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Chapter 9: Problem 24

Suppose five cards are drawn from a deck. Find the probability of obtaining the indicated cards. A royal flush (an ace, king, queen, jack, and 10 of the same suit)

Short Answer

Expert verified
The probability of drawing a royal flush is \( \frac{1}{649,740} \).

Step by step solution

01

Understand the Definition of Royal Flush

A royal flush consists of five cards: an ace, king, queen, jack, and 10, all from the same suit. There are exactly four possible royal flush hands in a deck of cards, one for each suit (spades, hearts, diamonds, clubs).
02

Calculate Total Possible Hands

A standard deck has 52 cards. When drawing 5 cards, the total number of possible 5-card hands from a deck can be calculated using the combination formula: \[ \binom{52}{5} = \frac{52!}{5!(52-5)!} \].
03

Evaluate Combinatorial Formula

Calculate the total number of possible hands: \[ \binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960 \] possible 5-card hands.
04

Probability of Drawing a Royal Flush

Since there are 4 royal flush combinations (one for each suit), the probability of drawing a royal flush is: \[ \frac{4}{2,598,960} \].
05

Simplify the Probability

Simplify the fraction \( \frac{4}{2,598,960} \) by dividing both the numerator and the denominator by 4. Thus, the probability is \( \frac{1}{649,740} \).

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Key Concepts

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

Royal Flush
In the world of card games, a Royal Flush is considered synonymous with luck and skill. It is the highest-ranking hand in poker, encompassing five specific cards: ace, king, queen, jack, and ten, all belonging to the same suit. There are four suits in a deck鈥攕pades, hearts, diamonds, and clubs鈥攕o there are exactly four possible Royal Flush hands, one for each suit. This rarity and high rank make it a sought-after combination in games, not only because it鈥檚 difficult to obtain, but also because it beats all other hands. Understanding a Royal Flush is fundamental to comprehending card game strategies, as its power often dictates high-stakes decisions.
Combination Formula
The combination formula is pivotal in solving various probability problems, particularly in card games. When you want to know how many different ways you can choose a subset from a larger set without caring about the order, combinations come into play. The formula, generally expressed as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), signifies choosing \(r\) objects from \(n\) objects. In the context of a 52-card deck, using \( \binom{52}{5} \) helps in determining the number of possible 5-card hands that can be drawn. This results in 2,598,960 different combinations. Embracing this formula is crucial for calculating probabilities and strategizing effectively in card games.
Combinatorial Probability
Combinatorial probability deals with determining the likelihood of a particular outcome happening out of all possible outcomes. It鈥檚 vital in understanding the mechanics of card games and predicting outcomes. In our example, to find the probability of drawing a Royal Flush, we consider the total number of Royal Flush combinations (which is 4) divided by the total number of possible 5-card hands, \( 2,598,960 \). Thus, the probability is calculated as \( \frac{4}{2,598,960} \), which simplifies to \( \frac{1}{649,740} \). Understanding combinatorial probability enables you to assess risks and probabilities in card games, leading to more informed decisions.
Deck of Cards
A standard deck of cards is composed of 52 cards, organized into four suits: spades, hearts, diamonds, and clubs. Each suit contains 13 cards: ace through ten, followed by jack, queen, and king. This uniform structure is crucial for calculating probabilities, as it provides a predictable way to measure potential outcomes. Each card鈥檚 suit and number play a role in the hierarchy and combinations in games, notably in poker, where specific groupings create powerful hands like Royal Flushes. Understanding a deck of cards is essential as it forms the basis for every calculation and strategic decision in card games.

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Most popular questions from this chapter

That girl-boy births are equally probable, find the probability that a family with five children has (a) all boys (b) at least one girl

The outcomes \(1,2, \ldots, 6\) of an experiment and their probabilities are listed in the table. $$\begin{array}{l|cccccc}\\\\\hline \text { Outcome } & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline \text { Probability } & 0.25 & 0.10 & 0.15 & 0.20 & 0.25 & 0.05 \\\\\hline\end{array}$$.For the Indicated events, find (a) \(P\left(E_{2}\right),\) (b) \(P\left(E_{1} \cap E_{2}\right)\) (c) \(P\left(E_{1} \cup E_{2}\right),\) and \((d) P\left(E_{2} \cup E_{3}^{\prime}\right)\). $$E_{1}=\\{1,2,3,6\\} ; \quad E_{2}=\\{3,4\\} ; \quad E_{3}=\\{4,5,6\\}$$

Letter and number experiment An experiment consists of selecting a letter from the alphabet and one of the digits 0 , \(\therefore 9\). (a) Describe the sample space \(S\) of the experiment, and find \(n(S)\) (b) Suppose the letters of the alphabet are assigned numbers as follows: \(A=1, B=2, \ldots, Z=26 .\) Let \(E_{1}\) be the event in which the units digit of the number assigned to the letter of the alphabet is the same as the digit selected. Find \(n\left(E_{1}\right), n\left(E_{1}^{\prime}\right),\) and \(P\left(E_{1}\right)\) Let \(E_{2}\) be the event that the letter is one of the five vowels and \(E_{3}\) the event that the digit is a prime number. Are \(E_{2}\) and \(E_{3}\) mutually exclusive? Are they independent? Find \(P\left(E_{2}\right), P\left(E_{3}\right), P\left(E_{2} \cap E_{3}\right),\) and \(P\left(E_{2} \cup E_{3}\right)\) Let \(E_{4}\) be the event that the numerical value of the letter is even. Are \(E_{2}\) and \(E_{4}\) mutually exclusive? Are they independent? Find \(P\left(E_{2} \cap E_{4}\right)\) and \(P\left(E_{2} \cup E_{4}\right)\)

If a single die is tossed, find the probability of obtaining an odd number or a prime number.

Ten horses are entered in a race. If the possibility of a tie for any place is ignored, in how many ways can the first-, second-, and third-place winners be determined?
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