Question

Question: What is the probability that a five-card poker hand contains cards of five different kinds and does not contain a flush or a straight.

Short answer

Answer

The probability that a five-card poker hand contains cards of five different kinds and does not contain a flush or a straight is $P\left(E\right)=0.501$.

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 cards of five different kinds and does not contain a flush or 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

${}^{52}{C}_5$. 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 of five different kinds and does not contain a flush or a straight this can be accomplished as follows:

The number of ways of selecting 5 cards of different kind

${=}^{13}{C}_5{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1$

It can be simplified as follows:

$$\begin{gathered} {}^{13}{C}_5{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1=\frac{13!}{5!8!}\times \frac{4!}{1!3!}\times \frac{4!}{1!3!}\times \frac{4!}{1!3!}\times \frac{4!}{1!3!}\times \frac{4!}{1!3!} \\ {}^{13}{C}_5{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1=1287\times 4\times 4\times 4\times 4\times 4 \\ {}^{13}{C}_5{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1=1,317,888 \end{gathered}$$

The total number of straight

$$\begin{gathered} {}^{10}{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1{\times }^4{C}_1-40=10\times 4\times 4\times 4\times 4\times 4-40 \\ =10,200 \end{gathered}$$

Total number of flushes

$$\begin{gathered} {}^{13}{C}_5{\times }^4{C}_1-40=1287\times 4-40 \\ =5108 \end{gathered}$$

The probability that a five-card poker hand contains cards of five different kinds and does not contain a flush or a straight is

$$\begin{gathered} P\left(E\right)=\frac{1317888-(10200+5108+40)}{{}^{52}{C}_5} \\ =\frac{1302540}{{}^{52}{C}_5} \\ =\frac{1277}{2548} \\ =0.501 \end{gathered}$$

The probability that a five-card poker hand contains cards of five different kinds and does not contain a flush or a straight is$P\left(E\right)=0.501$ .