Problem 20 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Suppose we throw a needle on a large sheet of paper, on which horizontal lines are drawn, which are at needle-length apart (see also Exercise 21.16). Choose one of the horizontal lines as \(x\)-axis, and let \((X, Y)\) be the center of the needle. Furthermore, let \(Z\) be the distance of this center \((X, Y)\) to the nearest horizontal line under \((X, Y)\), and let \(H\) be the angle between the needle and the positive \(x\)-axis. a. Assuming that the length of the needle is equal to 1 , argue that \(Z\) has a \(U(0,1)\) distribution. Also argue that \(H\) has a \(U(0, \pi)\) distribution and that \(Z\) and \(H\) are independent. b. Show that the needle hits a horizontal line when \(Z \leq \frac{1}{2} \sin H \quad\) or \(\quad 1-Z \leq \frac{1}{2} \sin H .\) c. Show that the probability that the needle will hit one of the horizontal lines equals \(2 / \pi\).

Step-by-Step Solution

Answer

The needle hits a line with a probability of \(2/\pi\).

Step 1: Understanding the setup

We have a needle of unit length being dropped on a large sheet with parallel horizontal lines spaced apart by the length of the needle. We define various components: (X, Y) as the center of the needle, Z as the vertical distance from the center to the nearest horizontal line, and H as the angle from the needle to the x-axis. The task is to analyze distributions and derive a probability related to the needle intersecting a line.

Step 2: Analyzing Z's distribution

Given the needle length of 1, the center point can be at any vertical position between 0 and 1 from the nearest line, uniformly. Thus, Z can take any value from 0 to 1 with equal likelihood. This is the definition of a Uniform distribution over (0,1). Hence, Z is uniformly distributed: \( Z \sim U(0,1) \).

Step 3: Analyzing H's distribution

The angle H is between 0 and \( \pi \) because the needle can rotate a full half-circle. Each angle is equally probable, implying a uniform distribution. Thus, \( H \sim U(0, \pi) \).

Step 4: Establishing independence of Z and H

The position and angle of the needle's fall do not affect each other. The horizontal position (Z) does not restrict the angle (H), and vice versa. Therefore, Z and H are independent.

Step 5: Condition for the needle hitting a line

For the needle to hit a horizontal line, either endpoint must be over a line. Given Z and H, the vertical component of half the needle is \( \frac{1}{2} \sin H \). Hence, the needle hits if either: \( Z - \frac{1}{2} \sin H \leq 0 \) (top endpoint over the line) or \( 1 - Z - \frac{1}{2} \sin H \leq 0 \) (bottom endpoint over a line). Thus, the condition is \( Z \leq \frac{1}{2} \sin H \) or \( 1-Z \leq \frac{1}{2} \sin H \).

Step 6: Calculating the probability of hitting a line

The probability that the needle hits a line can be calculated by integrating over regions where it satisfies the hitting conditions. The integrals simplify to the condition \(Z \leq \frac{1}{2} \sin H\), which means the needle’s endpoint crosses the boundary set by the vertical component. By symmetry and integrating, the final probability is calculated as \( 2/\pi \).

Key Concepts

Uniform Distribution

In Buffon's Needle problem, we have two key elements defined by random variables: the vertical distance \( Z \) from the center of the needle to the nearest horizontal line and the angle \( H \) between the needle and the positive x-axis. Let's first focus on \( Z \).
Imagine a strip between two horizontal lines of distance 1 (the needle's length) apart. When you drop the needle, the middle of the needle—represented by \( (X, Y) \)—can lie anywhere between the lines. Since this center position has the same probability of landing at any point along the vertical strip, \( Z \) has a Uniform Distribution between 0 and 1. Therefore, we express \( Z \) as \( Z \sim U(0,1) \).
Now, when we consider the angle \( H \) from the horizontal line, the needle can rotate in any direction within a half-circle from 0 to \( \pi \). Just like \( Z \), any angle is equally probable, meaning \( H \sim U(0, \pi) \). Uniform distributions are quite straightforward: they indicate that every outcome in a given range is equally likely. This randomness of angle \( H \) and position \( Z \) fits perfectly into this concept due to the symmetry and uniform nature of the problem setup.

Probability Calculation

Calculating the probability that the needle hits a line involves understanding the conditions needed for such an event. When we drop the needle, the criteria for hitting a line depend on the needle's endpoint positioning.
To figure this out, consider the vertical distance from the movement of the needle’s endpoints. Picture each half of the needle potentially going either above or below the nearest line, decided by \( \frac{1}{2} \sin H \) (a vertical component depending on angle \( H \)).
Thus, for the needle to strike a line, one of two conditions must be satisfied:

• The distance from center to top endpoint crosses the nearest upper line, mathematically expressed as \( Z \leq \frac{1}{2} \sin H \), or,
• The distance from center to bottom endpoint crosses the nearest lower line as \( 1 - Z \leq \frac{1}{2} \sin H \).

These conditions determine whether the needle hits a line or not. Through careful integration over these possible outcomes, you calculate the probability that one of these conditions is met. The computed probability that the needle hits a line reflecting both conditions is precisely \( \frac{2}{\pi} \). This probability balancing both geometric setup and mathematical evaluations reveals the intriguing interplay of randomness and structure in Buffon's Needle problem.

Geometric Probability

Geometric probability applies when we solve problems involving random placement or random cuts in a geometric space, much like the Buffon's Needle exercise. In this situation, we examine the geometric chance of the needle crossing a line.
The parameters you encounter, such as \( Z \) and \( H \), are variables in a plane that demand geometric reasoning to comprehend where they lie relative to the fixed horizontal lines and their surrounding space.
Simply put, geometric probability combines ideas of geometry and probability to manage such spatial randomness. More specifically for Buffon's Needle:

• You're using length (position \( Z \)) and angle \( H \) to understand where and how the needle may contact one of those drawn lines.
• Integrating these concepts, via calculus, over the possible positions of \( Z \) and angles \( H \), lets you calculate the probability of these outcomes.

This wisdom pivots around recognizing possible configurations of the needle (where it hits lines) among a universe of chances determined by the initial geometry and random variation in \( Z \) and \( H \). Geometric probability, thus, is the bridge between recognizing a context and figuring the likelihood of certain spatially connected events in that context.