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A die is rolled until the first time that a six turns up. We shall see that the probability that this occurs on the \(n\) th roll is \((5 / 6)^{n-1} \cdot(1 / 6)\). Using this fact, describe the appropriate infinite sample space and distribution function for the experiment of rolling a die until a six turns up for the first time. Verify that for your distribution function \(\sum_{\omega} m(\omega)=1\).

Short Answer

Expert verified
The sample space is infinite sequences ending with a six, with probabilities \( m(n) = \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \). The probabilities sum to 1, making it a valid distribution.

Step by step solution

01

Understand the Experiment

The exercise is about rolling a die until a six appears for the first time. We need to recognize that each roll of the die is an independent event, and each roll has a 1/6 probability of showing a six. The task is to describe the probability distribution for the roll number on which the first six appears.
02

Identify the Probability Formula

We are given that the probability of obtaining the first six on the \(n\)-th roll is given by \( \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \). This means that the first \(n-1\) rolls are not six (probability \(\frac{5}{6}\)), and the \(n\)-th roll is a six (probability \(\frac{1}{6}\)).
03

Define the Sample Space

The sample space consists of sequences of die rolls where we keep rolling until we roll a six for the first time. Formally, it can be denoted as \( { (x_1, x_2, \ldots, x_{n-1}, 6) \mid x_i eq 6 \text{ for } i = 1, 2, \ldots, n-1 } \). Each outcome corresponds to getting a six on a specific roll \(n\).
04

Define the Distribution Function

The distribution function is defined by the probability that the first six appears on the \(n\)-th roll. This probability is summed over all possible values of \(n\), specifically: \( m(n) = \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \). This is known as a geometric distribution.
05

Verify the Total Probability

For a valid probability distribution, the sum of all probabilities must equal 1. Calculate this sum: \( \sum_{n=1}^{\infty} \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} = \frac{1}{6} \sum_{n=0}^{\infty} \left( \frac{5}{6} \right)^n \). Recognize this as a geometric series with the first term 1 and ratio \(\frac{5}{6}\). The sum of an infinite geometric series is calculated as \( \frac{1}{1-r} \), where \(r = \frac{5}{6}\).
06

Calculate the Geometric Series Sum

Apply the formula for the sum of a geometric series: \( \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} \). Here, \( r = \frac{5}{6} \), so the sum is \( \frac{1}{1-\frac{5}{6}} = 6 \). Multiply by \( \frac{1}{6} \) to obtain the final sum: \( \frac{1}{6} \cdot 6 = 1 \). This confirms the probabilities sum to 1.

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Key Concepts

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

Probability Distribution
In the context of rolling a die, a probability distribution is a function that describes the likelihood of each outcome within the sample space. For our specific exercise, we are rolling a die until a six appears for the first time. This means the probability distribution focuses on which roll a six appears. The formula, \( \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \), tells us about the likelihood of getting the first six on the \(n\)-th roll. Each roll has a success chance of \(\frac{1}{6}\), while the previous rolls (\(n-1\)) were non-successes with a probability \(\frac{5}{6}\).
This setup is known as a geometric distribution, where the trials are repeated until the first success is recorded. Understanding this type of distribution is crucial for managing probabilistic events where the timing of the first success is essential.
Infinite Sample Space
Infinite sample space in probability refers to an experiment where theoretically, there is no limit to the number of possible outcomes. In our die-rolling scenario, we can keep rolling indefinitely until a six appears for the first time. Hence, the number of possible rolls before the appearance of a six creates an infinite sample space. Each outcome consists of a sequence of rolls ending in a six, something like (no sixes, ... , no six, six).
When dealing with infinite sample spaces, it's vital to define events properly and understand how probabilities are distributed over these endless possibilities. In our case, although each trial is finite, the potential number of trials is infinite, thus broadening the sample space significantly.
Geometric Series
The concept of a geometric series is central to solving the problem of summing probabilities in a geometric distribution. A geometric series is a series of terms where each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. In our exercise, the series you observe is derived from the probability distribution \( \sum_{n=1}^{\infty} \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \).
Recognize that this series is geometric with a common ratio, \(\frac{5}{6}\). The sum of an infinite geometric series can be found using the formula \( \frac{1}{1-r} \), where \(r\) is the common ratio, provided the magnitude of \(r\) is less than 1. For our scenario, applying this formula verifies that the sum of probabilities equals 1, thus confirming a valid probability distribution.
Independent Events
Understanding independent events is crucial when interpreting the die-rolling exercise. Independent events in probability theory are events where the outcome of one trial doesn't affect the outcome of another. In rolling a die, each roll is independent, meaning the result of one roll doesn't influence the next roll.
This independence is foundational for understanding why we can apply the formula \( \left( \frac{5}{6} \right)^{n-1} \cdot \frac{1}{6} \). Each time a die is rolled, there's a fresh start, with a \(\frac{1}{6}\) chance of landing a six, and a \(\frac{5}{6}\) chance of not. The independence ensures that the sequence (e.g., roll 4 brings the first six) is valid and conforms to the probability distribution described.

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Most popular questions from this chapter

(from vos Savant \(^{26}\) ) A reader of Marilyn vos Savant's column wrote in with the following question: My dad heard this story on the radio. At Duke University, two students had received A's in chemistry all semester. But on the night before the final exam, they were partying in another state and didn't get back to Duke until it was over. Their excuse to the professor was that they had a flat tire, and they asked if they could take a make-up test. The professor agreed, wrote out a test and sent the two to separate rooms to take it. The first question (on one side of the paper) was worth 5 points, and they answered it easily. Then they flipped the paper over and found the second question, worth 95 points: 'Which tire was it?' What was the probability that both students would say the same thing? My dad and I think it's 1 in 16\. Is that right?" (a) Is the answer \(1 / 16 ?\) (b) The following question was asked of a class of students. "I was driving to school today, and one of my tires went flat. Which tire do you think it was?" The responses were as follows: right front, \(58 \%\), left front, \(11 \%\), right rear, \(18 \%\), left rear, \(13 \%\). Suppose that this distribution holds in the general population, and assume that the two test-takers are randomly chosen from the general population. What is the probability that they will give the same answer to the second question?

Here is an attempt to get around the fact that we cannot choose a "random integer." (a) What, intuitively, is the probability that a "randomly chosen" positive integer is a multiple of \(3 ?\) (b) Let \(P_{3}(N)\) be the probability that an integer, chosen at random between 1 and \(N,\) is a multiple of 3 (since the sample space is finite, this is a legitimate probability). Show that the limit $$ P_{3}=\lim _{N \rightarrow \infty} P_{3}(N) $$ exists and equals \(1 / 3\). This formalizes the intuition in (a), and gives us a way to assign "probabilities" to certain events that are infinite subsets of the positive integers. (c) If \(A\) is any set of positive integers, let \(A(N)\) mean the number of elements of \(A\) which are less than or equal to \(N\). Then define the "probability" of \(A\) as $$ P(A)=\lim _{N \rightarrow \infty} A(N) / N $$ provided this limit exists. Show that this definition would assign probability 0 to any finite set and probability 1 to the set of all positive integers. Thus, the probability of the set of all integers is not the sum of the probabilities of the individual integers in this set. This means that the definition of probability given here is not a completely satisfactory definition. (d) Let \(A\) be the set of all positive integers with an odd number of digits. Show that \(P(A)\) does not exist. This shows that under the above definition of probability, not all sets have probabilities.

The Labouchere system for roulette is played as follows. Write down a list of numbers, usually \(1,2,3,4 .\) Bet the sum of the first and last, \(1+4=5,\) on red. If you win, delete the first and last numbers from your list. If you lose, add the amount that you last bet to the end of your list. Then use the new list and bet the sum of the first and last numbers (if there is only one number, bet that amount). Continue until your list becomes empty. Show that, if this happens, you win the sum, \(1+2+3+4=10\), of your original list. Simulate this system and see if you do always stop and, hence, always win. If so, why is this not a foolproof gambling system?

A student must choose one of the subjects, art, geology, or psychology, as an elective. She is equally likely to choose art or psychology and twice as likely to choose geology. What are the respective probabilities that she chooses art, geology, and psychology?

Another well-known gambling system is the martingale doubling system. Suppose that you are betting on red to turn up in roulette. Every time you win, bet 1 dollar next time. Every time you lose, double your previous bet. Suppose that you use this system until you have won at least 5 dollars or you have lost more than 100 dollars. Write a program to simulate this and play it a number of times and see how you do. In his book The Newcomes, W. M. Thackeray remarks "You have not played as yet? Do not do so; above all avoid a martingale if you do." 10 Was this good advice?

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