Question

Question: What is the probability that a five-card poker hand contain at least one ace.

Short answer

Answer

The probability that a five-card poker hand contain at least one ace is $P\left(E\right)\approx 0.34$.

Step-by-step solution

 Given  

A pack of cards. In a pack of cards there are $52$cards.

The Concept of Probability

If$\mathit{S}$represents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$.

Determine the probability

As per the problem we have been asked to find the probability that a five-card poker hand contain at least one ace.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$.

As we know that there are four Aces and a total of fifty-two cards.

We are supposed to draw five cards out of fifty-two cards but at least one ace out of four aces and this can be accomplished as follows:

At least one card is ace $=1-$(No card is ace)

$$\begin{gathered} P\left(E\right)=1-\frac{{}^{48}{C}_5}{{}^{52}{C}_5} \\ P\left(E\right)=1-\frac{48\times 47\times 46\times 45\times 44}{52\times 51\times 50\times 49\times 48} \\ P\left(E\right)=1-\frac{35673}{54145} \\ P\left(E\right)\approx 0.34 \end{gathered}$$

The probability that a five-card poker hand contain at least one ace is$P\left(E\right)\approx 0.34$ .