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Chapter 3: Problem 34

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2 .\) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\)

Short Answer

Expert verified
In summary, we derived the recursive formula for the expected number of throws needed for all dice to land on six as \(m_n = 6+\frac{m_{n-1}}{6}\). Using this formula, we calculated \(m_2 \approx 6.1667\), \(m_3 \approx 7.194\), \(m_4 \approx 10.0417\), and \(m_5 \approx 13.024\). Additionally, we found that the expected total number of dice rolled over all throws is given by \(E\left[\sum_{i=1}^N X_i\right] = n\cdot m_n\).

Step by step solution

01

Setting up the recursive formula for \(m_n\)

We can represent the given problem using probabilities. We know that the probability of rolling a six on a single dice is \(\frac{1}{6}\). Thus, the probability of not rolling a six on a single dice is \(1 - \frac{1}{6} = \frac{5}{6}\). Now, let's set up a recursive formula for the expected number of throws, \(m_n\). For \(n\) dice, we can write the expectation as follows: \(m_n = 1 + \frac{1}{6}m_{n-1} + \frac{5}{6}m_n\) This formula represents the average number of throws needed for \(n\) dice, which includes the current throw (1) as well as the expected number of throws for the remaining dice. The chances of rolling a six in the current throw are \(\frac{1}{6}\), so we consider the expectation for \(n-1\) dice, while the chance of not rolling a six is \(\frac{5}{6}\), so we consider the expectation for \(n\) dice.
02

Computing \(m_i\) for \(i=2,3,4\) and approximating \(m_5\)

We have the recursive formula now, as follows: \(m_n = 1 + \frac{1}{6}m_{n-1} + \frac{5}{6}m_n\) We can rearrange this equation and express \(m_n\) concerning \(m_{n-1}\): \(m_n(\frac{1}{6}) = 1+ \frac{1}{6}m_{n-1}\) \(m_n = 6+\frac{m_{n-1}}{6}\) Now we can use this equation to compute \(m_i\) for \(i=2,3,4\) and approximate \(m_5\): For \(n=1\) (base case), \(m_1=1\), as there is only one throw needed in this case. For \(n=2\), \(m_2= 6 + \frac{m_1}{6} = 6 + \frac{1}{6} = 6 + \frac{1}{6}\) For \(n=3\), \(m_3 = 6+\frac{m_2}{6} = 6 + \frac{1}{6}(7)=6+\frac{7}{6}\) For \(n=4\), \(m_4 = 6+\frac{m_3}{6} = 6 +\frac{1}{6}(6+\frac{7}{6})= 6+\frac{6}{6}+\frac{7}{36}\) For \(n=5\), \(m_5 \approx 6+\frac{m_4}{6} = 6+\frac{1}{6}(6+\frac{6}{6}+\frac{7}{36})\approx 13.024\) Here is the list of the computed expectations: \(m_2 \approx 6.1667\) \(m_3 \approx 7.194\) \(m_4 \approx 10.0417\) \(m_5 \approx 13.024\)
03

Finding the Expectation of the total number of dice rolled

Let \(X_i\) denote the number of dice rolled on the \(i\)th throw. We need to find the expected total number of dice rolled over all throws, given by the sum: \(E\left[\sum_{i=1}^N X_i\right]\) Since expectation is linear, the sum of expectations is equal to the expectation of the sum: \(E\left[\sum_{i=1}^N X_i\right] = \sum_{i=1}^N{E[X_i]} = n\cdot E\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right]\) From the previous calculations, we know that the expected number of throws for \(n\) dice is given by: \(m_n = 6 + \frac{m_{n-1}}{6}\) Since the expectation of the number of dice thrown is equal to the number of dice times the expected number of throws, we have: \(E\left[\sum_{i=1}^N X_i\right] = n\cdot m_n\) So, the expected total number of dice rolled over all throws is the number of dice times the expected number of throws for \(n\) dice.

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Key Concepts

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

Expected Value
The concept of expected value is central to the theory of probability and statistics, as it provides a measure of the center or average outcome we can expect from a random process. In simpler terms, it's what you would anticipate being the average result if you could repeat an experiment over and over again indefinitely.

Calculating the expected value involves multiplying each possible outcome by its corresponding probability and summing all of these products together. For a discrete random variable, the expected value, often denoted by E(X), is given by the sum:
\[ E(X) = \text{sum} (x_i \times p(x_i)) \text{where} x_i \text{are the outcomes and} p(x_i) \text{are their probabilities.} \]
Returning to our exercise, the expected value helps us understand the long-term behavior of rolling dice and setting them aside when they land on six. The term 'recursive' in the exercise indicates that the calculation of the expected value at a given stage depends on the expected values at previous stages, hence the 'recursive formula'.
Probability Models
A probability model is a mathematical representation that encapsulates all possible outcomes of a random phenomenon and their associated probabilities. It serves as a structured framework for predicting the probabilities of various results.

In our textbook problem, the probability model is used to describe the process of throwing a set of dice, observing which come up as sixes, and then repeating this process with the remaining dice. For each throw, the probability of any given die landing on a six is constant (\(\frac{1}{6}\)), but the number of dice changes depending on the previous throw, creating a dynamic and iterative structure reflective of a real-world situation.

By applying a probability model alongside recursive formulas, we can predict outcomes more efficiently by breaking down complex processes into simpler, interrelated steps. It is this modeling that enables us to derive expectations, such as the average number of throws needed (\(m_n\)) for all dice to show six.
Dice Probability
The term dice probability refers to the chance of rolling a specific number on a die, which is crucial for games of chance and for our problem as well. A standard six-sided die has an equal probability of landing on any of its faces, making the probability of any single outcome 1/6.

However, when multiple dice are involved, the calculations may become more complex. Understanding dice probability is not only key to games but is also used in various statistical models and simulations. As seen in our exercise, the use of dice probability can be extended to scenarios where dice are rolled multiple times, with certain outcomes influencing subsequent events.

The problem provides an excellent opportunity to apply real-world probability concepts to solve a recursive problem: calculating the average number of throws needed for a group of dice to all land on six. Thanks to the interplay of probability and recursive formulas, we're able to deduce this information logically and systematically.

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Most popular questions from this chapter

Two players alternate flipping a coin that comes up heads with probability \(p\). The first one to obtain a head is declared the winner. We are interested in the probability that the first player to flip is the winner. Before determining this probability, which we will call \(f(p)\), answer the following questions. (a) Do you think that \(f(p)\) is a monotone function of \(p ?\) If so, is it increasing or decreasing? (b) What do you think is the value of \(\lim _{p \rightarrow 1} f(p)\) ? (c) What do you think is the value of \(\lim _{p \rightarrow 0} f(p)\) ? (d) Find \(f(p)\).

Suppose there are \(n\) types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of the \(n\) types. Suppose one continues collecting until a complete set of at least one of each type is obtained. (a) Find the probability that there is exactly one type \(i\) coupon in the final collection. Hint: Condition on \(T\), the number of types that are collected before the first type \(i\) appears. (b) Find the expected number of types that appear exactly once in the final collection.

In a knockout tennis tournament of \(2^{n}\) contestants, the players are paired and play a match. The losers depart, the remaining \(2^{n-1}\) players are paired, and they play a match. This continues for \(n\) rounds, after which a single player remains unbeaten and is declared the winner. Suppose that the contestants are numbered 1 through \(2^{n}\), and that whenever two players contest a match, the lower numbered one wins with probability \(p\). Also suppose that the pairings of the remaining players are always done at random so that all possible pairings for that round are equally likely. (a) What is the probability that player 1 wins the tournament? (b) What is the probability that player 2 wins the tournament? Hint: Imagine that the random pairings are done in advance of the tournament. That is, the first-round pairings are randomly determined; the \(2^{n-1}\) firstround pairs are then themselves randomly paired, with the winners of each pair to play in round 2 ; these \(2^{n-2}\) groupings (of four players each) are then randomly paired, with the winners of each grouping to play in round 3, and so on. Say that players \(i\) and \(j\) are scheduled to meet in round \(k\) if, provided they both win their first \(k-1\) matches, they will meet in round \(k\). Now condition on the round in which players 1 and 2 are scheduled to meet.

A coin that comes up heads with probability \(p\) is flipped \(n\) consecutive times. What is the probability that starting with the first flip there are always more heads than tails that have appeared?

Two players take turns shooting at a target, with each shot by player \(i\) hitting the target with probability \(p_{i}, i=1,2\). Shooting ends when two consecutive shots hit the target. Let \(\mu_{i}\) denote the mean number of shots taken when player \(i\) shoots first, \(i=1,2\). (a) Find \(\mu_{1}\) and \(\mu_{2}\). (b) Let \(h_{i}\) denote the mean number of times that the target is hit when player \(i\) shoots first, \(i=1,2\). Find \(h_{1}\) and \(h_{2}\).
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