Difference between revisions of "User talk:Bobthesmartypants"
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==Bobthesmartypants's question collection== | ==Bobthesmartypants's question collection== | ||
| − | 1. Bob is rolling a <math>6</math>-sided die. Every time he rolls a number that he has already rolled before, he rolls again. He stops when he has rolled all the numbers. What is the expected number of rolls it will take Bob? | + | '''Problem 1.''' Bob is rolling a <math>6</math>-sided die. Every time he rolls a number that he has already rolled before, he rolls again. He stops when he has rolled all the numbers. What is the expected number of rolls it will take Bob? |
| − | 2. Suppose you have a rectangular box, with side lengths <math>a</math> and <math>b</math>, where <math>a,b\in\mathbb{Z}</math>. We launch a point-like ball from one of the vertices with an angular degree of <math>60^{\circ}</math>. The ball bounces off the sides of the box. Pretend there is no friction, drag, or anything else to slow down the ball. Prove or disprove that the ball won't ever hit a vertex again. | + | '''Problem 2.''' Suppose you have a rectangular box, with side lengths <math>a</math> and <math>b</math>, where <math>a,b\in\mathbb{Z}</math>. We launch a point-like ball from one of the vertices with an angular degree of <math>60^{\circ}</math>. The ball bounces off the sides of the box. Pretend there is no friction, drag, or anything else to slow down the ball. Prove or disprove that the ball won't ever hit a vertex again. |
| − | 3. In a country, there is a perticular way the cities inside are connected. One city has only one road leading out of it. One city has two roads leading out of it. Two cities have three roads leading out of it. Three cities have 5 roads leading out of it. In general, <math>F_n</math> have <math>F_{n-1}</math> cities leading out of it. What values of <math>n</math> are there such that this setup is possible? | + | '''Problem 3.''' In a country, there is a perticular way the cities inside are connected. One city has only one road leading out of it. One city has two roads leading out of it. Two cities have three roads leading out of it. Three cities have 5 roads leading out of it. In general, <math>F_n</math> have <math>F_{n-1}</math> cities leading out of it. What values of <math>n</math> are there such that this setup is possible? |
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| + | ==Answers== | ||
| + | '''Solution 1.''' For the first number, it will always be <math>1</math> roll. For the second number, there is a <math>\frac{5}{6}</math> chance of getting a new number, so the expected number of rolls is <math>\frac{6}{5}</math>. The expected number of rolls for the third number is <math>\frac{6}{4}=\frac{3}{2}</math>. Continuing the pattern, the expected number of rolls for the 4th, 5th, and final number is <math>2,3,6</math> respectively. So the total expected number of rolls is <math>1+\frac{6}{5}+\frac{3}{2}+2+3+6=\boxed{\frac{147}{10}, \text{ or }14.7}</math> | ||
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| + | '''Solution 2.''' | ||
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| + | '''Solution 3.''' | ||
Revision as of 13:51, 21 May 2013
Bobthesmartypants's question collection
Problem 1. Bob is rolling a 
-sided die. Every time he rolls a number that he has already rolled before, he rolls again. He stops when he has rolled all the numbers. What is the expected number of rolls it will take Bob?
Problem 2. Suppose you have a rectangular box, with side lengths 
and 
, where 
. We launch a point-like ball from one of the vertices with an angular degree of 
. The ball bounces off the sides of the box. Pretend there is no friction, drag, or anything else to slow down the ball. Prove or disprove that the ball won't ever hit a vertex again.
Problem 3. In a country, there is a perticular way the cities inside are connected. One city has only one road leading out of it. One city has two roads leading out of it. Two cities have three roads leading out of it. Three cities have 5 roads leading out of it. In general, 
have 
cities leading out of it. What values of 
are there such that this setup is possible?
Answers
Solution 1. For the first number, it will always be 
roll. For the second number, there is a 
chance of getting a new number, so the expected number of rolls is 
. The expected number of rolls for the third number is 
. Continuing the pattern, the expected number of rolls for the 4th, 5th, and final number is 
respectively. So the total expected number of rolls is 
Solution 2.
Solution 3.
