Chapter 10.1, Problem 58E

Solution

(a).

To find: The probability that a hand will contain at least one king.

The probability that a hand will contain at least one king is $0.34$ .

Given information:

In the original version of poker known as “straight” poker, a $5$ - card hand is dealt from a standard deck of $52$ cards.

Calculation:

Probability of no kings and probability of at least one king must add up to one so probability of at least one king would be one minus probability of no kings.

  $P\left(\text{At least one king}\right)=1-P\left(\text{No kings}\right)$

The probability of getting no kings is $48$ choose $5$ because there are $48$ non-king cards and you are choosing $5$ out of those $48$ and you divide that by $52$ choose $5$ because you are choosing $5$ cards out of a normal $52$ -card deck. You then subtract that from one to get the probability of at least one king in your hand.

  $1-\frac{C{}_5}{C{}_5}$

Now, calculate $1-\frac{C{}_5}{C{}_5}$ by using combination formula:

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Substitute the values:

  $\begin{array}{l}1-\frac{C{}_5}{C{}_5}=1-\frac{\frac{48!}{5!\left(48-5\right)!}}{\frac{52!}{5!\left(52-5\right)!}}\\ =1-\frac{\frac{48\times 47\times 46\times 45\times 44\times 43!}{5!\times 43!}}{\frac{52\times 51\times 50\times 49\times 48\times 47!}{5!\times 47!}}\\ =1-\frac{\frac{48\times 47\times 46\times 45\times 44}{5!}}{\frac{52\times 51\times 50\times 49\times 48}{5!}}\\ =1-\frac{18472}{54145}\\ \approx 0.34\end{array}$

Hence, the probability that a hand will contain at least one king is $0.34$ .

(b).

To find: The probability a hand will be a “full house” (any three of one kind and a pair of another kind).

The probability that a hand will be a “full house” is $0.0014$ .

Given information:

In the original version of poker known as “straight” poker, a $5$ - card hand is dealt from a standard deck of $52$ cards.

Calculation:

A full house is getting $3$ of a kind and a pair of a different card in a $5$ -card hand. There are $13$ permute $2$ ways to get $2$ different types of cards in your hand (like an ace and a $2$ ) you multiply that by $4$ choose $3$ which are the $3$ cards you pull of the same type ( $3$ aces for example) and $4$ choose two for the pair(a pair of $2$ s for example). That gives you the number of ways a full house can occur in a $5$ -card hand. You divide that amount by the number of hands possible which is $52$ choose $5$ ( $52$ cards in a deck and $5$ -card hands) .

  $P(\text{Full House})=\frac{P{}_2\cdot C{}_3\cdot C{}_2}{C{}_5}$

Now, calculate $\frac{P{}_2\cdot C{}_3\cdot C{}_2}{C{}_5}$ by using combination formula:

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Permutation formula: $P{}_r=\frac{n!}{\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Substitute the values:

  $\begin{array}{l}\frac{P{}_2\cdot C{}_3\cdot C{}_2}{C{}_5}=\frac{\frac{13!}{\left(13-2\right)!}\times \frac{4!}{3!\left(4-3\right)!}\times \frac{4!}{2!\left(4-2\right)!}}{\frac{52!}{5!\left(52-5\right)!}}\\ =\frac{6}{4165}\approx 0.0014\end{array}$

Hence, the probability a hand will be a “full house” (any three of one kind and a pair of another kind) is $0.0014$ .