Question

Question: What is the probability that a five-card hand contains two pairs (that is, two of each of two different kinds and a fifth card of a third kind)?

Short answer

Answer

The probability that a five-card hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) is$P\left(E\right)=0.0475$.

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 two pairs (that is, each two of different kind and a fifth card of a third kind).

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${}^{52}{C}_5$.

$\therefore n\left(S\right){=}^{52}{C}_5$

We are supposed to draw a five-card poker hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) this can be accomplished as follows:

Number of ways to choose two pairs (that is, each two of different kind and a fifth card of a third kind$)=C(13,2)\times C(4,2)\times C(4,2)\times C(44,1)$

$$\begin{gathered} P\left(E\right)=\frac{C(13,2)\times C(4,2)\times C(4,2)\times C(44,1)}{{}^{52}{C}_5} \\ P\left(E\right)=\frac{\frac{13!}{2!\times 11!}\times \frac{4!}{2!\times 2!}\times \frac{4!}{2!\times 2!}\times \frac{44!}{1!\times 43!}}{\frac{52!}{47!\times 5!}} \\ P\left(E\right)=\frac{198}{4165} \\ P\left(E\right)=0.0475 \end{gathered}$$

The probability that a five-card hand contains two pairs (that is, each two of different kind and a fifth card of a third kind) is$P\left(E\right)=0.0475$ .