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   New birthday Paradox question
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   Author  Topic: New birthday Paradox question  (Read 1015 times)
aicoped
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New birthday Paradox question  
« on: Apr 13th, 2010, 12:19am »
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Ok the normal birthday paradox yields 23 people.
 
Here is my new similar question.
 
Assuming you pair up 2 people and check to see if they have the same birthday. if they do, you are done. If not throw out those people and get 3 new people and check birthdays. If they do then you are done if not throw out those 3 and get 4 new people and keep repeating this process until 2 people within a set have the same birthday.
 
How many "sets" of people do you need on average to have a greater than 50% chance that some 2 within a set have the same birthday?
 
And then which set would you want to be in to be have the highest chance of being within the set of the first birthday.
 
If all of that wasn't clear, here is an example.
 
A,B: Jan12 and Mar23
C,D,E: Jun08, Apr30, Sep 14
F,G,H,ICheesyec09, Dec18, Oct22, Jan27
J,K,L,M,NShockedct 15, May11, May11, Feb14, Jul13
 
So in this example it took until the fifth set of people. This was a contrived example, but the point remains if they wouldn't have been in the 5th, then we would have just checked 6 new people and so on until we hit a birthday pair within a stack.
 
 
 
My only thoughts on this are it seems as if it would be around 23 still. The fact that their are going to be duplicates outside of sets is more than offset by the fact that there is a great extra number of total people, but then again my mind oculd just be being stupid.
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Re: New birthday Paradox question  
« Reply #1 on: Apr 13th, 2010, 1:01am »
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I think you need to find the n such that i=2..n (1- 365!/(365-i)!/365i) > 0.5; that should give one more than the number of sets of people you need (because you start with a set a pair).
 
With the 11th set you get a greater than 50% chance (over 61% in fact) that at least one set contains at least one pair with the same birthday

 
[edit]Actually, reconsidering, that's not exactly what you asked for. Oh well.[/edit]
« Last Edit: Apr 13th, 2010, 1:12am by towr » IP Logged

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