idelovski, April 25

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- 06-13-2008idelovski
idelovski, April 25

- 06-13-2008brewbuck
- 06-13-2008zacs7
> But it makes it easier to remember the date

Associative compression ;)

Always thought you were a computer :rolleyes: - 06-13-2008idelovski
So you dated Renée Zellweger? How was she?

- 06-13-2008dwks
Assuming Jackie Chan's list in post #27 was correct, we now have:

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

- 06-13-2008SlyMaelstrom
Or we can just buy into the theory that half the time... a 50% chance won't happen...

- 06-13-2008dwks
Hmm, well, yes. :p

Spoiler: http://cboard.cprogramming.com/calen...th=1&year=2008 - 06-13-2008zacs7
Ahh great, all you've done now is confused me for my statistics exam :s

- 06-14-2008brewbuck
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) :) - 06-14-2008brewbuck
- 06-15-2008indigo0086