Question

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

Answer 1

Step-1: Find the total number of possible hands-in tools:

A standard deck of cards contains $52$ cards, of which $26$ are red and $26$ are black, $13$ of each are hearts, diamonds, spades, and clubs.

The order of the cards does not matter (since a different order results in the same hand) and thus we need to use the definition of a combination.

A standard deck of cards contains $52$ cards. A five-card poker hand selects $5$ of the $52$ cards.

Use the combination formula:

$\Rightarrow \mathrm{C}(\mathrm{n},\mathrm{r})=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$.

Where $\mathrm{n}$ represents the total number of items and $\mathrm{r}$ represents the number of items being chosen at a time.

Substitute $\mathrm{n}=52$ and $\mathrm{r}=5$ in the above formula:

$\begin{array}{c}C(52,5)=\frac{52!}{5!(52-5)!}\\ =\frac{52!}{5!\left(47\right)!}\\ =\frac{52\times 51\times 50\times 49\times 48\times 47!}{5!\times 47!}\\ =\frac{52\times 51\times 50\times 49\times 48}{5!}[\mathrm{cancel}\mathrm{the}\mathrm{common}\mathrm{term}]\\ =\frac{311875200}{5!}\\ =\frac{311875200}{5\times 4\times 3\times 2\times 1}\\ =\frac{311875200}{120}\\ =2,598,960\end{array}$

So, the total number of possible outcomes are $\begin{array}{rcl}& & 2,598,960\end{array}$.

Step-2: Find the number of hands containing an ace:

The standard deck of cards contains four aces, so set $\mathrm{n}=4,\mathrm{r}=1$:
$\Rightarrow C_{4,1}$
As well as the five cards in our hands, we should consider the other four.

There are $48$ cards to pick from, so we can set $\mathrm{n}=48,\mathrm{r}=4$.
$\Rightarrow C_{48,4}$
Multiplying the above two equations together, we can get exactly one Ace:

$\begin{array}{c}{C}_{4,1}\times C_{48,4}=\frac{4!}{\left(1\right)!(4-1)!}\times \frac{48!}{\left(4\right)!(48-4)!}\\ =\frac{4\times 3!}{3!}\times \frac{48\times 47\times 46\times 45\times 44!}{4!44!}\\ =4\times \frac{48\times 47\times 46\times 45}{4!}[\mathrm{cancel}\mathrm{the}\mathrm{common}\mathrm{term}]\\ =4\times \frac{4669920}{4!}\\ =4\times \frac{4669920}{4\times 3\times 2\times 1}\\ =4\times \frac{4669920}{24}\\ =4\times 194580\\ =778320\end{array}$

So, the number of successful outcomes $778320$.

Step-3: Find the probability that the deal of a five-card hand provides exactly one ace:

The probability is the number of favorable outcomes divided by the number of possible outcomes.
$\begin{array}{rcl}\mathrm{P}(\mathrm{Hand}\mathrm{contains}\mathrm{ace})& =& \frac{\mathrm{Number}\mathrm{of}\mathrm{successful}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}\end{array}$

Substitute the values in the above formula:

$\begin{array}{rcl}\mathrm{P}(\mathrm{Hand}\mathrm{contains}\mathrm{ace})& =& \frac{778,320}{2,598,960}\\ & =& \frac{3243}{10829}\\ & \approx & 0.2995\end{array}$

Hence, the probability that the deal of a five-card hand provides exactly one ace is $\frac{\mathbf{3243}}{\mathbf{10829}}\mathbf{}\mathbf{or}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{2995}$.