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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:

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

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 $0.00665$.

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