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Topic: Generalized Birthday "Paradox" (M) (Read 14770 times) 

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


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ThudnBlunder
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Re: Generalized Birthday "Paradox" (M)
« Reply #27 on: Dec 10^{th}, 2006, 4:17am » 
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on Dec 9^{th}, 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 6^{th}, 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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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #29 on: Jan 6^{th}, 2007, 11:26am » 
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on Jan 6^{th}, 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.) :( people  1  2  3  4  5  6  7  8  9  10  11  12  13  14  16  17  18  19  20  21  22  23  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 6^{th}, 2007, 11:28am by Locke64 » 
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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #30 on: Jan 6^{th}, 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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Ghost Sniper
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Re: Generalized Birthday "Paradox" (M)
« Reply #31 on: Jan 6^{th}, 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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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #32 on: Jan 6^{th}, 2007, 3:03pm » 
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on Jan 6^{th}, 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 6^{th}, 2007, 5:22pm by Locke64 » 
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Ghost Sniper
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Re: Generalized Birthday "Paradox" (M)
« Reply #33 on: Jan 6^{th}, 2007, 5:16pm » 
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on Jan 6^{th}, 2007, 3:03pm, Locke64 wrote: Trust me, that's wrong. 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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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #34 on: Jan 6^{th}, 2007, 5:23pm » 
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on Jan 6^{th}, 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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Whiskey Tango Foxtrot
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Re: Generalized Birthday "Paradox" (M)
« Reply #35 on: Jan 6^{th}, 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.


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ThudnBlunder
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Re: Generalized Birthday "Paradox" (M)
« Reply #36 on: Jan 7^{th}, 2007, 5:10am » 
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on Jan 6^{th}, 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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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #37 on: Jan 7^{th}, 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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Whiskey Tango Foxtrot
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Re: Generalized Birthday "Paradox" (M)
« Reply #38 on: Jan 7^{th}, 2007, 9:45am » 
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on Jan 7^{th}, 2007, 5:10am, THUDandBLUNDER wrote:Er...no, I implied that grungy was a cheeky monkey. 



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ThudnBlunder
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Re: Generalized Birthday "Paradox" (M)
« Reply #39 on: Jan 7^{th}, 2007, 11:43am » 
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on Jan 7^{th}, 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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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #40 on: Jan 7^{th}, 2007, 11:49am » 
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on Jan 7^{th}, 2007, 11:43am, THUDandBLUNDER 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.


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Locke64
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Re: Generalized Birthday "Paradox" (M)
« Reply #41 on: Jan 7^{th}, 2007, 12:14pm » 
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on Jan 7^{th}, 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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StallionMang
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Re: Generalized Birthday "Paradox" (M)
« Reply #42 on: Jan 5^{th}, 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!/((365n)!(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!(n3)!))*365 / 365^n and two people: (n!/(2!(n2)!))*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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ThudnBlunder
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Re: Generalized Birthday "Paradox" (M)
« Reply #43 on: Jan 5^{th}, 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 5^{th}, 2008, 11:46pm by ThudnBlunder » 
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StallionMang
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Re: Generalized Birthday "Paradox" (M)
« Reply #44 on: Jan 6^{th}, 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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skeptic1000
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Re: Generalized Birthday "Paradox" (M)
« Reply #45 on: May 1^{st}, 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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