Chapter 15, Problem 46P

(a)

To determine

The rank of four aces and a king.

Answer to Problem 46P

The four aces and a king ranks second among all the option.

Explanation of Solution

Given:

The given that rank of the following five-card hand in order of increasing the probability.

Calculation:

Calculate the rank of the four aces and a king.

There are four different combination that satisfy this particular microstate.

Therefore, the four aces and a king ranks second among all the option.

(b)

To determine

The rank of six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades probability.

Answer to Problem 46P

The six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades ranks first because it has the lowest probability.

Explanation of Solution

Given:

The given that rank of the following five-card hand in order of increasing probability.

Calculation:

Calculate the probability,

Since there is only one of very card this microstate has one microstate.

Therefore, the six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades ranks first because it has the lowest probability.

(c)

To determine

The rank of two jacks, two queen and an ace.

Answer to Problem 46P

The two jacks, two queen and an ace is $144$ microstate.

Explanation of Solution

Given:

The given that rank of the following five-card hand in order of increasing probability.

Calculation:

Calculate the probability,

There are four possible ace combination and six possible combination for both and queen.

The two jacks, two queen, and an ace is,

  $4\times 6\times 6=144$

Therefore, the two jacks, two queen, and an ace has $144$ microstate.

(d)

To determine

The rank of any hand having no two equal-value cards.

Answer to Problem 46P

The hand having no two equal-value cards ranks last.

Explanation of Solution

Given:

The given that rank of the following five-card hand in order of increasing probability.

Calculation:

Calculate the probability,

This has the greatest possibility because the first card can be any card in the deck, $52$ option and the second card could be any of $48$ .

Therefore, the hand having no two equal-value cards ranks last.