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A player is dealt six cards out of a normal deck of cards. He looks at the first four and notices there is no ace among them. What is the probability that he does not have an ace at all.

This sounds like a conditional probability problem, but I reckon it just as well can be a counting problem: the required probability would be one minus the probability of a hand with four non-aces, one ace, and one arbitrary card.

So by that reasoning, $$p=1-\frac{{48\choose 4}{4\choose 1}{46\choose 1}}{ 52\choose{6}}$$

Is this right?

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  • $\begingroup$ Looks good to me! $\endgroup$ – ml0105 Nov 25 '14 at 20:40
  • $\begingroup$ I'd say $\frac{24}{28} \frac{23}{27} $ ? $\endgroup$ – Christian Nov 25 '14 at 20:42
  • $\begingroup$ This can't be right as the numerator is larger than the denomiator. $\endgroup$ – Spine Feast Nov 25 '14 at 20:46
  • $\begingroup$ Just a note: "four non-aces, one ace, and one arbitrary card" is not quite the same as "four non-aces and at least one ace", since the first double-counts the two-ace situations. Also, choosing a hand with some four non-aces is a lot easier than choosing a hand in which the first four are non-aces, which is what needs to be done. $\endgroup$ – Zubin Mukerjee Nov 25 '14 at 20:48
  • $\begingroup$ "He looks at the first four and notices there is no ace among them." indicates more of an event than a conditon. And the event has happened already. We need to think of other 2 non-ace cards from remaining cards. Better to handle this from Counting perspective. $\endgroup$ – Rajkumar Nov 25 '14 at 21:01
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Rajkumar's direct solution works nicely. You mentioned that the problem could be considered as conditional probability, and I'll show here that that method gives the same answer.

As I mentioned in a comment, the main reason for the inaccuracy of your original answer is overcounting the number of ordered $6$-card hands with the first four being non-aces, and at least one of the last two being aces.


Let $A$ be the event that the player is dealt $6$ non-ace cards that are randomly drawn from a standard deck of $52$ cards.

Let $B$ be the event that the first four of the player's $6$ cards are not aces.


$P(B \mid A$) is clearly $1$, since the first four must be non-aces if the six are all non-aces.

Since each card is chosen independently without replacement, we can find the probability of never getting an ace as

$$P(A) = \frac{48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47}$$

Likewise

$$P(B) = \frac{48 \cdot 47 \cdot 46 \cdot 45}{52 \cdot 51 \cdot 50 \cdot 49}$$


Now we can use Bayes' rule:

$$P(A\mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$

But $P(B \mid A)$ is $1$, so

$$P(A\mid B) = \frac{P(A)}{P(B)}$$

$$P(A\mid B) = \frac{44 \cdot 43}{48 \cdot 47}$$

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I would say.

When 4 cards are not aces, then remaining 48 cards are 4 ace.

So pull of 48 cards 2 cards without replacement --> hypergeometric probability distribution. The probability that between these two cards is not ace will:

$\displaystyle P=\frac{{4 \choose 0}{48 - 4\choose 2 - 0}}{{48 \choose 2}}=\frac{473}{564}$

Edit: I see that I reached the same finding another way user Yubin Mukerjee $\left(\frac{44 \cdot 43}{48 \cdot 47}=\frac{473}{564}\right)$.

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  • $\begingroup$ It's always nice when you can do a problem in many different ways :) $\endgroup$ – Zubin Mukerjee Nov 26 '14 at 0:47
  • $\begingroup$ @ZubinMukerjee Yes, yes - your calculation me too very pleased. $\endgroup$ – georg Nov 26 '14 at 9:46
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We know that given 4 cards are not aces. So remaining 48 cards = 4 ace cards + 44 non-ace cards.

Probability of getting 5th card as non-ace card = 44/48
Probability of getting 6th card as non-ace card(given 5th is non-ace card) = 43/47

So probability of getting no-ace card at all is = 44/48 times 43/47.

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    $\begingroup$ "The probability of getting 5th card as non-ace card & probability of getting 6th card as non-ace card are independent." - no they're not! The probability of getting an ace on the 6th card if you got an ace on the 5th is much less than if the 5th wasn't an ace. Your calculation correctly uses the conditional probability for an ace on the 6th dependent on not getting an ace on the 5th; your explanation is wrong. $\endgroup$ – user2357112 Nov 25 '14 at 23:15

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