Probability of getting 4 Aces What is the probability of drawing 4 aces from a standard deck of 52 cards. Is it:
$$
\frac{1}{52} \times \frac{1}{51} \times \frac{1}{50} \times \frac{1}{49} \times 4!
$$
or do I simply say it is:
$$
\frac{4}{52} = \frac{1}{13}
$$
 A: Do you mean getting 4 aces in a row when drawing them one by one from a full deck without replacement? 
If it is the case, then it is simply multiplication of successive probabilities:
$\frac{4}{52}$ * $\frac{3}{51}$ * $\frac{2}{50}$ * $\frac{1}{49}$ = 
 4! * $\frac{48!}{52!}$ =  3.6938e-006. 
A: The first answer you provided ($\frac{1}{52}\times \frac{1}{51}\times \frac{1}{50}\times \frac{1}{49}\times 4!$) is correct.
If we draw four cards from 52 cards, then the total possible outcomes are $C_4^{52} 4!$.
The number of outcomes that have four aces in a row is $4!$ 
Thus the probability of drawing 4 aces from a standard deck of 52 cards is 
$$
\frac{4!}{C_4^{52} 4!} = \frac{1}{C_4^{52}} = \frac{4!}{52\times  51\times  50\times  49}
$$
A: 
What is the probability of drawing 4 aces from a standard deck of 52 cards

The correct answer to the question posed is: The probability is 1.
The other solutions posited on this page are solutions to a different question than that posed here.
A: In a deck of cards what is the probability of getting 4 aces in the first 4 draws?
Answer: $4\times 4 = 16$


*

*$P(4\ {\rm aces}) = E=?\quad 16$

*$S=?\quad 52\times 51\times  50\times  49\quad (6,497,400) $

*$\frac{16}{6,497,400} = 0.000246\%$

