we want left
Fourth card:
Similarly the fourth card has probability $\frac{1}{49}$ of being chosen since there are is only one card of the rank we want left in those 49 cards left
Fifth card:
The fifth card can be any of the 48 cards left so $\frac{48}{48}=1$
Now this 5 cards can be ordered in $P(5,5)=120 \text{ ways}$
So the $P(\text{four_of_a_kind})=1\cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} \cdot 1 \cdot 120$
Which is wrong. What part I'm I doing wrongly?
 A: You’ve overcounted when you multiply by $5!$: your calculation of the probability of drawing a four-of-a-kind in the first four cards already takes into account all possible permutations of the four cards of the same rank. Thus, you need only take into account the possible positions of the odd card: it can appear in any of five positions, so you should be multiplying by $5$, not by $5!$. 
As confirmation, you can calculate the probability by counting outcomes. There are $13$ ranks, and each may be extended with any of $48$ other cards to make a hand, so there are $13\cdot48$ hands that qualify as a four-of-a-kind. There are altogether $\binom{52}5$ hands, so the probability of getting a four-of-a-kind is
$$\begin{align*}
\frac{13\cdot48}{\binom{52}5}&=\frac{13\cdot48\cdot5!}{52!}\\
&=\frac{13\cdot5\cdot4\cdot3\cdot2}{52\cdot51\cdot50\cdot49}\\\\
&=\frac{5\cdot3\cdot2}{51\cdot50\cdot49}\\\\
&=1\cdot\frac3{51}\cdot\frac2{50}\cdot\frac1{49}\cdot1\cdot5\;.
\end{align*}$$
A: You have overcounted by a factor $24$. The cards that match can be selected in $24$ different orders and you have counted each of them, then reordered them with the factor $120$. Either the final $120$ should be $5$, just choosing which place the odd card will occupy, or you must define a specific order to select the matching cards.
A: In your initial calculation, you've already counted all the possible permutations of the first four cards.  So you should not multiply by 120 at the end; you should only multiply by 5, to account for the fact that the remaining card could be in any of the 5 slots.
A: Instead of multiplying by $5!$, you should just multiply by 5 because by your logic, every 4 of a kind is already being counted $4!$ times (you are not considering the order in which you pick these 4 cards). So you just need to multiply by 5( the number of ways in which you can place the 5th card).
