idelovski, April 25
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idelovski, April 25
> But it makes it easier to remember the date
Associative compression ;)
Always thought you were a computer :rolleyes:
So you dated Renée Zellweger? How was she?
Assuming Jackie Chan's list in post #27 was correct, we now have:
Only three more to go and then we can buy into the theory that those sharing birthdays just aren't posting. :)Code:1) Jackie Chan April 7
2) Prelude November 11
3) Neo1 October 11
4) brewbuck February 7
5) ahluka June 7
6) ping December 15
7) anon October 25
8) indigo0086 June 13
9) Mario F. August 23
10) BMJ May 29
11) maxorator March 25
12) SlyMaelstrom August 10
13) abh!shek January 10
14) P4R4N01D February 28
15) QuantumPete August 11
16) mike_g October 14
17) zacs7 October 17
18) foxman October 15
19) dra September 28
20) idelovski April 25
Or we can just buy into the theory that half the time... a 50% chance won't happen...
Hmm, well, yes. :p
Spoiler: http://cboard.cprogramming.com/calen...th=1&year=2008
Ahh great, all you've done now is confused me for my statistics exam :s
It is of course, random. The fact still remains that from a mathematical standpoint, the chance of a hit becomes greater than 0.5 when there are 23 or more people. This doesn't mean the hit will happen, but it becomes more and more likely.
In other words, given an infinite set of sets of 23 people, approximately half of such sets (only approximate, because the actual halfway point is not an integral number of people, but a non-integral number of people is impossible) will contain at least one coincident birthday. But you shouldn't be that surprised if any GIVEN set does not contain a hit, any more than you should be surprised by flipping 8 heads in 10 flips.
Now, if you had a set of 367 people and got no hits, you should be VERY surprised, since there are only 366 days in the year (worst case, counting leap years) :)