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

Mar 6, 2017

(48 choose 5) / (52 choose 5) = 0.6588

Explanation:

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 "no aces" combinations by all possible combinations of the full deck to determine the chance that any given hand dealt won't contain any aces.

Mar 24, 2017

0.659 (3dp)

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)

$P \left(\text{Card 1 not Ace}\right) = \frac{48}{52}$

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

$P \left(\text{Card 2 not Ace}\right) = \frac{47}{51}$

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

$P \left(\text{Card 3 not Ace}\right) = \frac{46}{50}$

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

$P \left(\text{Card 4 not Ace}\right) = \frac{45}{49}$

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

$P \left(\text{Card 5 not Ace}\right) = \frac{44}{48}$

So then:

$P \left(\text{all five cards non aces}\right) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{45}{49} \cdot \frac{44}{48}$

$\text{ } = \frac{35673}{54145}$

$\text{ } = 0.658841 \ldots$



Method 2

Using the combination formula:

 ""_nC^r = ( (n), (r) ) = (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") = ""_48C^5 = ( (48), (5) )
 " " = (48!)/(5!(48-5)!)
 " " = (48!)/(5!43!)

And the total number of all combinations of choosing any five cards

 n("any five cards") = ""_52C^5 = ( (52), (5) )
 " " = (52!)/(5!(52-5)!)
 " " = (52!)/(5!47!)

 P("all five cards non aces") = (n("all five cards non aces"))/(n("any five cards"))

 " " = ((48!)/(5!43!))/((52!)/(5!47!))

 " " = (48!)/(5!43!) * (5!47!)/(52!)

 " " = (47! * 48!)/(43! * 52!)

 " " = (43! * 44*45*46*47 * 48!)/(43! * 48! * 49 * 50 * 51 * 52)

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

$\text{ } = \frac{4280760}{6497400}$

$\text{ } = \frac{35673}{54145}$, as above

Or we could compute the combinations using a calculator;

$P \left({\text{all five cards non aces") = (""_48C^5) /( }}_{52} {C}^{5}\right)$

$\text{ } = \frac{1712304}{2598960}$

$\text{ } = \frac{35673}{54145}$, as above