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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: $$ \frac{4}{52} = \frac{1}{13} $$

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  • $\begingroup$ The OP seems to give the right answer but then makes a mistake when simplifying. This assumes you take the first 4 cards without replacement from a well mixed deck. $\endgroup$ Jan 14, 2017 at 19:24

4 Answers 4

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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.

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  • $\begingroup$ removed comment duplicating what KC_CAL said $\endgroup$
    – user20637
    Sep 11, 2014 at 8:05
  • $\begingroup$ I get 1/270725 which is approximately 3.6 e-5. I say approximately because I did this by hand and not with a calculator. This agrees with the above answer except for the far RHS. $\endgroup$ Jan 14, 2017 at 19:41
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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 $$ \frac{4!}{C_4^{52} 4!} = \frac{1}{C_4^{52}} = \frac{4!}{52\times 51\times 50\times 49} $$

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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.

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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) $
  • $\frac{16}{6,497,400} = 0.000246\%$
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    $\begingroup$ The other answers all claim that the number of ways to get four aces in the first four draws is $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$. If you disagree on that point and somehow think it is actually $4 \cdot 4 = 16$ you need to back that statement up with some reasoning. $\endgroup$ Nov 7, 2016 at 6:28
  • $\begingroup$ The other answers are all wrong. The correct answer is: the Probabiltiy is 1 $\endgroup$
    – wolfies
    Jan 16, 2017 at 7:13

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