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Chapter 5: Problem 17

The probability of a royal flush in a poker hand is \(p=1 / 649,740 .\) How large must \(n\) be to render the probability of having no royal flush in \(n\) hands smaller than \(1 / e ?\)

Short Answer

Expert verified
Approximately 449,232 hands are needed.

Step by step solution

01

Understanding the Problem

We need to find the number of poker hands, denoted as \(n\), where the probability of not getting a royal flush in all those \(n\) hands is less than \(\frac{1}{e}\), given that the probability of a royal flush in a single hand is \(p = \frac{1}{649,740}\).
02

Define the Probability of No Royal Flush

The probability of not getting a royal flush in one poker hand is \(1 - p\). In \(n\) hands, the probability of no royal flush occurring is \((1 - p)^n\).
03

Set the Condition

The problem asks us to find \(n\) such that the probability of having no royal flush in \(n\) hands is less than \(\frac{1}{e}\). Therefore, we need to solve the inequality:\[(1 - p)^n < \frac{1}{e}\]
04

Applying the Logarithm

Taking the natural logarithm of both sides gives us:\[\ln((1 - p)^n) < \ln\left(\frac{1}{e}\right) = -1\] This simplifies to:\[ n \ln(1 - p) < -1 \]
05

Solving for n

Since \(\ln(1 - p)\) is negative (because \(1 - p < 1\)), we can divide both sides by it to isolate \(n\):\[n > \frac{-1}{\ln(1 - p)}\] Now, substitute \(p = \frac{1}{649,740}\) to calculate \(n\):\[ n > \frac{-1}{\ln\left(1 - \frac{1}{649,740}\right)}\]
06

Compute the Value

Using a calculator, evaluate \(\ln\left(1 - \frac{1}{649,740}\right)\), and then find \(n\). This value is approximately:\[ n > 449,231.55\] Since \(n\) must be a whole number, we round up to \(n = 449,232\).

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Key Concepts

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

Poker Probability
Poker, a popular card game, involves combining different cards to make winning hands. One of the most challenging hands to achieve in poker is the royal flush. A royal flush consists of an Ace, King, Queen, Jack, and Ten, all of the same suit.
  • There are 4 suits in a deck (hearts, diamonds, clubs, spades), meaning 4 possible royal flushes.
  • Since there are 52 cards in a deck and 5 cards in a hand, the total number of possible hands is calculated by the combination formula \( \binom{52}{5} \).
  • The actual probability of a royal flush in one hand is thus \( \frac{4}{649,740} \), which simplifies to approximately \( 1/649,740 \).
Understanding the probability of drawing specific hands in poker helps players make better decisions during the game, anticipating their moves based on odds.
Inequality Solving
Inequalities are mathematical statements indicating that two expressions are not equal. In probability exercises like the one about poker, solving inequalities helps determine the necessary conditions for achieving certain outcomes.
  • The inequality from the poker problem is \((1 - p)^n < \frac{1}{e}\), where \(p\) is the probability of getting a royal flush.
  • To solve this inequality and find \(n\), we take the logarithm of both sides, which transforms the inequality into: \(n \ln(1 - p) < -1\).
  • Since \(\ln(1 - p)\) is negative (because probabilities less than 1 have negative logs), dividing by it reverses the inequality sign, leading to \(n > \frac{-1}{\ln(1 - p)}\).
By solving such inequalities, we discover the necessary number of trials (hands) to achieve a certain probability.
Logarithmic Functions
Logarithms, especially natural logs, are useful in solving problems involving exponential growth, decay, or probabilities. In the royal flush problem, we encounter natural logarithms while finding the number of poker hands required.
  • Natural logarithms are denoted by \(\ln\) and have a base of \(e\), where \(e\) is approximately 2.718.
  • Transforming probabilities using logarithms simplifies multiplicative relationships to additive ones, making calculations more manageable.
  • For example, \(\ln((1 - p)^n)\) becomes \(n \ln(1 - p)\), turning a power operation into multiplication, which is easier to handle.
Utilizing logarithmic functions in problem-solving allows us to manage complex exponential relationships, providing clearer results.
Exponential Functions
Exponential functions are vital in understanding growth processes in many fields, including probability. In the context of the poker problem, exponential functions represent repeated trials with a consistent probability of success or failure.
  • An exponential function is generally of the form \(a^n\), where \(a\) is a constant and \(n\) is a variable exponent.
  • In the royal flush context, \((1 - p)^n\) represents the probability of not drawing a royal flush across several hands.
  • This exponential decrease as \(n\) increases showcases the diminishing likelihood of not encountering at least one royal flush over time.
Understanding exponential functions helps us model scenarios where repeated actions (like poker hands) affect outcomes over time.

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Most popular questions from this chapter

Suppose that \(X\) is a random variable which represents the number of calls coming in to a police station in a one-minute interval. In the text, we showed that \(X\) could be modelled using a Poisson distribution with parameter \(\lambda\), where this parameter represents the average number of incoming calls per minute. Now suppose that \(Y\) is a random variable which represents the number of incoming calls in an interval of length \(t\). Show that the distribution of \(Y\) is given by $$P(Y=k)=e^{-\lambda t} \frac{(\lambda t)^{k}}{k !}i.e., \(Y\) is Poisson with parameter \(\lambda t\). Hint: Suppose a Martian were to observe the police station. Let us also assume that the basic time interval used on Mars is exactly \(t\) Earth minutes. Finally, we will assume that the Martian understands the derivation of the Poisson distribution in the text. What would she write down for the distribution of $Y ?$$$

Write a computer algorithm that simulates a hypergeometric random variable with parameters \(N, k,\) and \(n\).

Let \(U\) be a uniformly distributed random variable on \([0,1] .\) What is the probability that the equation $$x^{2}+4 U x+1=0$$ has two distinct real roots \(x_{1}\) and \(x_{2} ?\)

A census in the United States is an attempt to count everyone in the country. It is inevitable that many people are not counted. The U. S. Census Bureau proposed a way to estimate the number of people who were not counted by the latest census. Their proposal was as follows: In a given locality, let \(N\) denote the actual number of people who live there. Assume that the census counted \(n_{1}\) people living in this area. Now, another census was taken in the locality, and \(n_{2}\) people were counted. In addition, \(n_{12}\) people were counted both times. (a) Given \(N, n_{1},\) and \(n_{2},\) let \(X\) denote the number of people counted both times. Find the probability that \(X=k,\) where \(k\) is a fixed positive integer between 0 and \(n_{2}\). (b) Now assume that \(X=n_{12}\). Find the value of \(N\) which maximizes the expression in part (a). Hint: Consider the ratio of the expressions for successive values of \(N\).

(Ross \(^{11}\) ) An expert witness in a paternity suit testifies that the length (in days) of a pregnancy, from conception to delivery, is approximately normally distributed, with parameters \(\mu=270, \sigma=10 .\) The defendant in the suit is able to prove that he was out of the country during the period from 290 to 240 days before the birth of the child. What is the probability that the defendant was in the country when the child was conceived?
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