User:Boris Tsirelson/Sandbox1: Difference between revisions

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: ''In this article [[leap year]]s and other details (such as a difference between winter and summer days) are ignored.''
: ''In this article [[leap year]]s and other details (such as a difference between winter and summer days) are ignored.''


To show how to calculate the [[probability]] of a group including such a match, it is simpler to first find the probability of all the birthdays being different.  Consider a group of two people.  The first person can have been born on any of the 365 days of the year, while the second must have been born on one of the other 364 days in order to not match.  For the first person alone, a match cannot happen, thus no-match has a probability 1, which is <math>\tfrac {365}{365}</math> in the sense that all the 365 possibilities lead to no-match. Given the first person's birthday, the second person's birthday differs in 364 cases, with a conditional probability of <math>\tfrac {364}{365}</math> which is 0.9973.  Multiplying these probabilities together gives a net (unconditional) probability of 0.9973 for having different birthdays.  Subtracting this number from 1.0 gives a 0.0027 probability of having the same birthday.
To show how to calculate the [[probability]] of a group including such a match, it is simpler to first find the probability of all the birthdays being different.  Consider a group of two people.  The first person can have been born on any of the 365 days of the year, while the second must have been born on one of the other 364 days in order to not match.  For the first person alone, a match cannot happen, thus the no-match probability is 1, which is <math>\tfrac {365}{365}</math> in the sense that all the 365 possibilities lead to no-match. Given the first person's birthday, the second person's birthday differs in 364 cases, with a conditional probability of <math>\tfrac {364}{365}</math> which is 0.9973.  Multiplying these probabilities together gives a net (unconditional) probability of 0.9973 for having different birthdays.  Subtracting this number from 1.0 gives a 0.0027 probability of having the same birthday.





Revision as of 01:31, 9 July 2010

In this article leap years and other details (such as a difference between winter and summer days) are ignored.

To show how to calculate the probability of a group including such a match, it is simpler to first find the probability of all the birthdays being different. Consider a group of two people. The first person can have been born on any of the 365 days of the year, while the second must have been born on one of the other 364 days in order to not match. For the first person alone, a match cannot happen, thus the no-match probability is 1, which is in the sense that all the 365 possibilities lead to no-match. Given the first person's birthday, the second person's birthday differs in 364 cases, with a conditional probability of which is 0.9973. Multiplying these probabilities together gives a net (unconditional) probability of 0.9973 for having different birthdays. Subtracting this number from 1.0 gives a 0.0027 probability of having the same birthday.


I have always felt that, if one day someone came up with a contradiction in mathematics, I would just say, "Well, those crazy logicians are at it again," and go about my business as I was going the day before.[1]

  1. Vaughan Jones. See Casacuberta & Castellet 1992, page 91.

References

Feynman, Richard (1995), The character of physical law (twenty second printing ed.), the MIT press, ISBN 0 262 56003 8.

Gowers, Timothy, ed. (2008), The Princeton companion to mathematics, Princeton University Press, ISBN 978-0-691-11880-2.

Mathias, Adrian (2002), "A term of length 4,523,659,424,929", Synthese 133 (1/2): 75–86. (Also here.)

Casacuberta, C & M Castellet, eds. (1992), Mathematical research today and tomorrow: Viewpoints of seven Fields medalists, Lecture Notes in Mathematics, vol. 1525, Springer-Verlag, ISBN 3-540-56011-4.