Question

Question: What is the probability that a five-card poker hand contains a royal flush, that is, the $\mathbf{10}$, jack, queen, king and ace of one suit.

Short answer

The probability that a five-card poker hand contains a royal flush, that is, the $10$,jack, queen, king and ace of one suit is $P\left(E\right)=0.0000015$.

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

Here we have to find the probability that a five-card poker hand contain contains a contains a straight, that is, five cards that have consecutive kinds.

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 a total of fifty-two cards and five cards can be chosen in${}^{52}{C}_5$. and so, we have$n\left(S\right){=}^{52}{C}_5$.

As we know that there are only four Royal flushes for each suit. So, we can write

$$\begin{gathered} P\left(E\right)=\frac{4}{{}^{52}{C}_5} \\ =\frac{4}{\frac{52!}{47!\times 5!}} \\ =\frac{4}{2598960} \\ =0.0000015 \end{gathered}$$

The probability that a five-card poker hand contains a royal flush, that is, the 10, jack, queen, king and ace of one suite is $P\left(E\right)=0.0000015$.