Question

Question: To determine the probability that a five-card poker hand contains a straight, that is, five cards that have consecutive kinds(Note that an ace can be considered either the lowest card of an $\mathit{A}\mathbf{-}\mathbf{2}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{4}\mathbf{-}\mathbf{5}$straight or the highest card of a $\mathbf{10}\mathbf{-}\mathit{J}\mathbf{-}\mathit{Q}\mathbf{-}\mathit{K}\mathbf{-}\mathit{A}$straight.)

Short answer

Answer

The probability that a five-card poker hand contains a straight, that is, five cards that have consecutive kinds is $P\left(E\right)=0.000394$.

Step-by-step solution

 Given  

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

The Concept ofProbability

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.

An ace card can be considered either the lowest card of an$A-2-3-4-5$or the highest card of a $10-J-Q-K-A$straight.

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 localid="1668487992289" ${}^{52}{C}_5$and so, we have$n\left(S\right){=}^{52}{C}_5$

We are supposed to draw a five-card poker hand contains a straight, that is, five cards that have consecutive kinds this can be accomplished as follows:

There are four cards of each kind

The first one can be selected in${}^4{C}_1$ways.

The second one can be selected in${}^4{C}_1$ways.

The third one can be selected in ${}^4{C}_1$ways.

The fourth one can be selected in${}^4{C}_1$ways.

The fifth one can be selected in${}^4{C}_1$ways.

$$\begin{gathered} P\left(E\right)=\frac{10{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1}{{}^{52}{C}_5} \\ P\left(E\right)=\frac{10\times 4^5}{\frac{52!}{47!\times 5!}} \\ P\left(E\right)=\frac{10240}{2598960} \\ P\left(E\right)=0.000394 \end{gathered}$$

The probability that a five-card poker hand contains a straight, that is, five cards that have consecutive kinds is $P\left(E\right)=0.000394$$P\left(E\right)=0.000394$.