Remember
this one?
Well, this time, the question is:
Assuming that birthdays are evenly distributed around 365 days of the year...
what is the minimum number of people I must have in a room, such that the odds are that at least n people share the same birthday?
Let's limit this question to n values from 1 to 12.
We know that for n=1, 1 person is sufficient.
For n=2, as is described in Happy Birthday, 23 people are sufficient.
What are the minimum numbers for n=3 to 12?
(In reply to
2 More Links by Richard)
As Richard has noted in his previous posts, there is info on the web relating to this problem.
The minimal number of people to give a 50% probability of having at least n coincident birthdays is 1, 23, 88, 187, 313, 460, 623, 798, 985, 1181, 1385, 1596, 1813, ... (Sloane's A014088; Diaconis and Mosteller 1989).
(Source Mathworld}
These are exact results and the formulae to produce the sequence may be found at Mathworld but are not short and sweet so I have not reproduced them here. For the same reason I do not think it is likely that an elegant solution can be found.
Edited on February 27, 2006, 4:57 am
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Posted by goFish
on 2006-02-27 04:55:03 |
re:Best Solution
