91Ó°ÊÓ

In Exercise 2.2 .12 you proved the following: If you take a stick of unit length and break it into three pieces, choosing the breaks at random (i.e., choosing two real numbers independently and uniformly from [0,1]\()\), then the probability that the three pieces form a triangle is \(1 / 4\). Consider now a similar experiment: First break the stick at random, then break the longer piece at random. Show that the two experiments are actually quite different, as follows: (a) Write a program which simulates both cases for a run of 1000 trials, prints out the proportion of successes for each run, and repeats this process ten times. (Call a trial a success if the three pieces do form a triangle.) Have your program pick \((x, y)\) at random in the unit square, and in each case use \(x\) and \(y\) to find the two breaks. For each experiment, have it plot \((x, y)\) if \((x, y)\) gives a success. (b) Show that in the second experiment the theoretical probability of success is actually \(2 \log 2-1\)

Step by step solution

01

Understanding the Problem

We have two different stick-breaking experiments. In the first, we choose two break points randomly along the stick, while in the second, we break the stick once randomly and then break the longer piece randomly again. We need to compare both experiments by simulating them and computing the probability of forming a triangle with the resulting pieces, then show the theoretical probability for the second experiment.

02

Simulating Experiment 1

In the first experiment, we select two break points, \(x\) and \(y\), uniformly from the interval \[0,1\]. The three pieces have lengths \(x, y-x, 1-y\). They form a triangle if each piece's length is less than the sum of the other two (triangle inequality). The probability found previously is \frac{1}{4}\.

03

Simulating Experiment 2

In this experiment, we randomly select a break point \(x\). The stick is broken into two pieces of lengths \(x\) and \(1-x\). We then break the longer piece randomly by choosing another break point \(y\) along its length. Depending on which piece is longer, the second break alters the configuration.

04

Calculating Probability for Experiment 2

For the second experiment, we first break the stick at \(x\) and then break the longer remaining piece, altering how \(x\) and \(y\) influence the triangle formation. The probability that the resulting three pieces form a triangle is theoretically \(2 \log 2 - 1\), due to the different conditions imposed by breaking the longer piece.

05

Implementing the Simulation

Develop a program to simulate both experiments over 1000 trials, repeating this process ten times. For each experiment, when the sums of any two of the three resulting stick pieces exceed the third, the trial is successful. Compare and print the proportions of successful trials against the theoretical values.

06

Plotting Results

Use the results of the simulation to plot the points \(x, y\) when a successful triangle is formed, distinguishing between the two experiments visually while using statistics to identify areas of success for \(x, y\).

07

Analyzing Results

Analyze the results of the simulations, noting how the probability differs in the two cases and reflecting on how the theoretical probability of \(2 \log 2 - 1\) compares for experiment 2 versus the known \(\frac{1}{4}\) for experiment 1.

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Key Concepts

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

Triangle Inequality

At the heart of understanding whether three sticks can form a triangle is the Triangle Inequality Theorem. This theorem states that, for three line segments to form a triangle, the sum of the lengths of any two segments must be greater than the length of the third. In mathematical terms, if you have segments of lengths \( a \), \( b \), and \( c \), they can form a triangle if and only if:

• \( a + b > c \)
• \( a + c > b \)
• \( b + c > a \)

These conditions ensure that the sides can "reach around" to meet and form a closed shape. When breaking a stick into three pieces, applying the triangle inequality helps determine whether those pieces can form a triangle.
In our scenario, with pieces from breaks \( x \) and \( y \) along a unit-length stick, the pieces would be \( x \), \( y-x \), and \( 1-y \). Only when each of these pieces satisfies all triangle inequality conditions can they form a triangle.

Stick-Breaking Problem

This fascinating problem involves breaking a stick of unit length into segments and asks about the conditions under which those segments form a triangle. The original version of the problem suggests breaking the stick at two random points \( x \) and \( y \), chosen uniformly from \([0, 1]\). In this setup, you calculate probabilities using the lengths \( x \), \( y-x \), and \( 1-y \). Here, the probability that these form a triangle is known to be \( \frac{1}{4} \). In a more complex variation, you first break the stick at a random point \( x \), then choose a second break randomly along the longer piece. This changes the stick lengths and their relations, given the altering condition of breaking the longer piece rather than simply at two points \( x \) and \( y \). The differing route of breaking in these two experiments leads to distinct probabilistic outcomes. This reflects the nuanced changes in conditions which impact the likelihood of forming a triangle.

Uniform Distribution

A uniform distribution is a type of probability distribution where all outcomes are equally likely within a given range. For the stick-breaking problem, both break points are selected independently and uniformly from the interval \([0, 1]\). This means any point within the interval could be the break with equal probability.
In mathematical terms, for a uniform distribution along \([0, 1]\), every subinterval of length \( l \) has a probability equal to \( l \). When applying uniform distribution to the problem, it ensures that every possible way the stick can break is considered equally, without bias towards a particular section.
Such a distribution plays a critical role in simulating the experiments, as each scenario tested assumes an unbiased random selection. This unbiased approach is key to accurately estimating the probability of different outcomes, such as whether the breaks form a triangle.