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| − | ==Bobthesmartypants's question collection== | + | ==Introduction== |
| − | '''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?
| + | Hello, I am bobthesmartypants. This is my ever-expanding problem set that I made up completely, and am contwinuing to make up. Please refrain from editing the problems page, but if you have an alternate solution to a problem or disagree with a solution, feel free to PM me. Hope you have fun solving these problems! |
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| − | '''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. | + | ''Note: I may be transferring this to my Blog. Check it out when I do so!'' |
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| − | '''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? | + | ==Bobthesmartypants's problem set== |
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| − | ==Answers==
| + | [[User talk:Bobthesmartypants/Problems]] |
| − | '''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.'''
| + | [[User talk:Bobthesmartypants/Solutions]] |
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| − | '''Solution 3.'''
| + | ==Sandbox== |
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| | + | [[User talk:Bobthesmartypants/Sandbox]] |
Latest revision as of 21:51, 8 October 2013
Introduction
Hello, I am bobthesmartypants. This is my ever-expanding problem set that I made up completely, and am contwinuing to make up. Please refrain from editing the problems page, but if you have an alternate solution to a problem or disagree with a solution, feel free to PM me. Hope you have fun solving these problems!
Note: I may be transferring this to my Blog. Check it out when I do so!
Bobthesmartypants's problem set
User talk:Bobthesmartypants/Problems
User talk:Bobthesmartypants/Solutions
Sandbox
User talk:Bobthesmartypants/Sandbox
