/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 In five-card poker, a straight c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 42

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10 ? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

Short Answer

Expert verified
The probability of a straight with high card 10 is \( 0.00039 \), any straight is \( 0.00392 \), and a straight flush is \( 0.000015 \).

Step by step solution

01

Understand the Problem

We need to find the probability of being dealt specific types of poker hands: a straight with high card 10, any straight, and a straight flush. A straight consists of five consecutive cards (like 6, 7, 8, 9, 10), and a straight flush is a straight where all five cards are of the same suit. Aces can be high (above King) or low (below 2).
02

Calculate Total Number of 5-Card Hands

We need to calculate the total number of possible 5-card hands from a standard 52-card deck. This is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of cards and \( k \) is the number of cards to choose.\[ \binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960 \]
03

Calculate the Probability of a Straight with High Card 10

For a straight with high card 10, the cards would be 6, 7, 8, 9, and 10. Each card can be any of the four suits, so the number of ways to form this straight is \( 4^5 \). The four suits multiply for each card in the straight.\[ 4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 1024 \]The probability is given by dividing the number of favorable outcomes by the total number:\[ P(\text{straight with high card 10}) = \frac{1024}{2,598,960} \approx 0.00039 \]
04

Calculate the Probability of Any Straight

There are 10 possible straights (Ace-to-5, 2-to-6, ..., 10-to-King). Each straight has \( 4^5 \) combinations because each card can be of any of the four suits. The total number of ways to form any straight is:\[ 10 \times 4^5 = 10 \times 1024 = 10240 \]The probability is:\[ P(\text{any straight}) = \frac{10240}{2,598,960} \approx 0.00392 \]
05

Calculate the Probability of a Straight Flush

A straight flush requires all cards to be in the same suit, so for each straight, there is only one way to arrange them in the same suit per suit. With 10 possible straights and 4 suits, there are:\[ 10 \times 4 = 40 \]ways to form a straight flush. The probability is:\[ P(\text{straight flush}) = \frac{40}{2,598,960} \approx 0.000015 \]
06

Conclusion and Compilation of Results

The probability of being dealt a straight with high card 10 is approximately 0.00039. The probability of being dealt any straight is approximately 0.00392. The probability of having a straight flush is approximately 0.000015.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics in Poker
In the world of poker, understanding combinatorics is essential. Combinatorics helps you calculate the different possible card combinations in a deck. When we say combinations, we mean the different ways you can draw cards from a 52-card deck without considering the order of the draw.
In our poker exercise, we're interested in finding how many possible ways you can have a five-card hand out of a full deck. This is calculated using the combination formula, represented as \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of selected items.
  • For our exercise, \( n = 52 \) (the total cards in a deck) and \( k = 5 \) (the number of cards in a poker hand).
  • The calculation gives us \( \binom{52}{5} = 2,598,960 \), indicating there are over 2.5 million possible five-card hands.
This large number sets the groundwork for then calculating specific poker hands probabilities, such as straights.
Probability Calculation
Once you understand the total possible hands, you can move on to calculating the probability of specific poker hands. Probability in poker tells you the chance of drawing particular card combinations, such as a straight or a straight flush.
A straight in poker is a sequence of consecutive denominations. If you want to calculate the probability of being dealt a straight with a high card 10, you would first find the number of favorable outcomes, i.e., how many ways can you draw this hand of consecutive cards out of the deck?
  • For a straight with high card 10, the cards would be 6, 7, 8, 9, 10. Each of these cards can be in any of the four suits, thus giving \( 4^5 = 1024 \) combinations.
  • To find the probability, divide the number of successful outcomes (1024) by the total possible hands (2,598,960): \( P(\text{straight with high card 10}) = \frac{1024}{2,598,960} \approx 0.00039 \).
Similar steps are followed for calculating the probability of other straights and straight flushes.
Poker Hands Analysis
Understanding different poker hands and their likelihoods is crucial for any poker strategy. Analyzing poker hands involves identifying which hand ranks higher or is more probable than others.
Let's break this down for the different types of straights:
  • A straight requires the cards to be in a sequence, without all being in the same suit. There are 10 possible straights, each divisible into \( 4^5 = 1024 \) combinations, giving 10,240 ways to have any straight. The probability of any straight is \( \frac{10240}{2,598,960} \approx 0.00392 \).
  • A straight flush is even more rare, needing both the sequential order and an identical suit. There are only 40 ways to have a straight flush from the deck, assigning it a minute probability of \( \frac{40}{2,598,960} \approx 0.000015 \).
Ultimately, knowing these probabilities helps players make informed decisions on their strategies, understanding the risk and reward ratio involved in aiming for certain hands during gameplay.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For customers purchasing a refrigerator at a certain appliance store, let \(A\) be the event that the refrigerator was manufactured in the U.S., \(B\) be the event that the refrigerator had an icemaker, and \(C\) be the event that the customer purchased an extended warranty. Relevant probabilities are. $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2 , another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

An engineering construction firm is currently working on power plants at three different sites. Let \(A_{f}\) denote the event that the plant at site \(i\) is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of \(A_{1}, A_{2}\), and \(A_{3}\), draw a Venn diagram, and shade the region corresponding to each one. a. At least one plant is completed by the contract date. b. All plants are completed by the contract date. c. Only the plant at site 1 is completed by the contract date. d. Exactly one plant is completed by the contract date. e. Either the plant at site 1 or both of the other two plants are completed by the contract date.

An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration. a. If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible? b. How many experimental runs are there that involve use of the lowest temperature and two lowest pressures? c. Suppose that five different experimental runs are to be made on the first day of experimentation. If the five are randomly selected from among all the possibilities, so that any group of five has the same probability of selection, what is the probability that a different catalyst is used on each run?

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.