# What is the probability that the deal of a five-card hand provides at least one ace?

$P = \frac{886 , 656}{2 , 598 , 960} \cong 0.341$

#### Explanation:

I'm going to assume that we're working with a standard 52-card deck.

Probability is a ratio - the numerator is the number of ways a given condition can be met and the denominator is the total number of ways things can be.

For instance, for figuring out the denominator in this question, how many different 5-card hands can we get from a standard 52-card deck? This is a combinations question - we don't care in what order the cards are dealt, only that we have a certain set of cards in our hand.

The general equation for finding a combination is:

C_(n,k)=(n!)/((k)!(n-k)!) with $n = \text{population", k="picks}$

C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))

Let's evaluate it!

(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960

Now let's do the numerator - the number of ways we can have a hand that has at least one Ace. There are two ways to calculate this - we can add up hands that have 1, 2, 3, and 4 Aces, and we can also find the number of hands that have no Aces and subtract it from all hands. Let's do it both ways:

Method 1: All hands minus 0 Ace hands

We know the total number of hands (2, 598, 960), so now let's find the number of hands possible with no aces. We can find that number by noticing that we can have any of 48 cards chosen for our 5-card hand, so that's:

C_(48,5)=(48!)/((5)!(48-5)!)=(48!)/((5!)(43!))=>

(cancelcolor(brown)48^2xx47xx46xxcancelcolor(orange)45^9xx44xxcancelcolor(red)(43!))/(cancelcolor(orange)5xxcancelcolor(brown)(4xx3xx2)xxcancelcolor(red)(43!))=2xx47xx46xx9xx44=1,712,304

And so the number of hands with at least one Ace is:

$2 , 598 , 960 - 1 , 712 , 304 = 886 , 656$

This gives the probability as:

$\frac{886 , 656}{2 , 598 , 960} \cong 0.341$

Method 2: Adding up hands with Aces

This method involves adding up the number of hands that have 1, 2, 3, and 4 Aces. Each separate calculation involves allowing a certain number of Aces and calculating that combination and filling in the remaining part of the hand with non-Ace cards and calculating that combination, and multiplying the two:

1 Ace

C_(4,1)xxC_(48,4)=(4!)/((1!)(4-1)!)(48!)/((4!)(48-4)!)=(4!48!)/(3!4!44!)=>

(cancelcolor(red)(4!)xxcancelcolor(orange)(48)^8xx47xx46xx45xxcancelcolor(brown)(44!))/(cancelcolor(orange)(3xx2)xxcancelcolor(red)(4!)xxcancelcolor(brown)(44!))=8xx47xx46xx45=778,320

2 Aces

C_(4,2)xxC_(48,3)=(4!)/((2!)(4-2)!)(48!)/((3!)(48-3)!)=(4!48!)/(2!2!3!45!)=>

(cancelcolor(orange)(4)xxcancelcolor(red)(3xx2)xx48xx47xx46xxcancelcolor(brown)(45!))/(cancelcolor(orange)(2xx2)xxcancelcolor(red)(3xx2)xxcancelcolor(brown)(45!))=48xx47xx46=103,776

3 Aces

C_(4,3)xxC_(48,2)=(4!)/((3!)(4-3)!)(48!)/((2!)(48-2)!)=(4!48!)/(3!2!46!)=>

(cancelcolor(brown)4xxcancelcolor(orange)3xx2xx48xx47xxcancelcolor(red)(46!))/(cancelcolor(orange)3xxcancelcolor(brown)(2xx2)xxcancelcolor(red)(46!))=2xx48xx47=4,512

4 Aces

C_(4,4)xxC_(48,1)=(4!)/((4!)(4-4)!)(48!)/((1!)(48-1)!)=(4!48!)/(4!47!)=>

(cancelcolor(red)(4!)xx48xxcancelcolor(brown)(47!))/(cancelcolor(red)(4!)xxcancelcolor(brown)(47!))=48

Adding them all up:

$778 , 320 + 103 , 776 + 4 , 512 + 48 = 886 , 656$

Again, this gives the probability as:

$\frac{886 , 656}{2 , 598 , 960} \cong 0.341$