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Roulette-betting on red. A roulette wheel has 38 slots, numbered 0,00 , and 1 to 36 . The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) If you bet on "red," you win if the ball lands in a red slot. What is the probability of winning with a bet on red in a single play of roulette? (b) You decide to play roulette four times, each time betting on red. What is the distribution of \(X\), the number of times you win? (c) If you bet the same amount on each play and win on exactly two of the four plays, then you will "break even." What is the probability that you will break even? (d) If you win on fewer than two of the four plays, then you will lose money. What is the probability that you will lose money?

Short Answer

Expert verified
(a) \(\frac{9}{19}\); (b) \(X\sim B(4, \frac{9}{19})\); (c) \(\frac{48600}{130321}\); (d) \(\frac{46000}{130321}\).

Step by step solution

01

Understand the Event

We need to find the probability of winning when betting on "red." There are 18 red slots out of 38 total slots (including 0 and 00).
02

Calculate Single Win Probability

The probability of winning when betting on "red" is given by the ratio of red slots to total slots: \( P(\text{win on red}) = \frac{18}{38} = \frac{9}{19} \).
03

Define the Random Variable

Define \(X\) as the number of times you win in 4 plays. \(X\) follows a binomial distribution with parameters \(n = 4\) (number of trials) and \(p = \frac{9}{19}\) (probability of winning each trial).
04

Binomial Distribution Probability Function

The probability mass function for a binomial distribution is: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \(n = 4\) and \(p = \frac{9}{19}\).
05

Probability of Breaking Even

To break even, you need to win exactly 2 out of 4 plays. Calculate \( P(X = 2) = \binom{4}{2} \left(\frac{9}{19}\right)^2 \left(\frac{10}{19}\right)^2 \).
06

Calculate Specific Probability

Evaluate \( \binom{4}{2} = 6 \), \(\left(\frac{9}{19}\right)^2 = \frac{81}{361}\), and \(\left(\frac{10}{19}\right)^2 = \frac{100}{361}\). Thus, \( P(X = 2) = 6 \times \frac{81}{361} \times \frac{100}{361} = \frac{48600}{130321} \).
07

Probability of Losing Money

For losing money, you win fewer than 2 out of 4 plays: \( P(X < 2) = P(X = 0) + P(X = 1) \).
08

Calculate Probability for Losing

Calculate \( P(X = 0) = \binom{4}{0} \left(\frac{9}{19}\right)^0 \left(\frac{10}{19}\right)^4 = \left(\frac{10}{19}\right)^4 = \frac{10000}{130321} \). \( P(X = 1) = \binom{4}{1} \left(\frac{9}{19}\right)^1 \left(\frac{10}{19}\right)^3 = 4 \times \frac{9}{19} \times \left(\frac{10}{19}\right)^3 = \frac{36000}{130321} \).
09

Combine Probabilities for Losing

Add \( P(X = 0) \) and \( P(X = 1) \) to get \( P(X < 2) = \frac{10000}{130321} + \frac{36000}{130321} = \frac{46000}{130321} \).

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Key Concepts

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

Roulette Probability
When you play roulette, the goal is often to predict where the small ball will land on the spinning wheel. One common bet is on the color red. The roulette wheel has 38 slots, and among these, 18 are red. This means that when you bet on red, the probability of winning is determined by dividing the number of red slots by the total number of slots. In this case, it is \( \frac{18}{38} \), which simplifies to \( \frac{9}{19} \). Therefore, every time you bet on red, you have a bit less than a 50% chance of winning.

Understanding this probability is crucial. It demonstrates that although betting on red may appear straightforward, the house (casino) always has a slight edge because of the green slots (0 and 00), which are neither red nor black.
Probability Calculation
Probability calculation in roulette, especially when using a binomial distribution, involves more steps. If we consider multiple bets, like betting on red for four plays, we need to calculate the probability distribution of winning using these multiple bets. This is where the binomial distribution comes into play.

The binomial distribution is used to find the probability of having a certain number of successes (winning bets) in a specified number of trials (games played). Each trial is independent, and the probability of success remains the same for each play.

In our example, if you bet on red four times, you need to define a random variable \(X\) that represents winning a certain number of times. With the probability of winning each trial as \( \frac{9}{19} \), the distribution of \(X\) can be calculated using the formula:
  • \( P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \)
where \(n\) is the number of games (4 in this case), \(k\) is the number of times you win, \(p\) is the probability of winning one game, and \( (1-p) \) is the probability of losing. This framework helps to quantify your likelihood of winning a specified number of times, important for planning your betting strategy.
Gambling Odds
Understanding gambling odds, like those in roulette, helps assess the likelihood of various outcomes and potential winnings. Odds give a different perspective compared to probability. In roulette, the odds are heavily influenced by the probability of each outcome and the house edge.

Let's say you aim to "break even"—winning exactly as much money as you bet—by winning two out of four games. You can calculate the probability of this scenario with the binomial formula, yielding \( P(X = 2) \), which turns out to be \( \frac{48600}{130321} \) or about 0.373. Knowing this helps you understand the likelihood of recouping your investments.

On the other hand, if you are more concerned about the risk of losing your money, you can calculate the probability of winning fewer than two out of four games. By summing the probabilities \( P(X = 0) \) and \( P(X = 1) \), you find there's approximately a 0.353 chance you'll lose money.
  • For fewer than two wins: \( P(X < 2) = \frac{46000}{130321} \)
Understanding these odds not only frames your gambling strategy but also emphasizes the significant impact of chance in games of luck like roulette. The real takeaway is that while you can improve your understanding of the odds and the nature of the game, gambling outcomes remain unpredictable.

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Most popular questions from this chapter

College Admissions. A small liberal arts college in Ohio would like to have an entering class of 450 students next year. Past experience shows that about \(37 \%\) of the students admitted will decide to attend. The college is planning to admit 1175 students. Suppose that students make their decisaions independently and that the probability is \(0.37\) that a randomly chosen student will accept the offer of admission. (a) What are the mean and standard deviation of the number of students who accept the admissions offer from this college? (b) Use the Normal approximation to approximate probability that the college gets more students than it wants. Be sure to check that you can safely use the approximation. (c) Use software or an online binomial calculator to compute the exact probability that the college gets more students than it wants. How good is the approximation in part (b)? (d) To decrease the probability of getting more students than are wanted, does the college need to increase or decrease the number of students it admits? Using software or an online bìnomial calculator, what is the largest number of students that the college can admit if administrators want the exact probability of getting more students than they want to be no larger than \(5 \%\) ?

Larry reads that one out of four eggs contains salmonella bacteria. So he never uses more than three eggs in cooking. If eggs do or don't contain salmonella independent of each other, the number of contaminated eggs when Larry uses three chosen at random has the distribution (a) binomial with \(n=4\) and \(p=1 / 4\). (b) binomial with \(n=3\) and \(p=1 / 4\). (c) binomial with \(n=3\) and \(p=1 / 3\).

Multiple-choice tests. Here is a simple probability model for multiple-choince tests. Suppose each student has probability p of correctly answering a question chosen at random from a universe of possible questìons. (A strong student has a higher \(p\) than a weak student.) Answers to different questions are independent. (a) Stacey is a good student for whom \(p=0.75\). Use the Normal approximation to find the probability that Stacey scores between \(70 \%\) and \(80 \%\) on a 100 -question test. (b) If the test contains 250 questions, what is the probability that Stacey will score between \(70 \%\) and \(80 \%\) ? You see that Stacey's score on the longer test is more likely to be close to her "true score."

Using Benford's Law. According to Benford's law (Example 12.7, page 2B7) the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is \(0.477\). You examine 90 invoices from a vendor and find that 29 have first digits 1 or 2 . If Benford's law holds, the count of 1 s and \(2 s\) will have the binomial distribution with \(n=90\) and \(p=0.477\). Too few \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) suggests fraud. What is the approximate probability of 29 or fewer 15 and \(2 \mathrm{~s}\) if the invoices follow Benford's law? Do you suspect that the invoice amounts are not genuine?

Working Cell Numbers. When an opinion poll selects cell phone numbers at random to dial, the cell phone exchange is first selected and then random digits are added to form a complete telephone number (see Example 8.5). When using this procedure to generate random cell phone numbers, approximately \(55 \%\) of the cell numbers generated correspond to working numbers. You watch a pollster dial 15 cell numbers that have been selected in this manner. (a) What is the probability that exactly three calls reach working cell numbers? (b) What is the probability that at most three calls reach working cell numbers? (c) What is the probability that at least three calls reach working cell numbers? (d) What is the probability that fewer than three calls reach working cell numbers? (e) What is the probability that more than three calls reach working cell numbers?

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