Prof. Templeton Posted August 9, 2008 Report Share Posted August 9, 2008 There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet. How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents 366, a full calendar year, plus me. Although in reality, it rarely takes that many people Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents 181? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 unreality said: Reveal hidden contents 181? Reveal hidden contents it's fewer than that (by quite a bit) Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 I figured it would be... hang on a sec Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 can we safely say there are 365 days in a year and each day is equally likely for a birthday? Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 9, 2008 Author Report Share Posted August 9, 2008 Justsnapd8 said: Reveal hidden contents 366, a full calendar year, plus me. Although in reality, it rarely takes that many people Nope. Reality and the answer are less. unreality said: Reveal hidden contents 181? Nope And as a note, we'll disregard leap years Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents 50 Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 can I get a definitive answer to this: unreality said: can we safely say there are 365 days in a year and each day is equally likely for a birthday? Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 9, 2008 Author Report Share Posted August 9, 2008 unreality said: can we safely say there are 365 days in a year and each day is equally likely for a birthday? Yes and yes. Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents A, B: 1/365 A, B, C: three different pairs, 3/365 A, B, C, D: six different pairs, 6/365 5 choose 2: 10 pairs 6 choose 2: 15 pairs 19 choose 2: 171/365 20 choose 2: 190/365 the answer is 20 Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 9, 2008 Author Report Share Posted August 9, 2008 unreality said: Reveal hidden contents A, B: 1/365 A, B, C: three different pairs, 3/365 A, B, C, D: six different pairs, 6/365 5 choose 2: 10 pairs 6 choose 2: 15 pairs 19 choose 2: 171/365 20 choose 2: 190/365 the answer is 20 This is VERY close to the answer I have. It doesn't seem possible does it? Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 yes it does... ANY pairs, not just you and someone if one of the people had to be you, it would be something like 365 or w/e Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents I think the issue is to look at the probability of someone NOT sharing a birthday, which for the first person is 364/365, the second person is 363/365 and so on. If you multiply these together again and again you will eventually get a fraction which is less than 0.5 The number of multiplications will be the number of people to get a 50% probability. I think that after 22 such calculations the result is 0.49273 but I'm not very good at that. So my answer is . . . ABOUT 22 Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 9, 2008 Author Report Share Posted August 9, 2008 StuPot said: Reveal hidden contents I think the issue is to look at the probability of someone NOT sharing a birthday, which for the first person is 364/365, the second person is 363/365 and so on. If you multiply these together again and again you will eventually get a fraction which is less than 0.5 The number of multiplications will be the number of people to get a 50% probability. I think that after 22 such calculations the result is 0.49273 but I'm not very good at that. So my answer is . . . ABOUT 22 The first person in the room would be 365/365 and the second person would be 364/365 so the answer Reveal hidden contents 23 The first time I saw the answer it seemed incredible that so few people were needed to get at least a 50/50 outcome Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 It was on here some where - i think it was a cinema queue Reveal hidden contents about two football teams i think Quote Link to comment Share on other sites More sharing options...
0 unreality Posted August 9, 2008 Report Share Posted August 9, 2008 Prof. Templeton said: The first person in the room would be 365/365 and the second person would be 364/365 so the answer Reveal hidden contents 23 The first time I saw the answer it seemed incredible that so few people were needed to get at least a 50/50 outcome why doesn't Reveal hidden contents 20 work? I think I can prove it: Reveal hidden contents 20C2 or C(20,2) or Combination(20,2) or 20 choose 2 or however you want to say it is 190, ie, there are 190 unique pairs (380 if in either order, but that doesn't matter, so 190) of people from within 20 people. In each of those 190 pairs, if person A has a set birthday, and person's B chance of having the same birthday is 1/365, so each pair has a 1/365 chance of getting a match. With 190 pairs present, there is a 190/365 chance of a match happening, which is a little over 52% 20 is less than 23 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 OMG all my HS math is flooding back to me! This is a cool one, but I definitely heard it before. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 two, including yourself. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 I remember my math teacher talking about something about this back in grade 8... He went, I will bet that someone here will have the same birthday as someone else. (class size is ~30) The funniest thing I remember was that my class, was the one of two classes that he taught (around 20) that didn't have any same birthdays. Great puzzle Templeton Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 Reveal hidden contents This is easy. The answer is 366. If everyone had a different birthday for every day of the 365 days of the year the 366th person would have to match someone. (I am pretending its not a leap year which I don't really think would make a difference) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 jpisanic said: Reveal hidden contents This is easy. The answer is 366. If everyone had a different birthday for every day of the 365 days of the year the 366th person would have to match someone. (I am pretending its not a leap year which I don't really think would make a difference) You're forgetting the question says 50/50 chance. =) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 I realized that right after I posted. Doh! Here is the corrected answer. Reveal hidden contents The answer is 366 for 100% chance. If everyone had a different birthday for every day of the 365 days of the year the 366th person would have to match someone. If one wants a perfect 50:50 chance you would reduce that number by half so the answer is 182. I have actually chosen not to do statistic and probability, which could possibly reduce that number significantly, however my answer is still sound. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 I don't think there is an answer to this Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 jpisanic said: I realized that right after I posted. Doh! Here is the corrected answer. Reveal hidden contents The answer is 366 for 100% chance. If everyone had a different birthday for every day of the 365 days of the year the 366th person would have to match someone. If one wants a perfect 50:50 chance you would reduce that number by half so the answer is 182. I have actually chosen not to do statistic and probability, which could possibly reduce that number significantly, however my answer is still sound. Reveal hidden contents I think it should be 183, not 182. Half of 365 (50% chance) plus one for remainder. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 52 Quote Link to comment Share on other sites More sharing options...
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Prof. Templeton
There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet.
How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ).
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