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If a die is to be rolled until all sides have appeared at least once, find the expected number of times that outcome 1 appears.

Short Answer

Expert verified
The expected number of times that outcome 1 appears when rolling a die until all sides have appeared at least once is approximately \(2.45\) times.

Step by step solution

01

Calculate the probability of side 1 appearing in one roll

First, let's find the probability of side 1 appearing in a single roll. Since there are 6 sides, the probability of rolling side 1 in a single roll is: \[P(1) = \frac{1}{6}\] Remember this value, as it will be used later in the calculations.
02

Calculate the expected number of rolls to get all 6 sides

Let E be the expected number of rolls needed to get all 6 sides. When the first roll is made, there are 6 possible sides that can appear, which we denote as S. After rolling S for the first time, there are 5 remaining sides yet to appear, with the probability of one of the remaining sides appearing being 5/6. Similarly, after rolling one of the remaining sides (5S) for the first time, there will be 4 sides left to appear, with the probability of one of the remaining sides appearing being 4/6. This pattern continues until all 6 sides appear. Therefore, the expected number of rolls required to get all 6 sides can be expressed as: \[E = 1 + \frac{1}{\frac{5}{6}} + \frac{1}{\frac{4}{6}} + \frac{1}{\frac{3}{6}} + \frac{1}{\frac{2}{6}} + \frac{1}{\frac{1}{6}}\]
03

Calculate E

Now, let's calculate E using the formula obtained in the previous step: \[E = 1 + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1}\] \[E = 1 + \frac{6}{5} + \frac{3}{2} + 2 + 3 + 6\] \[E = 1 + 1.2 + 1.5 + 2 + 3 + 6\] \[E = 14.7\] This means that on average, it takes approximately 14.7 rolls to have all 6 sides appear at least once.
04

Calculate the expected number of times side 1 appears

We will now use the probability calculated in step 1 and the expected number of rolls calculated in step 3 to find the expected number of times that side 1 appears. We multiply the probability of rolling a 1 by the expected number of rolls: \[Expected\ number\ of\ 1's = P(1) \times E\] \[Expected\ number\ of\ 1's = \frac{1}{6} \times 14.7\] \[Expected\ number\ of\ 1's = 2.45\] So, in approximately 14.7 rolls needed to get all 6 sides, side 1 is expected to appear about 2.45 times.

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Key Concepts

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

Probability Theory
At the heart of many statistical calculations, we find probability theory. It is a branch of mathematics concerned with analyzing random phenomena and quantifying the chances of various outcomes. In our daily lives, we use probability theory when we talk about 'chances', 'likelihoods', or 'odds' without realizing it. In the case of rolling a die, which is a classic probability problem, there are a fixed number of outcomes. Since a standard die has six faces, the probability of any particular face appearing on a single roll is \( \frac{1}{6} \). Probability theory is what allows us to make predictions over the long term, like estimating how many rolls it will take for all sides to appear at least once.

To apply probability theory effectively, we must first understand the concepts of experiments, outcomes, and events. An experiment is any procedure that can be infinitely repeated and has a well-defined set of outcomes, like rolling a die. An individual result of an experiment is called an outcome, while an event is a set of outcomes that we assign a probability to. For instance, the event 'rolling a 1' on a die has the probability of \( \frac{1}{6} \) because there is one outcome (rolling a 1) out of six possible outcomes.
Discrete Random Variables
Discrete random variables are a fundamental concept in probability theory. They can take on a countable number of distinct values, like flipping a coin (heads or tails), rolling a die (1 through 6), or counting the number of times an event occurs. Each possible value the variable can assume is associated with a probability. For example, when rolling a fair six-sided die, the probability of rolling any number between 1 and 6 is \( \frac{1}{6} \).

The term 'random' indicates that the outcome cannot be predicted with certainty before we actually perform the roll. However, we can describe the overall behavior of this variable using a probability distribution, which assigns probabilities to each of the discrete outcomes. When we talk about the 'expected number of rolls' needed for an event to occur, we are talking about the average number of rolls we would expect if we repeated the rolling process many times. Mathematically, this expected value is found by taking the summation of all possible values multiplied by their respective probabilities.
Geometric Distribution
The geometric distribution is a probability distribution that models the number of trials needed for the first success in a sequence of independent and identical Bernoulli trials (random trials with only two possible outcomes). Here, 'success' defines one specific outcome of interest. It is discrete because the number of trials until success is countable (1st trial, 2nd trial, etc.).

In our die-rolling scenario, if we are looking to roll a '1', the question of 'how many rolls until we get a 1' can be modeled using a geometric distribution with the probability \( p = \frac{1}{6} \). The expected value or mean of a geometrically distributed random variable is \( \frac{1}{p} \), which is the average number of trials it takes to get the first success.

It is important to note that the geometric distribution is memoryless, meaning the probability of success on a given trial does not depend on the number of failed trials preceding it. Each roll of the die is independent of the previous rolls. However, in the exercise, we are looking at a more complex situation where the aim is to roll all different sides, which doesn't conform to a simple geometric distribution. Despite that, understanding the geometric distribution gives us valuable insights into problems involving sequences of independent trials, especially when we are focused on a single outcome appearing for the first time.

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Most popular questions from this chapter

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$

The number of people that enter an elevator on the ground floor is a Poisson random variable with mean 10. If there are \(N\) floors above the ground floor and if each person is equally likely to get off at any one of these \(N\) floors, independently of where the others get off, compute the expected number of stops that the elevator will make before discharging all of its passengers.

Prove that if \(E[Y \mid X=x]=E[Y]\) for all \(x\), then \(X\) and \(Y\) are uncorrelated, and give a counterexample to show that the converse is not true. HINT: Prove and use the fact that \(E[X Y]=E[X E[Y \mid X]]\).

Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having a common mean \(\mu\). Suppose also that \(\operatorname{Var}\left(X_{1}\right)=\sigma_{1}^{2}\) and \(\operatorname{Var}\left(X_{2}\right)=\sigma_{2}^{2} .\) The value of \(\mu\) is unknown and it is proposed to estimate \(\mu\) by a weighted average of \(X_{1}\) and \(X_{2}\). That is, \(\lambda X_{1}+(1-\lambda) X_{2}\) will be used as an estimate of \(\mu\), for some appropriate value of \(\lambda\). Which value of \(\lambda\) yields the estimate having the lowest possible variance? Explain why it is desirable to use this value of \(\lambda\).

A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, then she wins twice, and if tails, then one-half of the value that appears on the die. Determine her expected winnings.

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