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I am vaguely aware of the Pigeonhole principle and I understand that you would need 367 people to ensure that two people have the same birthday. I think that it may be required to have 734 people in a room to ensure 100% probability, since you can have 366 birthdays repeated and not have three people. Is this a correct assumption, and if so, how would you solve it without just guessing/checking?

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    $\begingroup$ You mean 733. $ $ $\endgroup$ – Did Apr 12 '15 at 16:08
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    $\begingroup$ I think the use of "probability" in your question is misleading. There are situations in which "occurs with 100% probability" is weaker than "must occur", but this is not one of them. This is indeed a direct application of the Pigeonhole Principle. It sounds like you should become more than vaguely aware of the Pigeonhole Principle: certainly it is simple enough for anyone to understand, and once you get it you'll regard questions like this as being 100% (!) straightforward. $\endgroup$ – Pete L. Clark Apr 12 '15 at 17:16
  • $\begingroup$ Any space large enough to contain that many people would probably not be considered to be a "room"; the word "hall" would seem more appropriate. $\endgroup$ – Marc van Leeuwen Apr 13 '15 at 10:35
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Your reasoning looks pretty much right (you're off by one) and looks to me like a direct approach more than guessing and checking. Maybe if we just crystalize your argument a little, you'll understand it more deeply:

There are $366$ possible birthdays; if we have $366\times 2=732$ people, then it is possible for each day, exactly two people call it their birthday. This is the maximum number of people where we can have at most $2$ per birthday. If we have one more person, then at least three people must share a birthday. Thus, in a room of $733$ people, at least three share a birthday.

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  • $\begingroup$ Oh okay! Your explanation cleared up my confusion. I was thinking it may have been +2 for some reason, instead of just +1. $\endgroup$ – Ampage Green Apr 12 '15 at 16:15

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