/*! 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 37 A flush in the card game of poke... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A flush in the card game of poker occurs if a player gets five cards that are all the same suit (clubs, diamonds, hearts, or spades). Answer the following questions to obtain the probability of being dealt a flush in five cards. (a) We initially concentrate on one suit, say clubs. There are 13 clubs in a deck. Compute \(P(\) five clubs \()=P(\) first card is clubs and second card is clubs and third card is clubs and fourth card is clubs and fifth card is clubs). (b) A flush can occur if we get five clubs or five diamonds or five hearts or five spades. Compute \(P\) (five clubs or five diamonds or five hearts or five spades). Note that the events are mutually exclusive.

Short Answer

Expert verified
The probability of being dealt a flush in five cards is 0.00198.

Step by step solution

01

Understand the Problem

We need to find the probability of getting a flush in poker. A flush means getting five cards all of the same suit from a deck of 52 cards.
02

Probability of Drawing Five Clubs

There are 13 clubs in a deck of 52 cards. We want to find the probability of drawing 5 clubs consecutively. The probability of drawing the first club is \( \frac{13}{52} \). After drawing one club, there are now 12 clubs left out of 51 remaining cards for the second draw, and so on. Thus, the probability for all 5 cards being clubs is calculated as:\[P(\text{five clubs}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \times \frac{10}{49} \times \frac{9}{48}\]
03

Simplify the Calculation

Simplify the fraction step-by-step to get the exact probability:\[P(\text{five clubs}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \times \frac{10}{49} \times \frac{9}{48} \ = \frac{13}{52} \times \frac{4}{17} \times \frac{11}{50} \times \frac{5}{49} \times \frac{3}{16} \ \approx 0.000495 \]
04

Probability of Flush in Any Suit

A flush can occur in any suit (clubs, diamonds, hearts, spades). These are mutually exclusive events. Therefore, the probability of a flush in any suit is four times the probability of a flush in clubs.\[P(\text{flush in any suit}) = 4 \times P(\text{five clubs}) \ = 4 \times 0.000495 \ = 0.00198\]

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 91Ó°ÊÓ!

Key Concepts

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

mutually exclusive events
In probability theory, mutually exclusive events are events that cannot happen at the same time. For example, when flipping a coin, getting heads and getting tails are mutually exclusive because you can't get both at once.
In the context of poker, the events of drawing five clubs, five diamonds, five hearts, or five spades are mutually exclusive. This is because a single hand of five cards can only belong to one suit.
Understanding mutually exclusive events helps in calculating the overall probability of a flush across different suits. When calculating probabilities for such events, you can simply add the individual probabilities, since they do not overlap.
probability calculation
Probability calculation involves determining the likelihood of a given outcome. In our poker example, we calculated the probability of drawing five clubs using conditional probabilities.
The key steps were as follows:
  • Probability of drawing the first club: \( \frac{13}{52} \)
  • Probability of drawing the second club after the first club: \( \frac{12}{51} \)
  • Probability of drawing the third club after the second club: \( \frac{11}{50} \)
  • Probability of drawing the fourth club after the third club: \( \frac{10}{49} \)
  • Probability of drawing the fifth club after the fourth club: \( \frac{9}{48} \)

We then multiplied these probabilities together:
\ P(\text{five clubs}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \times \frac{10}{49} \times \frac{9}{48} \ Simplifying this gives a probability of approximately 0.000495.
Finally, since a flush can occur in any of the four suits and these are mutually exclusive events, the total probability of a flush in any suit is four times the probability of a flush in clubs:
\ P(\text{flush in any suit}) = 4 \times P(\text{five clubs}) \ \ \ = 4 \times 0.000495 \
0.00198.
combinatorics in card games
Combinatorics is a branch of mathematics concerned with counting, arrangement, and combination of objects. It plays a crucial role in card games like poker.
For example, calculating the number of ways to get five cards of the same suit involves understanding combinations.
Consider that initially, we calculated the sequential probability of drawing five clubs. This involves understanding arrangements where order matters because the probability changes with each draw.
In poker, however, order generally doesn't matter for a final hand, just the combination of cards. Thus, combinatorics helps determine the total possible hands and specific desired outcomes.
  • The total number of ways to choose 5 cards from a deck of 52 is given by the combination formula:
    \[\binom{52}{5}\]
  • The number of ways to choose 5 clubs from 13 is:
    \[\binom{13}{5}\]

Combinatorial calculations make it easier to determine probabilities by counting the number of favorable outcomes and dividing by the total number of possible outcomes.

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

Suppose Dan is going to upload 12 songs to his digital music player. In how many ways can the 12 songs be played without repetition?

In a survey of 1100 female adults (18 years of age or older), it was determined that 341 volunteered at least once in the past year. (a) What is the probability that a randomly selected adult female volunteered at least once in the past year? (b) Interpret this probability.

Birthdays Exclude leap years from the following calculations and assume each birthday is equally likely: (a) Determine the probability that a randomly selected person has a birthday on the 1 st day of a month. Interpret this probability. (b) Determine the probability that a randomly selected person has a birthday on the 31 st day of a month. Interpret this probability. (c) Determine the probability that a randomly selected person was born in December. Interpret this probability. (d) Determine the probability that a randomly selected person has a birthday on November 8 . Interpret this probability. (e) If you just met somebody and she asked you to guess her birthday, are you likely to be correct? (f) Do you think it is appropriate to use the methods of classical probability to compute the probability that a person is born in December?

Find the value of each factorial. \(10 !\)

Four members from a 20-person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice-chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?

See all solutions

Recommended explanations on Math Textbooks

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.