This **article aims to determine the probability of holding** $3$ aces in a **poker hand** of $5$. The **article** uses the background concept of probability and combination. To** solve** problems like this, the idea of combinations should be clear. A **combination** combines $n$ things $k$ at once **without repetition.** The formula to find the **combination** is:

\[\binom {n}{k} = \dfrac{n!}{k!(n-k)!}\]

**Expert Answer**

A **poker hand **have $5$ cards, and we need to have $3$ aces.

In the standard deck of $52$ cards, there are $4$ aces from which we have to choose $3$. To **find the number of ways to choose** $3$ out of $4$ aces, we have to use **combinations since the order is unimportant.**

\[ \binom {4}{3} = \dfrac{4! }{3! (4-3)!} = 4\:ways \]

Now we have to pick $2$** cards from the remaining** $48$ cards ($52$ cards minus $4$ aces). The **number of ways to pick these** $2$ cards out of $48$ cards is

\[ \binom {48}{2} = \dfrac {48!}{2! (48-2)! } = \dfrac{48 * 47}{2} = 1128\:ways \]

If **first operation can be performed** in $4$ ways (the number of ways to select $3$ of the $4$ aces), and for each of these ways, the **second operation can be performed** in $1128\: ways $ (the number of ways to select the remaining $2$ cards), then these $2$ **operations can be performed** together in

\[4*1128 = 4512\:ways\]

So there is $4512\: ways $ **to choose** $3$ aces in a **poker hand**.

**Number of ways to **pick $5$ out of $52$ cards:

\[ \binom {52}{5} = \dfrac{52!}{5! (52-5)!} = \dfrac{52.51.50.49.48.47}{5.4.3.2.1} = 2598960\: ways\]

So there are $2598960 \: ways $ to **choose for a poker hand.**

So the **probability of choosing** $3 $ **aces in a poker hand**.

\[P = \dfrac{the\: number\: of \:ways\:to \:choose\: 3\:aces\: in\:a \:poker \:hand}{the\:number\:of\:ways \:to\:choose\: a \:poker\:hand} = \dfrac{4512}{2598960} = 0.00174 \]

Hence, **probability of choosing** $3 $ **aces in a poker hand** is $0.00174$.

**Numerical Result**

**Probability of choosing** $3$ **aces in a poker hand is** $0.00174$.

**Example**

**In a $5$-card poker game, find the probability of holding $2$ aces.**

**Solution**

To **find number of ways to choose** $ 2 $ out of $ 4 $ aces, we have to use **combinations since the order is unimportant.**

\[ \binom {4}{2} = \dfrac{4! }{2! (4-2)!} = 6\:ways \]

The **number of ways to pick these** $ 3 $ cards out of $ 48 $ cards is

\[ \binom {48}{3} = \dfrac {48!}{3! (48-3)! } = 17296 \:ways \]

\[4*17296 = 69184\:ways\]

So there are $ 69184\: ways $ **to choose** $ 2 $ aces in a **poker hand**.

**Number of ways to **pick $5$ out of $52$ cards

So there are $2598960 \: ways $ to **choose for a poker hand.**

So the **probability of choosing** $ 2 $ **aces in a poker hand**.

\[P = \dfrac{the\: number\: of \:ways\:to \:choose\: 2\:aces\: in\:a \:poker \:hand}{the\:number\:of\:ways \:to\:choose\: a \:poker\:hand} = \dfrac{17296}{2598960} = 0.00665 \]

The **probability of choosing** $ 2 $ **aces in a poker hand** is $0.00665$.