Still neck division
01162014, 09:06 AM
Post: #31




RE: Still neck division
(01162014 04:23 AM)amoffett11 Wrote: This is the same as the classic Birthday problem: given 25 people in a room, what are the odds that at least 2 of them have the same birthday. People are usually quite surprised to find that the odds are slightly higher than 50% that 2 people will be born on the same day. Are you sure? 

01162014, 11:41 PM
Post: #32




RE: Still neck division
(01162014 09:06 AM)jchris98 Wrote:(01162014 04:23 AM)amoffett11 Wrote: This is the same as the classic Birthday problem: given 25 people in a room, what are the odds that at least 2 of them have the same birthday. People are usually quite surprised to find that the odds are slightly higher than 50% that 2 people will be born on the same day. Sure about what? The birthday thing? GC name: amoffett11 1v1 2v2 with .Memories. and 2v2 random Until you stalk and overrun you can't devour anyone 

01172014, 02:10 AM
Post: #33




RE: Still neck division
Yeah, that doesn't sound right


01172014, 02:14 AM
(This post was last modified: 01172014 02:15 AM by lawtai.)
Post: #34




RE: Still neck division


01172014, 02:33 AM
Post: #35




RE: Still neck division
(01172014 02:14 AM)lawtai Wrote: http://en.wikipedia.org/wiki/Birthday_problem Ya, whenever I hear the example, it's always 25, but probably because that's just an even number. I remember having this discussion with my brother years and years ago, he didn't believe it either, so we started looking at baseball teams (every team 25 players on its roster), we went through about 16 teams and 10 or 11 teams already had two players with the same birthday, so we stopped. GC name: amoffett11 1v1 2v2 with .Memories. and 2v2 random Until you stalk and overrun you can't devour anyone 

01172014, 02:54 AM
Post: #36




RE: Still neck division
I read an article about it, and there was the mathematic proof of this, so I could only believe it
You can ask a... (drawing by Chemoeum) 

01172014, 02:59 AM
Post: #37




RE: Still neck division
Pretty sure it's about probabilities:
Need to find the percentage of no people have birthdays together (this is continuing to assume that no two people have the same birthday) For 1 person, there is 365/365 For 2 people, I would say 364/365 to keep it simple, although the fraction is much weirder than that due to leap years. 3 people: 364/365 * 363/365 4 people: 364/365 * 363/365 * 362/365 5 people: 364/365 * ... * 361/365 And so on. Even for 10 people: already ~12% chance that 2 people will have the same birthday. (because ~88% chance that they will not have the same birthday) Doowhat? Doodat? Okey:3 1v1  Peak #32 // Current #56 as AlliDooisRush 2v2  w/jesusfuentesh #7 2v2  w/baustin #17 // w/Juslas #85 // Randoms #38 

01172014, 03:06 AM
Post: #38




RE: Still neck division
Yes, Doodat has it right.


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