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   Generalized Birthday "Paradox" (M)
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   Author  Topic: Generalized Birthday "Paradox" (M)  (Read 13938 times)
ThudnBlunder
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Re: Generalized Birthday "Paradox" (M)  
« Reply #25 on: Dec 9th, 2006, 10:28am »
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on Dec 9th, 2006, 8:37am, Whiskey Tango Foxtrot wrote:
What site did you find that solution on?  Maybe I'll explore it for some inspiration.  Wink

Please feel free: http://brainyplanet.com/index.php/Line%20Solution?PHPSESSID=98133811aa83 97f93ceb0a8893bb7775
 
 Roll Eyes
« Last Edit: Dec 9th, 2006, 10:33am by ThudnBlunder » IP Logged

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Re: Generalized Birthday "Paradox" (M)  
« Reply #26 on: Dec 9th, 2006, 10:56am »
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Brainy planet, eh?  Thanks.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #27 on: Dec 10th, 2006, 4:17am »
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on Dec 9th, 2006, 10:56am, Whiskey Tango Foxtrot wrote:
Brainy planet, eh?  Thanks.

Give a cheeky monkey a typewriter and....
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Locke64
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Re: Generalized Birthday "Paradox" (M)  
« Reply #28 on: Jan 6th, 2007, 10:56am »
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I've heard this one before.  You have to work backwards (find the chances that they don't have the same birthday).
I'm working on it now.  (I forgot how to do it. >_<)
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Re: Generalized Birthday "Paradox" (M)  
« Reply #29 on: Jan 6th, 2007, 11:26am »
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on Jan 6th, 2007, 10:56am, Locke64 wrote:
I've heard this one before.  You have to work backwards (find the chances that they don't have the same birthday).
I'm working on it now.  (I forgot how to do it. >_<)

The number of people required for there to be at least a 50% chance of two of them sharing the same birthday is 23.
 
In order to solve it, you must first find the probability of people not sharing the same birthday as anyone else.  For one person, it's 365/365 (We're ignoring leap years BTW), since it is impossible to share a birthday with anyone else because there is no one else to share a birthday with.
If you add a second person, the probability would be 364/365, since he could have a birthday on any day but the first person.  The third person would have a probability of 363/365, since he must not share a birthday with either of the first two.  Obviously, the numerator decreases by one each time a person is added.  With this knowledge, you can make a table, multiplying 365/365x364/365x363/365...:

 
 
IGNORE THE TABLE IF YOU WANT TO DO IT ON YOUR OWN!!!
(you can't put tables in hides.) :(
 
people12345678910111213141617181920212223
probablity.997.991.984.973.96.944.926.905.883.859.833.806.777.747.716.685.653.621.589.556.524.493

 
 
After you found the number of people needed to bring the probability of no two of them sharing the same birthday down below 50, just subtract it from one to find the probability of two people sharing the same birthday. ;)
« Last Edit: Jan 6th, 2007, 11:28am by Locke64 » IP Logged

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Re: Generalized Birthday "Paradox" (M)  
« Reply #30 on: Jan 6th, 2007, 11:39am »
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nvm, I misunderstood the question.  I'm not mathematically advanced enough to solve that one. :(
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Re: Generalized Birthday "Paradox" (M)  
« Reply #31 on: Jan 6th, 2007, 2:51pm »
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Personally, I think I have an easier answer, unless someone already posted it. I'm not completely sure about this, though.
 
365/2=182.5, so I would say 183+k number of people.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #32 on: Jan 6th, 2007, 3:03pm »
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on Jan 6th, 2007, 2:51pm, hiyathere wrote:
Personally, I think I have an easier answer, unless someone already posted it. I'm not completely sure about this, though.
 
365/2=182.5, so I would say 183+k number of people.
Trust me, that's wrong. ;)
It is known that the answer for 2 people is 23. 183+2 is not equal to 23.
« Last Edit: Jan 6th, 2007, 5:22pm by Locke64 » IP Logged

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Re: Generalized Birthday "Paradox" (M)  
« Reply #33 on: Jan 6th, 2007, 5:16pm »
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on Jan 6th, 2007, 3:03pm, Locke64 wrote:

Trust me, that's wrong. Wink
It is known that the answer for 2 people is 23. 183+2 is not equal to 23.

What do you mean? How do you know that?
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Re: Generalized Birthday "Paradox" (M)  
« Reply #34 on: Jan 6th, 2007, 5:23pm »
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on Jan 6th, 2007, 5:16pm, hiyathere wrote:

What do you mean? How do you know that?

Well, if you go through the steps in my previous post, you'll find out for yourself, plus most other people here could explain it even better than me, and there are many websites that you can find that have that as the correct answer.  But that's just for 2, not for k.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #35 on: Jan 6th, 2007, 9:54pm »
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I just noticed T 'n' B called me a cheeky monkey a month ago!  How rude!
 
At least I got a typewriter though.  Grin
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Re: Generalized Birthday "Paradox" (M)  
« Reply #36 on: Jan 7th, 2007, 5:10am »
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on Jan 6th, 2007, 9:54pm, Whiskey Tango Foxtrot wrote:
I just noticed T 'n' B called me a cheeky monkey a month ago!  

Er...no, I implied that grungy was a cheeky monkey.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #37 on: Jan 7th, 2007, 9:37am »
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To find the number of people needed so that the probability of 3 people having the same birthday is at least 50%, would you do 365/365*365/365*364/365*363/365*362/365...?
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Re: Generalized Birthday "Paradox" (M)  
« Reply #38 on: Jan 7th, 2007, 9:45am »
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on Jan 7th, 2007, 5:10am, THUDandBLUNDER wrote:
Er...no, I implied that grungy was a cheeky monkey.
Embarassed Cheesy
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Re: Generalized Birthday "Paradox" (M)  
« Reply #39 on: Jan 7th, 2007, 11:43am »
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on Jan 7th, 2007, 9:37am, Locke64 wrote:
To find the number of people needed so that the probability of 3 people having the same birthday is at least 50%, would you do 365/365*365/365*364/365*363/365*362/365...?

Exactly 3 or at least 3?
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Re: Generalized Birthday "Paradox" (M)  
« Reply #40 on: Jan 7th, 2007, 11:49am »
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on Jan 7th, 2007, 11:43am, THUDandBLUNDER wrote:

Exactly 3 or at least 3?

Well, to find how many people you need for 2 ppl to have the same birthday, you use 365/365*364/365..., so If I add one to the requirement, I'd think I'd have to add one more impossible (365/365) to the list of factors.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #41 on: Jan 7th, 2007, 12:14pm »
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on Jan 7th, 2007, 11:49am, Locke64 wrote:

Well, to find how many people you need for 2 ppl to have the same birthday, you use 365/365*364/365..., so If I add one to the requirement, I'd think I'd have to add one more impossible (365/365) to the list of factors.

Actually, that can't be right.  All that would be doing is adding one person, since it is multiplying it by one and not changing any of the rest of the calculation.
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Re: Generalized Birthday "Paradox" (M)  
« Reply #42 on: Jan 5th, 2008, 9:58pm »
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OK.  I think I've got it.
 
In the original "birthday" problem, we found the probability that no two people have the same birthday, and then took the complement of that fraction to obtain the probability that at least two have the same birthday.
 
Now let's set k to the value of 4, for example.
 
So in order to obtain the probability that at least 4 people have the same birthday, let's take the probability that we derived in the original riddle - the probability that at least 2 people have the same birthday: (365^n) - (365!/((365-n)!(365^n)))
 
We simply take this probability and subtract from it the probabilities that EXACTLY three people have the same birthday and that EXACTLY two people have the same birthday.
 
The probability that EXACTLY three people have the same birthday is:  
 
(n!/(3!(n-3)!))*365 / 365^n
 
and two people:
 
(n!/(2!(n-2)!))*365 / 365^n
 
(Probability of At Least Two) - (Probability of EXACTLY 3 )- (Probability of EXACTLY 2) = Probability of At Least 4.
 
So in order to scale this up for all values of k, we simply say that for any value of k greater than 2, just take Probability of At Least Two and subtract from it the sum of all probabilities that EXACTLY p number of people have the same birthday for all values of p where p>1 and p<k.
 
Did I get it right?
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Re: Generalized Birthday "Paradox" (M)  
« Reply #43 on: Jan 5th, 2008, 11:30pm »
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StallionMang, I think you fail to take into account the fact that more than one pair/trio/etc can share different birthdays.
 
Probability(3 or More) 1 - Probability(None) - Probability(1 Pair)
 
Probability(3 or More) = 1 - Probability(None) - Probability(1 Pair) - Probability(2 Pairs Share 2 Different Birthdays) -  ...  - Probability(n/2 Pairs Share n/2 Different Birthdays)
« Last Edit: Jan 5th, 2008, 11:46pm by ThudnBlunder » IP Logged

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Re: Generalized Birthday "Paradox" (M)  
« Reply #44 on: Jan 6th, 2008, 12:53am »
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Alas, you are right.
 
And on top of that, my equations for calcuating the number of possible combinations for EXACTLY 3 and EXACTLY 2 people with the same birthday were completely wrong.  I was doing C(n,3) * 365, which is not right.
 
Oh well....back to the drawing board.
 
Thanks!
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Re: Generalized Birthday "Paradox" (M)  
« Reply #45 on: May 1st, 2008, 10:43am »
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Has anyone already tried finding the answer for the first several values of k, and then attempted to detect some sort of pattern that might make the answer more apparent?
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