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Riddle/Quiz thread by request

 
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Fuzzy
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Riddle/Quiz thread by request
« 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.
 
 
 
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stroh
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Re: Riddle/Quiz thread by request
« Reply #1 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.
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TFT
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Re: Riddle/Quiz thread by request
« Reply #2 on: February 07, 2007, 07:00:46 AM »

Less than 5 to one?

Greater than 80%?
« Last Edit: February 07, 2007, 07:09:56 AM by TFT » Logged Return to Top

gleek
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Re: Riddle/Quiz thread by request
« Reply #3 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.
« Last Edit: February 07, 2007, 07:23:05 AM by cbae » Logged Return to Top

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Re: Riddle/Quiz thread by request
« Reply #4 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.
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Re: Riddle/Quiz thread by request
« Reply #5 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*.  Disgusted
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Re: Riddle/Quiz thread by request
« Reply #6 on: February 07, 2007, 07:56:11 AM »

I'm guessing 'paradox' means 'cheaty lying bugger.'

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Aske
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Re: Riddle/Quiz thread by request
« Reply #7 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.
 Smiley
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Re: Riddle/Quiz thread by request
« Reply #8 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.
 Smiley

But I divided it by 366. So it's the average number of combinations of birthdays with 23 people over an entire year. Tongue
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Re: Riddle/Quiz thread by request
« Reply #9 on: February 07, 2007, 08:32:01 AM »

i didn't read the fine print on not answering
 Cry
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Re: Riddle/Quiz thread by request
« Reply #10 on: February 07, 2007, 08:33:23 AM »

i didn't read the fine print on not answering
 Cry

I forgive you.

Fuzzy holds grudges for years though. 

 Wink

Why are the numbers so high?

Is it real/tested or just some geek crap?
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Re: Riddle/Quiz thread by request
« Reply #11 on: February 07, 2007, 08:35:00 AM »

It's all geek crap, when you get right down to it.
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Aske
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Re: Riddle/Quiz thread by request
« Reply #12 on: February 07, 2007, 08:35:16 AM »

i didn't read the fine print on not answering
 Cry

I forgive you.

fuzzy holds grudges for years though. 

 Wink

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) Wink


it's not real in several ways (fundamental assumptions on how birthdays distribute) but the statistics/math are correct
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Russia has invaded a sovereign neighboring state and threatens a democratic government elected by its people. Such an action is unacceptable in the 21st century.
--  Chimpy McFlightsuit, CEO of Bu$hco Industries of 'Merka
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Re: Riddle/Quiz thread by request
« Reply #13 on: February 07, 2007, 08:37:48 AM »

i removed my post, so you should too (quoting the answer) Wink


Done.
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Clive
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Re: Riddle/Quiz thread by request
« Reply #14 on: February 07, 2007, 08:38:22 AM »

TFT, you should also remove your last hundred or so posts, just to be sure.
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