Poker Hand Probability

Question

You draw 4 cards from a standard deck of cards without replacement. Find the probability you get:

A Straight (4 consecutive cards, Ace can count as either highest or lowest)

The solution that's given is: $$\frac{(4^4)(11)(4!)}{52 \cdot 51 \cdot 50 \cdot 49}$$

I don't understand where the $4!$ came from.

Shouldn't the answer be $\dfrac {4^4 \cdot 11}{\binom{52}{4}}$ ?

Thank you!

Answer 1

As asdf says, these are the same

$$\dfrac {4^4 \cdot 11}{\binom{52}{4}} = \dfrac {4^4 \cdot 11}{\frac{52!}{48!\,4!}} = \dfrac {4^4 \cdot 11\cdot 4!}{\frac{52!}{48!}} =\dfrac {4^4 \cdot 11\cdot 4!}{52\cdot 51 \cdot 50 \cdot 49}$$

The $4!$ can be seen as the number of ways of ordering the four cards in the straight, the $11$ as the choices of the highest value in the straight and the $4^4$ as the suits of the four cards in the straight, and the $52\cdot 51 \cdot 50 \cdot 49$ as the ways of choosing four cards in order

Answer 2

Your answer $\frac{4^4\times11}{\binom{52}4}$ is the same as the given answer, because $\binom{52}{4}=\frac{52\times51\times50\times 49}{4!}$.

There are two ways to think about this. Either you draw four cards with order not mattering, and then there are $4^4\times 11$ different straights out of $\binom{52}4$ possibilities, or you draw four cards with order mattering, in which case there are $52\times51\times50\times 49$ possibilities, but since you can draw the four cards of each straight in any order, each straight counts $4!$ times. These come to the same thing.