91Ó°ÊÓ

(Banach's Matchbox \(^{16}\) ) A man carries in each of his two front pockets a box of matches originally containing \(N\) matches. Whenever he needs a match, he chooses a pocket at random and removes one from that box. One day he reaches into a pocket and finds the box empty. (a) Let \(p_{r}\) denote the probability that the other pocket contains \(r\) matches. Define a sequence of counter random variables as follows: Let \(X_{i}=1\) if the \(i\) th draw is from the left pocket, and 0 if it is from the right pocket. Interpret \(p_{r}\) in terms of \(S_{n}=X_{1}+X_{2}+\cdots+X_{n} .\) Find a binomial expression for \(p_{r}\) (b) Write a computer program to compute the \(p_{r},\) as well as the probability that the other pocket contains at least \(r\) matches, for \(N=100\) and \(r\) from 0 to 50 . (c) Show that \((N-r) p_{r}=(1 / 2)(2 N+1) p_{r+1}-(1 / 2)(r+1) p_{r+1}\). (d) Evaluate \(\sum_{r} p_{r}\) (e) Use (c) and (d) to determine the expectation \(E\) of the distribution \(\left\\{p_{r}\right\\}\). (f) Use Stirling's formula to obtain an approximation for \(E .\) How many matches must each box contain to ensure a value of about 13 for the expectation \(E ?\) (Take \(\pi=22 / 7\).)

Step by step solution

01

Understand Problem Definition

A person has two matchboxes, each with originally \( N \) matches. When one box empties, we need to find the probability of how many matches \( r \) are left in the other box.

02

Define Random Variables

We define \( X_i \) for match drawn from left (1) or right pocket (0). \( S_n = X_1 + X_2 + \cdots + X_n \) denotes the number of times the left pocket is chosen.

03

Probability Expression

\( p_r \) is the probability that \( r \) matches remain in the non-empty pocket. Since one pocket is empty, \( n-r = N \), the number of draws from one pocket must sum to \( N \). This is described by a binomial distribution: \( p_r = \Pr(S_{2N} = N + r) \).

04

Computer Program

Write a program (using Python, for example) that computes \( p_r \) for \( N=100 \) for all \( r \) from 0 to 50 by simulating the random draws between the two pockets.

05

Derive the Expression (c)

Use recurrence relations and binomial coefficients to show \( (N-r) \cdot p_r = \frac{1}{2}(2N+1) \cdot p_{r+1} - \frac{1}{2} (r+1) \cdot p_{r+1} \).

06

Sum Evaluation

Recognize \( \sum_{r=0}^{N} p_r = 1 \) since it represents a total probability distribution over all possible remaining match scenarios.

07

Find the Expectation

Calculate the expected value \( E \) using \( E = \sum_{r=0}^{N} r \cdot p_r \). Use results from steps (c) and (d) to express \( E \).

08

Approximating with Stirling's Formula

Approximate values using Stirling's approximation for factorials, especially when \( N \) becomes large. Determine \( N \) such that the expectation \( E \) is approximately 13. Possibly results from the relationship involving binomial coefficients.

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Key Concepts

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

Random Variables

In probability theory, random variables are crucial as they represent outcomes of random phenomena. A random variable can take on different values, each associated with a probability.
In the context of Banach's Matchbox problem, we define random variables to model the random process of drawing a match.
Specifically, we define:

• \(X_i\) for each draw, where \(X_i = 1\) if a match is drawn from the left pocket, and \(X_i = 0\) if drawn from the right pocket.

By summing these values, we get \(S_n = X_1 + X_2 + \cdots + X_n\), the total number of draws from the left pocket after \(n\) draws. This setup helps in analyzing the distribution of matches between the two pockets.

Binomial Distribution

The binomial distribution is a foundational concept in probability, applicable when an experiment has two possible outcomes. In this problem, each draw of a match is a trial with two outcomes: left pocket or right pocket.
A binomial distribution models the number of successes (drawing from the left) in a fixed number of trials (total draws).
Properties of a Binomial Distribution:

• Trials are independent
• Each trial has the same probability of success

In this problem, we model the situation with \(p_r = \Pr(S_{2N} = N + r)\), where \(S_{2N}\) is the total number of draws from either pocket. This gives the probability of having \(r\) matches left in the non-empty pocket when the other is empty, using the binomial distribution.

Expectation

The expectation (or expected value) of a distribution provides an average outcome if the random experiment is repeated many times.
This concept is crucial to calculate how many matches are likely to be left in the non-empty pocket once the other is empty.
In Binach's Matchbox problem, the expectation \(E\) is calculated as: \[E = \sum_{r=0}^{N} r \cdot p_r\] To find \(E\), it is especially useful to use recurrence relations and evaluate the probabilistic sums.
This gives a detailed view of the likely number of matches remaining by considering each possible number \(r\) of remaining matches and their probabilities \(p_r\). It is worth noting that \(E\) depicts the central tendency of this distribution.

Stirling's Approximation

Stirling's approximation is a mathematical formula used to approximate factorials. It becomes particularly useful when dealing with very large numbers, like in factorial expressions of the binomial coefficient when \(N\) is large.
The approximation is: \[n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\] In the Banach's Matchbox problem, Stirling’s approximation helps simplify the computation of expectation values, especially when determining the necessary box size to achieve a specific expectation, such as \(E = 13\).
This simplification aids in dealing with large \(N\), making factorial computations feasible and less computationally intensive. Through this, one can deduce the potential size for \(N\) where the matches balance at a given expected average.