 # In a poker hand consisting of 5 cards, find the probability of holding 3 aces.

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:

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

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 $0.00665$.

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