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You are not logged in. #1 20130716 03:20:56
Birthday problemHi, I've been looking at the birthday problem (which is a statistical problem which aims at finding out the how many people you would need in a random group to be certain that two of them shared a birthday. Obviously the vacuous answer is 367, but as it turns out, there is a probability of 99% that two people will share a birthday in a group of just 57 and 50% in a group of just 23 (see: http://en.wikipedia.org/wiki/Birthday_problem)). Last edited by Au101 (20130716 04:18:36) #2 20130716 15:19:14
Re: Birthday problemHi; These are done using a well known formula. Exactly 2 is different. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20130716 22:20:12
Re: Birthday problemThank you bobbym, that's great. I've been playing with the numbers myself and I'm completely with you on 1 & 2 . 3 & 4 are utterly beyond me, though. Someone suggested the formula in the wikipedia article under collision counting as a possible approach to me, but i've no idea, really, what to do with it #4 20130716 22:30:32
Re: Birthday problemYou want to see how to use the formula? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20130717 05:09:26
Re: Birthday problemI think his suggestion was about 3 & 4, yes. Where n is the number of people in the group, k is the range of days (so k = 7 in the case where I'm trying to calculate the number of people required in the group for me to be certain that two of them have birthdays within a week of each other) and m = 365 (the number of days, excluding the 29th of February.) Using this, I was able to get the same answers as you I'm completely stuck on 3 & 4, though. I can understand that there isn't an algebraic solution to this problem, but is there a numerical one? The only advice i've been given is to try and make use of that formula, but I wouldn't know what to do with it! #6 20130717 05:13:11
Re: Birthday problemThat formula is for the expected number of collisions as far as I know, not the probability. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20130717 05:17:33
Re: Birthday problemSorry, to quote: I can't see what to do with this, either, but I suppose you could incorporate the expected total number of collisions into your calculation, along with the probability of two collisions already calculated. I just don't know how you would do that. If you know of a computer model, though, I'd be very happy just to have a numerical answer, I can understand that it would be very difficult to calculate this algebraically #8 20130717 05:26:14
Re: Birthday problemHi Au101; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #10 20130717 05:35:25
Re: Birthday problemWill post it as soon as I compute them. Please have patience it is a big calculation and will take a long time. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #12 20130717 08:33:14
Re: Birthday problemHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #16 20130717 22:28:56
Re: Birthday problemHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #18 20130718 09:40:00
Re: Birthday problemSomeone who supplied me with a different way of solving the problem. I wanted to mention him because he did not get the credit he deserved. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 