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Title: Riddle/Quiz thread by request Post by: Fuzzy on February 07, 2007, 06:34:53 AM Quote from: stroh on February 06, 2007, 08:06:14 PM -------------------------------------------------------------------------------- Quote from: cbae on February 06, 2007, 08:04:44 PM Quote from: Fuzzmental on February 06, 2007, 07:11:48 PM If I may be so bold and join in the reindeer games.... For those of you who have heard of the Birthday paradox please allow the others a chance. If you have 23 randomly chosen people in a room, what are the odds that 2 of them share a birthday*? For extra credit, now you have 60 randomly chosen people in a room. What are the odds? *not the same year, just the day This looks like fun, but it should probably be in another thread for puzzles/riddles since it's not really trivia. How 'bout it? I might have a puzzle or two to contribute. Agreed. So K fuzz? Agreed. So I'll kick it off with my question(s) above. Title: Re: Riddle/Quiz thread by request Post by: stroh on February 07, 2007, 06:45:56 AM pure guess. Just thinking of the days of the year.
I'll go with 1 in 167. Title: Re: Riddle/Quiz thread by request Post by: TFT on February 07, 2007, 07:00:46 AM Less than 5 to one?
Greater than 80%? Title: Re: Riddle/Quiz thread by request Post by: gleek on February 07, 2007, 07:18:55 AM Just thinking outloud here.
If you take one person, and check against all others, you have 22 chances out of 366 that their birthdays will be the same. Then take the 2nd person and check against the remaining, you have 21 chances out of 366 that their birthdays will be the same. Therefore, the number of chances that you get for a match is the value of the arithmetic series from 1 to 22. Or (22+1) * 11 = 253. So, I'm guessing 253 out of 366 or 69% chance. Edit: Hmm...Now that I think about the extra credit one, something looks f'ed up in my methodology. Title: Re: Riddle/Quiz thread by request Post by: Fuzzy on February 07, 2007, 07:38:30 AM So far, for the 23 people, chin is the closest. Plus he's on the right track with his solution.
I'll let this go a little while longer. Title: Re: Riddle/Quiz thread by request Post by: gleek on February 07, 2007, 07:54:04 AM So far, for the 23 people, chin is the closest. Plus he's on the right track with his solution. I'll let this go a little while longer. I think I need to divide the 253 by something else OR I need to divide the individual components (22,21,20...) by something. I suck at calculating probability. I remember I had a tough time in my 11th grade Math Analysis class which covered this *feces*. [sm_disgust] Title: Re: Riddle/Quiz thread by request Post by: TFT on February 07, 2007, 07:56:11 AM I'm guessing 'paradox' means 'cheaty lying bugger.'
Title: Re: Riddle/Quiz thread by request Post by: Aske on February 07, 2007, 08:22:06 AM Just thinking outloud here. If you take one person, and check against all others, you have 22 chances out of 366 that their birthdays will be the same. Then take the 2nd person and check against the remaining, you have 21 chances out of 366 that their birthdays will be the same. Therefore, the number of chances that you get for a match is the value of the arithmetic series from 1 to 22. Or (22+1) * 11 = 253. So, I'm guessing 253 out of 366 or 69% chance. Edit: Hmm...Now that I think about the extra credit one, something looks f'ed up in my methodology. you basically just counted the number of combinations of birthdays with 23 people. :) Title: Re: Riddle/Quiz thread by request Post by: gleek on February 07, 2007, 08:30:36 AM Just thinking outloud here. If you take one person, and check against all others, you have 22 chances out of 366 that their birthdays will be the same. Then take the 2nd person and check against the remaining, you have 21 chances out of 366 that their birthdays will be the same. Therefore, the number of chances that you get for a match is the value of the arithmetic series from 1 to 22. Or (22+1) * 11 = 253. So, I'm guessing 253 out of 366 or 69% chance. Edit: Hmm...Now that I think about the extra credit one, something looks f'ed up in my methodology. you basically just counted the number of combinations of birthdays with 23 people. :) But I divided it by 366. So it's the average number of combinations of birthdays with 23 people over an entire year. :P Title: Re: Riddle/Quiz thread by request Post by: Aske on February 07, 2007, 08:32:01 AM i didn't read the fine print on not answering
:'( Title: Re: Riddle/Quiz thread by request Post by: TFT on February 07, 2007, 08:33:23 AM i didn't read the fine print on not answering :'( I forgive you. Fuzzy holds grudges for years though. ;) Why are the numbers so high? Is it real/tested or just some geek crap? Title: Re: Riddle/Quiz thread by request Post by: Clive on February 07, 2007, 08:35:00 AM It's all geek crap, when you get right down to it.
Title: Re: Riddle/Quiz thread by request Post by: Aske on February 07, 2007, 08:35:16 AM i didn't read the fine print on not answering :'( I forgive you. fuzzy holds grudges for years though. ;) Why are the numbers so high? Is it real/tested or just some geek crap? i removed my post, so you should too (quoting the answer) ;) it's not real in several ways (fundamental assumptions on how birthdays distribute) but the statistics/math are correct Title: Re: Riddle/Quiz thread by request Post by: TFT on February 07, 2007, 08:37:48 AM i removed my post, so you should too (quoting the answer) ;) Done. Title: Re: Riddle/Quiz thread by request Post by: Clive on February 07, 2007, 08:38:22 AM TFT, you should also remove your last hundred or so posts, just to be sure.
Title: Re: Riddle/Quiz thread by request Post by: Clive on February 07, 2007, 08:38:43 AM Aske, you too. But let's make it 1000 for you, cheater.
Title: Re: Riddle/Quiz thread by request Post by: TFT on February 07, 2007, 08:39:23 AM TFT, you should also remove your last hundred or so posts, just to be sure. I will nuke them from orbit, that's really the only way to be sure. Title: Re: Riddle/Quiz thread by request Post by: Aske on February 07, 2007, 08:40:59 AM :o
Title: Re: Riddle/Quiz thread by request Post by: spacey on February 07, 2007, 08:52:14 AM [Checking Neil Boortz for the answer]
Title: Re: Riddle/Quiz thread by request Post by: Aske on February 07, 2007, 08:56:03 AM [Checking Neil Boortz for the answer] it depends if hillary is in the 23. Title: Re: Riddle/Quiz thread by request Post by: JDerion on February 07, 2007, 08:57:31 AM The free market can sort out the answer to this.
Title: Re: Riddle/Quiz thread by request Post by: Fuzzy on February 07, 2007, 09:06:28 AM O.K. :)
The answer for 23: probability is more than 50% that 2 will have the same birthday. For 60, probability is greater than 99%. I thought about plagiarizing other web sites but decided against it. You can check here for the math: http://en.wikipedia.org/wiki/Birthday_paradox NEXT! Title: Re: Riddle/Quiz thread by request Post by: spacey on February 07, 2007, 09:16:31 AM Okay, stop me if you've heard this one:
A man and his son are in a car crash and the father dies, the son is rushed to the ER, where.... oh you know that one? Well, I got nothing then. Title: Re: Riddle/Quiz thread by request Post by: gleek on February 07, 2007, 09:19:06 AM Okay, stop me if you've heard this one: A man and his son are in a car crash and the father dies, the son is rushed to the ER, where.... oh you know that one? Well, I got nothing then. Does this story involve a farmer's daughter? If so, please proceed. Title: Re: Riddle/Quiz thread by request Post by: spacey on February 07, 2007, 09:20:55 AM Okay, stop me if you've heard this one: A man and his son are in a car crash and the father dies, the son is rushed to the ER, where.... oh you know that one? Well, I got nothing then. Does this story involve a farmer's daughter? If so, please proceed. |