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Neo · 11-30-2005, 11:28 PM

Ignoring the possibility that someone was born on Feb29...

P = (365)(364)(363)....(365 - n + 1)/(365)^n where n = 30.

P = .29368, which is actually the probability that no two people have the same birthday, so

1 - .29368 = .70632 => .71

So in a room with 30 people, chances are more likely than not that two people will have the same birthday. May seem counterintuitive but it's true.

11-30-2005, 11:28 PM	#8
Neo Former CEO Neo is offline Location: Longhorn country Now Playing: Silent Hill: Downpour Posts: 6,528	Re: How do I solve this? Ignoring the possibility that someone was born on Feb29... P = (365)(364)(363)....(365 - n + 1)/(365)^n where n = 30. P = .29368, which is actually the probability that no two people have the same birthday, so 1 - .29368 = .70632 => .71 So in a room with 30 people, chances are more likely than not that two people will have the same birthday. May seem counterintuitive but it's true. __________________ I write for Cracked. So can you! Last edited by Neo : 11-30-2005 at 11:37 PM.