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Chapter 10: Problem 1

a) Write a program that generates and prints out two random numbers between 1 and 6 , to simulate the rolling of two dice. b) Modify your program to simulate the rolling of two dice a million times and count the number of times you get a double six. Divide by a million to get the fraction of times you get a double six. You should get something close to, though probably not exactly equal to, \(\frac{1}{36}\).

Short Answer

Expert verified
The fraction of double sixes is approximately \frac{1}{36}.

Step by step solution

01

- Set Up the Random Number Generator

First, import the random module which will be used to generate random numbers. In Python, you can do this using the command: import random.
02

- Generate Two Random Numbers

Generate two random integers between 1 and 6 using random.randint(1, 6). Store these numbers in variables, for example, die1 and die2.
03

- Print the Random Numbers

Print the values of die1 and die2 to simulate the roll of two dice. For example: print(die1, die2).
04

- Initialize a Counter for Double Sixes

Create a counter variable double_sixes and initialize it to 0. This will keep track of the number of times double sixes occur.
05

- Simulate Rolling Dice a Million Times

Use a for loop to simulate rolling the dice one million times. Inside the loop, generate two random numbers and check if both are six. If they are, increment the double_sixes counter.
06

- Calculate the Fraction of Double Sixes

After the loop, calculate the fraction of double sixes by dividing double_sixes by 1,000,000. Store this result in a new variable, fraction_double_sixes.
07

- Print the Result

Print the value of fraction_double_sixes. This will give you the fraction of times double sixes occurred in a million rolls. For example: print(fraction_double_sixes).

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Key Concepts

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

Python Random Number Generation
Python's `random` module provides various functions for generating random numbers. To simulate rolling dice, we can use the function `random.randint(a, b)`, which generates random integers between `a` and `b`, inclusive.
  • First, import the random module using `import random`.
  • Then, use `random.randint(1, 6)` to generate a random number between 1 and 6, simulating a dice roll.
Here's an example snippet for generating and printing two dice rolls:
```python import random die1 = random.randint(1, 6) die2 = random.randint(1, 6) print(die1, die2) ```
The `import random` statement makes the module's functions available, and `random.randint(1, 6)` simulates rolling a dice. This is fundamental for any simulation requiring random numbers in Python.
Monte Carlo Simulation
Monte Carlo simulations rely on repeated random sampling to estimate statistical properties. Let's apply this to simulate dice rolls.
  • Start by initializing a counter, e.g., `double_sixes`, to zero.
  • Next, run a loop that rolls two dice a million times.
  • Inside the loop, check if the outcome is double sixes. If so, increment the counter.

Here's how you would set this up in Python:
```python double_sixes = 0 for _ in range(1000000): die1 = random.randint(1, 6) die2 = random.randint(1, 6) if die1 == 6 and die2 == 6: double_sixes += 1 ```
Monte Carlo methods allow us to model complex systems and predict probabilities by running simulations many times and analyzing the results. This technique is powerful in fields such as finance, engineering, and science.
Probability Calculation
After running the simulation, the next step is to calculate the fraction of times double sixes occur.
  • To find this fraction, divide the number of double sixes by the total number of trials (one million in this case).
  • Store the result in a variable, say `fraction_double_sixes`.
  • Print the fraction to obtain the probability.

Here's how this would look in code:
```python fraction_double_sixes = double_sixes / 1000000 print(fraction_double_sixes) ```
The result should be close to \(\frac{1}{36}\), as the probability of rolling a double six is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
This fraction represents the empirical probability derived from our simulation and gets closer to the true probability with more trials. Monte Carlo simulations often converge to theoretical probabilities as the number of trials increases.

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Most popular questions from this chapter

Calculate a value for the integral $$ I=\int_{0}^{1} \frac{x^{-1 / 2}}{\mathrm{e}^{x}+1} \mathrm{~d} x, $$ using the importance sampling formula, Eq. (10.42), with \(w(x)=x^{-1 / 2}\), as follows. a) Show that the probability distribution \(p(x)\) from which the sample points should be drawn is given by $$ p(x)=\frac{1}{2 \sqrt{x}} $$ and derive a transformation formula for generating random numbers between zero and one from this distribution. b) Using your formula, sample \(N=1000000\) random points and hence evaluate the integral. You should get a value around \(0.84\).

Brownian motion is the motion of a particle, such as a smoke or dust particle, in a gas, as it is buffeted by random collisions with gas molecules. Make a simple computer simulation of such a particle in two dimensions as follows. The particle is confined to a square grid or lattice \(L \times L\) squares on a side, so that its position can be represented by two integers \(i, j=0 \ldots L-1\). It starts in the middle of the grid. On each step of the simulation, choose a random direction-up, down, left, or right-and move the particle one step in that direction. This process is called a random walk. The particle is not allowed to move outside the limits of the lattice-if it tries to do so, choose a new random direction to move in. Write a program to perform a million steps of this process on a lattice with \(L=101\) and make an animation on the screen of the position of the particle. (We choose an odd length for the side of the square so that there is one lattice site exactly in the center.) Note: The visual package doesn't always work well with the randon package, but if you import functions from visual first, then from random, you should avoid problems.

A random point on the surface of the Earth: Suppose you wish to choose a random point on the surface of the Earth. That is, you want to choose a value of the latitude and longitude such that every point on the planet is equally likely to be chosen. In a physics context, this is equivalent to choosing a random vector direction in threedimensional space (something that one has to do quite often in physics calculations). Recall that in spherical coordinates \(\theta, \phi\) (where \(\theta\) is the angle from the north pole and \(\phi\) is the azimuthal or longitudinal angle) the element of solid angle is \(\sin \theta \mathrm{d} \theta \mathrm{d} \phi\), and the total solid angle in a whole sphere is \(4 \pi\). Hence the probability of our point falling in a particular element is $$ p(\theta, \phi) \mathrm{d} \theta \mathrm{d} \phi=\frac{\sin \theta \mathrm{d} \theta \mathrm{d} \phi}{4 \pi} $$ We can break this up into its \(\theta\) part and its \(\phi\) part thus: $$ p(\theta, \phi) \mathrm{d} \theta \mathrm{d} \phi=\frac{\sin \theta \mathrm{d} \theta}{2} \times \frac{\mathrm{d} \phi}{2 \pi}=p(\theta) \mathrm{d} \theta \times p(\phi) \mathrm{d} \phi . $$ a) What are the ranges of the variables \(\theta\) and \(\phi\) ? Verify that the two distributions \(p(\theta)\) and \(p(\phi)\) are correctly normalized-they integrate to 1 over the appropriate ranges. b) Find formulas for generating angles \(\theta\) and \(\phi\) drawn from the distributions \(p(\theta)\) and \(p(\phi)\). (The \(\phi\) one is trivial, but the \(\theta\) one is not.) c) Write a program that generates a random \(\theta\) and \(\phi\) using the formulas you worked out. (Hint: In Python the function acos in the math package returns the arc cosine in radians of a given number.) d) Modify your program to generate 500 such random points, convert the angles to \(x, y, z\) coordinates assuming the radius of the globes is 1 , and then visualize the points in three-dimensional space using the visual package with small spheres (of radius, say, \(0.02\) ). You should end up with a three- dimensional globe spelled out on the screen in random points.
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