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

Feb 12, 2017

$\frac{3243}{10829}$

#### Explanation:

First let us look at the probability that the first card dealt is an ace and the other four non-aces...

There are $4$ aces in $52$ cards, so the probability of the first card dealt being an ace is:

$\frac{4}{52} = \frac{1}{13}$

The remaining pack consists of $51$ cards of which $3$ are aces, and $48$ not. So the probability of the second card being a non-ace is:

$\frac{48}{51} = \frac{16}{17}$

The probability of the third card being a non-ace is:

$\frac{47}{50}$

The probability of the fourth card being a non-ace is:

$\frac{46}{49}$

The probability of the fifth card being a non-ace is:

$\frac{45}{48} = \frac{15}{16}$

So the probability of an ace followed by $4$ non-aces is:

$\frac{1}{13} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}}}{17} \cdot \frac{47}{50} \cdot \frac{46}{49} \cdot \frac{15}{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}}} = \frac{32430}{541450} = \frac{3243}{54145}$

The other four possible sequences of ace vs non-ace which result in exactly one ace being dealt are mutually exclusive, and will have the same probability as this first case.

So the probability of exactly one ace being dealt is:

$5 \cdot \frac{3243}{54145} = \frac{3243}{10829}$

P=(778,320)/(2,598,960)~=29.9%

#### Explanation:

An alternative way to do this is to use equations for combinations, the general formula for which is:

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

We want our hand to have exactly one Ace. Since there are four aces in a deck, we can set $n = 4 , k = 1$, and so:

${C}_{4 , 1}$

But we also need to account for the other four cards in our hand. There are 48 cards to pick from, so we can set $n = 48 , k = 4$, and so:

${C}_{48 , 4}$

We multiply them together to find the total number of ways we can get exactly one Ace in our hand:

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

(cancel(4!)(48!))/((3!)cancel(4!)(44!))=(cancel(48)^8xx47xx46xx45xxcancel44!)/(cancel(3xx2)xxcancel44!)=>

$8 \times 47 \times 46 \times 45 = 778 , 320$

~~~~~

To figure out the probability, we also need to know the total number of 5-card hands possible:

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

And so:

P=(778,320)/(2,598,960)~=29.9%