What is the probability of getting at least one ace when selecting 5 cards from a deck of 52 cards without replacement?

If we have a deck of 52 unique cards, which contains 4 aces (also unique), and we want to deal out a hand of 5 cards (that can be arranged in any order) then how many different hand combinations are there that contain at least one ace?

I can calculate how many hand combinations are there that involve all of the four aces, but I can't figure out how to deal with cases similar to the above described. E.g. calculating how many combinations contain at least 2 or 3 aces.

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="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:

#(886,656)/(2,598,960)~=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:

#(886,656)/(2,598,960)~=0.341#

I should perhaps note that the accepted solution appears incorrect.

The distribution of the number of ace cards in a 5-card hand is a problem in sampling without replacement ... which leads to a Hypergeometric distribution.

Imagine an urn contains a total of $T$ balls, $r$ of which are red $(r<T)$. The experiment proceeds by drawing one-by-one a sample of $n$ balls from the urn without replacement $(n<T)$. Interest lies in determining the pmf of $X$, where $X$ is the number of red balls drawn. Then, the pmf of $X$, $P(X=x)$ is:

$$f(x)=\frac{\left( \begin{array}{c} n \\ x \\ \end{array} \right) \left( \begin{array}{c} T-n \\ r-x \\ \end{array} \right)}{\left( \begin{array}{c} T \\ r \\ \end{array} \right)}$$

In our example, the 'urn' is the deck of $T=52$ playing cards, and the 'red balls' are the ace cards, so $r=4$. There are $n=5$ cards in a hand. Therefore, the pmf of the number of ace cards in a hand of 5 cards is given by:

$$f(x)=\frac{\left( \begin{array}{c} 5 \\ x \\ \end{array} \right) \left( \begin{array}{c} 47\\ 4-x \\ \end{array} \right)}{\left( \begin{array}{c} 52 \\ 4 \\ \end{array} \right)}$$

where $X = \{0,1,2,3,4\}$.In the case of just one ace, $P(X=1) = \frac{3243}{10829} \approx 0.299474$.

The following diagram plots the pmf of the number of aces in a hand of 5 cards: