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Chapter 4: Problem 66

Consider a roulette wheel consisting of 38 numbers-1 through 36,0 , and double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12 , what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his fourth bet?

Short Answer

Expert verified
The probability that Smith will lose his first 5 bets is approximately 20.57%, and the probability that his first win will occur on the fourth bet is approximately 12.43%.

Step by step solution

01

Determine the probability of Smith winning or losing a single bet

There are 38 numbers on the roulette wheel, and Smith bets on 12 numbers (1 through 12). Therefore, the probability of Smith winning a single bet is the ratio of the number of winning outcomes (12) to the total number of outcomes (38). The probability of Smith losing a single bet is the complement of the winning probability, which equals 1 minus the probability of winning. So, the probability of Smith winning a single bet is: \[P(win) = \frac{12}{38}\] The probability of Smith losing a single bet is: \[P(lose) = 1 - P(win) = 1 - \frac{12}{38} = \frac{26}{38}\]
02

Calculate the probability of Smith losing his first 5 bets

To find the probability of Smith losing his first 5 bets, we need to multiply the probability of Smith losing a single bet by itself 5 times since the events are independent. Thus, the probability of losing 5 bets in a row is: \[P(lose\ 5\ bets) = \left(\frac{26}{38}\right)^5\]
03

Calculate the probability of Smith winning his first bet on his fourth bet

For Smith's first win to occur on his fourth bet, he needs to lose the first three bets and win on his fourth. We can find the probability of this happening by multiplying the probability of losing three times by the probability of winning on the fourth bet. Thus, the probability of the first win on the fourth bet is: \[P(first\ win\ on\ fourth\ bet) = \left(\frac{26}{38}\right)^3 \times \left(\frac{12}{38}\right)\]
04

Calculate the probabilities

Using the steps above, we can now find the probabilities for the given scenarios: (a) The probability of Smith losing his first 5 bets: \[P(lose\ 5\ bets) = \left(\frac{26}{38}\right)^5 \approx 0.2057\] (b) The probability of Smith's first win occurring on his fourth bet: \[P(first\ win\ on\ fourth\ bet) = \left(\frac{26}{38}\right)^3 \times \left(\frac{12}{38}\right) \approx 0.1243\] Thus, the probability that Smith will lose his first 5 bets is approximately 20.57%, and the probability that his first win will occur on the fourth bet is approximately 12.43%.

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Key Concepts

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

Roulette Wheel
A roulette wheel is a popular gambling device found in casinos around the world. It typically consists of a spinning wheel with numbered slots. There are two main types: European with 37 slots (numbers 1-36 and a single zero), and American, which has 38 slots, including the numbers 1-36, a single zero (0), and a double zero (00). The wheel's design ensures randomness and is key to probability theory exercises.
In the context of our problem, the roulette wheel is American-style. Smith places a bet on 12 numbers it includes numbers 1 through 12. Each spin is an independent event, meaning past spins do not influence the outcome of future spins. Understanding the roulette setup is crucial to calculating odds, making it an exciting element in probability theory.
Winning and Losing Probabilities
Winning and losing probabilities are fundamental concepts in probability theory. They enable us to determine the likelihood of different outcomes. In this exercise, Smith bets on 12 numbers on a roulette wheel with a total of 38 numbers.
  • The probability of winning a single bet is the ratio of winning outcomes (12 numbers) to the total possible outcomes (38 numbers). Thus, Smith's probability of winning is \( P(win) = \frac{12}{38} \).

  • The probability of losing can be understood as the complement of winning, which is 1 minus the probability of winning, giving us \( P(lose) = 1 - P(win) = \frac{26}{38} \).
These calculations are essential for understanding subsequent events, like losing multiple times or winning after a sequence of losses.
Independent Events
Independent events in probability theory refer to situations where the outcome of one event does not affect the outcome of another. In the context of a roulette wheel, regardless of previous spins, each new spin is an independent event. This aspect is central in calculating the probability of sequences like losing several times in a row.
For example, when calculating the likelihood of Smith losing his first 5 bets, each spin's outcome does not depend on previous spins:
  • The probability of losing a single bet is \( \frac{26}{38} \).
  • The probability of losing all 5 bets is \( \left(\frac{26}{38}\right)^5 \).
This independence makes it straightforward to calculate probabilities by raising individual probabilities to the power of the number of events.
Geometric Distribution
The geometric distribution is a probability distribution used for modeling the number of trials until the first success in a series of independent and identically distributed Bernoulli trials. In simple terms, it helps in finding the probability that Smith wins for the first time on a specific bet.
In this exercise, we want to determine the probability of Smith winning for the first time on his fourth bet:
  • He must lose the first three bets and win on the fourth one.
  • The probability that Smith's first win occurs on the fourth bet is given by \( (\frac{26}{38})^3 \times (\frac{12}{38}) \).
This scenario neatly fits the geometric distribution model, emphasizing its relevance to problems involving sequences of independent trials and their outcomes.

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Most popular questions from this chapter

On a multiple-choice exam with 3 possible answers for each of the 5 questions, what is the probability that a student would get 4 or more correct answers just by guessing?

A newsboy purchases papers at 10 cents and sells them at 15 cents. However, he is not allowed to return unsold papers. If his daily demand is a binomial random variable with \(n=10, p=\frac{1}{3}\), approximately how many papers should he purchase so as to maximize his expected profit?

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through 80. A player can select from 1 to 15 numbers; a win occurs if some fraction of the player's chosen subset matches with any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of 20 , and the payoff is \(\$ 2.2\) won for every dollar bet. (As the player's probability of winning in this case is \(\frac{1}{4}\), it is clear that the "fair" payoff should be \(\$ 3\) won for every \(\$ 1\) bet.) When the player selects 2 numbers, a payoff (of odds) of \(\$ 12\) won for every \(\$ 1\) bet is made when both numbers are among the 20 , (a) What would be the fair payoff in this case? Let \(P_{n, k}\) denote the probability that exactly \(k\) of the \(n\) numbers chosen by the player are among the 20 selected by the house. (b) Compute \(P_{n, k}\). (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet the casino pays off as shown in the following table. Compute the expected payoff:

Let \(X\) be a Poisson random variable with parameter \(\lambda\). Show that \(P\\{X=i\\}\) increases monotonically and then decreases monotonically as \(i\) increases, reaching its maximum when \(i\) is the largest integer not exceeding \(\lambda\).

An urn initially contains one red and one blue ball. At each stage a ball is randomly chosen and then replaced along with another of the same color. Let \(X\) denote the selection number of the first chosen ball that is blue. For instance, if the first selection is red and the second blue, then \(X\) is equal to 2 . (a) Find \(P\\{X>i\\}, i \geq 1\). (b) Show that with probability 1 , a blue ball is eventually chosen. (That is, show that \(P\\{X<\infty\\}=1 .\) ) (c) Find \(E[X]\).
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