What is the probability that the deal of a five-card hand provides no aces?

Question

Robert Pina · Accepted Answer

(48 choose 5) / (52 choose 5) = 0.6588Explanation:All the combinations of hands that could contain no aces is the equivalent of removing the four aces from the deck and dealing all the combinations of hands from the remaining 48 cards. So, the numerator of the probability ratio should be (48 choose 5).If you want the probability that any possible 5 card hand is a hand that contains no aces, then the denominator should contain all possible hands that can be dealt from a full deck of cards, which is (52 choose 5).Then divide all the &#039;&#039;no aces&#039;&#039; combinations by all possible combinations of the full deck to determine the chance that any given hand dealt won&#039;t contain any aces.

Piosellisf · Accepted Answer

Explanation:
Method 1
If we start with a full deck there are 52 cards; 4 aces; 48 non-aces
Card 1 deal should not be an Ace (48 non aces; 52 cards)
P(Card 1 not Ace)

= \frac{48}{52}

Card 2 deal should not be an Ace (47 non aces; 51 cards)
P(Card 2 not Ace)

= \frac{47}{51}

Card 3 deal should not be an Ace (46 non aces; 50 cards)
P(Card 3 not Ace)

= \frac{46}{50}

Card 4 deal should not be an Ace (45 non aces; 49 cards)
P(Card 4 not Ace)

= \frac{45}{49}

Card 5 deal should not be an Ace (44 non aces; 48 cards)
P(Card 5 not Ace)

= \frac{44}{48}

So then:
P(all five cards non aces)

= \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{45}{49} \cdot \frac{44}{48}

= \frac{35673}{54145}

=0.658841...
Method 2
Using the combination formula:

_{{n}} C^{r} = (b e g \in {a r r a y} {c} n ∖ r e n d {a r r a y}) = \frac{n!}{r! (n - r)!}

The number of combination of choosing five non-aces from 52 cards (48 of which are not aces) is given by:
n(all five cards non aces)

=_{48} C^{5} = (b e g \in {a r r a y} {c} 48 ∖ 5 e n d {a r r a y})

= \frac{48!}{5! (48 - 5)!}

= \frac{48!}{5! 43!}

And the total number of all combinations of choosing any five cards
n(any five cards)

=_{52} C^{5} = (b e g \in {a r r a y} {c} 52 ∖ 5 e n d {a r r a y})

= \frac{52!}{5! (52 - 5)!}

= \frac{52!}{5! 47!}

P (all five cards non aces)

= \frac{n (a l l f i v e c a r d s n o n a c e s)}{n (a n y f i v e c a r d s)}

= \frac{\frac{48!}{5! 43!}}{\frac{52!}{5! 47!}}

= \frac{48!}{5! 43!} \cdot \frac{5! 47!}{52!}

= \frac{47! \cdot 48!}{43! \cdot 52!}

= \frac{43! \cdot 44 \cdot 45 \cdot 46 \cdot 47 \cdot 48!}{43! \cdot 48! \cdot 49 \cdot 50 \cdot 51 \cdot 52}

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What is the probability that the deal of a five-card

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