Pages

Tuesday, August 20, 2019

Monopoly PR2: Throwing Doubles and Probability of Turn Outcomes.

What is the probability of throwing doubles?

Each turn of monopoly starts with a roll of two dice.   If the player rolls doubles (the same value on each dice) the person rolls again.   If the player rolls three doubles in a row the person goes directly to jail.   The player who rolls three doubles in a row does not get to buy a property or use a community chest or chance card.

The monopoly turn ends after the first roll of the dice if the dice roll is not doubles and after the third roll of the dice regardless of the dice roll outcomes.


Question one:  The Chance Square exists seven squares away from GO.  What is the probability that a player starts as Go and lands on Chance on the first turn of the game?

Analysis

The person can roll a seven on the first roll or roll a double go again and hope the sum of all rolls adds to seven.   One also can get to chance on three rolls.

As noted in a previous post 7 is the most common outcome from the sum of two dice rolls.   There are six combos of two dice rolls that add to 7 – (1,6), (6, 1), (2,5), (5,2), (3,4), and (4,3).   Each of these outcomes has probability 1/36.   The sum of the six mutually exclusive outcomes is 1/6 or 0.16667.

The person can also reach 7 by throwing a (1,1) or a (2,2) on the first turn and throwing and taking a second turn.   (Note (3,3) even though sum is less than 7 is not an option.  Explain why.)

To get to seven on two throws of the dice after throwing (1,1) one must throw (4,1) (1,4), (2,3) or (3,2).    The probability of this occurring is 4/36 or 0.11111.  The probability of getting to seven after two throws of the dice after throwing (1,1) on the first throw of the dice is probability of throwing (1,1) (1/36) multiplied by probability of getting dice rolls to sum to 5 (4/36).  This product is (1/(9x36)) or 1/324 or 0.00308642.

To get to seven after throwing (2,2) on two throws of the dice one must throw (1,2) or (2,1).   The probability of throwing (2.2) is 1/36.   The probability of throwing dice that sum to 3 is 1/18.   The product of both occurring is the product of the two or (1/(36x18)) or 1/648 or 0.00154321.

There is one other way to get to seven on one turn.  Throw (1,1) on the first turn, (1,1) on the second turn and dice that sum to 3 on the third turn.   The probability of this sequence happening is (1/36) x (1/36) x (1/18) 1/23328  or 0.0000428669.

The probability of getting to seven on the first turn is the sum of the probabilities of all the ways to get to seven, which is 0.171339163

More monopoly probability problems will follow.
.


Friday, August 2, 2019

The earth's actual orbit



Issue:  In a previous post I calculated the speed of the earth under the assumption that the earth’s orbit was circular.


The earth’s orbit is elliptical, not circular.    The link below provides a description of the actual orbit of the earth around the sun.  At its nearest point the earth is

Describes the orbit of the earth around the sun


According to this web site at its nearest point the earth is 147 million kilometers away from the sun and at its furthest point the earth is 152 million kilometers away from the sun. 

Find an equation for an ellipse that approximates the earth’s orbit around the sun.

What is the perimeter of an object that travels around this ellipse?

How do estimates of the earth’s speed and distance traveled based on the elliptical-orbit estimate differ from the earth’s distance traveled and speed based on the circular orbit assumption. 

Answer:  The equation of an ellipse with the center at (0,0) is of the form

X2/a2 + Y2/b2 = 1

It has been a really long time since I thought about ellipses.  (The web site below has a nice explanation.)


If we set a=147 and b=152 we get an ellipse where the furthest point is 147 kilometers and the nearest point is 152 kilometers.

The formulas for the perimeter of an ellipse are actually very difficult.  See the link below for an explanation.



The approximation for the ellipse perimeter that we use is


PER = pi x [ 3 x (a+b) – ((3a+b)(a+3b)}0.5 ]

Plugging in values of a=147 and b=152 we find that the earth’s annual orbit around the sun is approximately 939.4 million kilometers.  Our estimate of the earth’s orbit assuming a circle (radius 149.7 kilometers) is 940.4 kilometers.

The circular orbit and elliptical orbit estimates are in fact quite similar.

Please feel free to check my work, to comment and to provide suggestions for more posts.



Thursday, August 1, 2019

Speed of an object going around the sun



This post evaluates the speed of an object going around the sun assuming a circular orbit.

Question:  An object travels in a circle around the sun.  The radius of the circle is 93 million miles.  It takes 365 ¼ days for the object to make one complete revolution around the sun.  What is the average speed in miles per hour for this object?

Answer:  The distance traveled is the circumference of the circle, which is


2 x 3.1416 x 93,000,000


or 584,337,600 miles.

(Note to self.  Remember to multiply by 2.)

The number of hours in 365 ¼ days is 8,766.



The average speed in miles per hour is (585,337,600)/(8766) or 66,661 miles per hour.

Note:  The earth is 93 million miles from the sun but I do not know the exact nature of its orbit.  Does it travel in a circle or ellipse?  I am clueless.   If science students wants to create more questions of this type I will publish guest posts.

Sunday, July 14, 2019

Radius of a one-ton gold sphere



This post considers the radius of a one ton sphere of gold.

Question:   A previous problem found that a ton of gold has a volume of 47,004.4 cubic centimeters.   Go here to better understand this interesting calculation.

Volume of one ton of gold.
http://www.dailymathproblem.com/2019/05/volume-of-one-ton-of-gold.html

What is the radius of a sphere consisting one ton of gold?


Answers:  

The volume of a sphere is V=(4/3)*pi*r3

Plugging in the value of the item and solving for r we get

r=(47,0004.4 * (3/4)/pi)1/3


This reduces to 48.323 cm3


Plugging this value of r into the formula for the volume of the sphere does in fact give us the correct volume.

"(4/3)
1.33333333
"pi
3.14159265
r
48.2322777
(4/3)pir3
470004.4