Possible outcomes of "four of kind"

Question

I have been working on my probability. I have "possibly" memorized how to solve but, I really like to understand it with my heart. But.. I am kinda stuck.

I am trying to find out the possible outcomes of four of kind (unordered) Here is my intuition:

$$\frac{52 \cdot 3 \cdot 2 \cdot 1 \cdot 48}{5!}$$

I do not see why this is wrong.

Please help me out!

Answer 1

Consider drawing 4 cards:

1st card: any card is fine

2nd card: only 3 cards can make 4of a kind: 3/51

3rd card: 2/50

4th card: 1/49

So the answer is $\frac{3\times 2 \times 1}{51 \times 50 \times 49}$

Answer 2

I suppose that you are playing poker with a deck of 52 cards and you're asking what is the probability of dealing a hand $$ xxxxy\qquad\qquad\qquad\qquad(\star) $$ where $x$ is just any card from $2$ to $A$ and $y$ is any (and $y\neq x$ since there are only four $x$ in the deck).

Since the order of dealing is unimportant, you're just counting how many subsets with $5$ elements of the form ($\star$) you can form out of $52$ cards.

To form any such set you have to choose $x$, for which you have $13$ choices, and any of these choice can be coupled with any $y$ for which you have $52-4=48$ possibilities.

Thus you have $13\cdot48=624$ subsets of the form ($\star$).

Since there are alltogether $\binom{52}5$ possible hands dealt (the number of all subsets with $5$ elements chosen out of $52$) the probability to be dealt a Four of a Kind in poker is $$ \frac{624}{\binom{52}5}=624\frac{5!\cdot47!}{52!}. $$ I'll leave to you the task to come with the actual number.

Answer 3

My interpretation of your question is that you are asking how many 5-card poker hands include 4 of a kind, where the order of the cards in the hand is not considered relevant (so, for example, the hand 2D, 2H, 2S, 2C, 8C is the same as 8C, 2C, 2S, 2H, 2D).

There are $13$ ways to pick the rank of the 4-of-a-kind. Once the rank is chosen, the hand must include all four cards of that rank. This can only be done in $1$ way, since we don't consider the order of the cards significant. Then the remaining card can be any one of the $48$ cards remaining. So in all, there are $$13 \times 1 \times 48$$ hands.

If you wanted to compute the probability of four of a kind, you would need to divide by the number of five-card hands, $\binom{52}{5} = 2,598,960$.