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Search: id:A091715
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| A091715 |
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Numerator Q of probability P=Q(n)/365^(n-1) that three or more out of n people share the same birthday. |
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+0
3
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| 1, 1457, 1326781, 966556865, 616113172585, 359063094171965, 196176047915944825, 102076077386001384485, 51120278427593115164425, 24824896058243745467563925, 11753675337747799989826426225
(list; graph; listen)
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OFFSET
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3,2
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COMMENT
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A 365 day year and a uniform distribution of birthdays throughout the year is assumed. The probability that 3 or more out of n people share a birthday equals the probability A091674(n)/365^(n-1) that 2 or more share a birthday minus the probability A091673(n)/365^(n-1) that exactly 2 share a birthday.
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LINKS
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Patrice Le Conte, Coincident Birthdays.
The Math Forum (AT) Drexel, Three Share a Birthday. Ask Dr. Math
Eric Weisstein's World of Mathematics, Birthday Problem. Section in World of Mathematics.
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FORMULA
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a(n)=A091674(n)-A091673(n)
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EXAMPLE
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The probability that 3 or more people in a group of 10 share the same birthday is a(10)/365^9=102076077386001384485/114983567789585767578125~=8.87744913*10^-4.
The probability exceeds 50% for n>A014088(3)=88.
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CROSSREFS
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Cf. A014088, A091673 Probabilities for exactly two, A091674 Probabilities for two or more.
Sequence in context: A020396 A052237 A145529 this_sequence A035764 A107560 A023074
Adjacent sequences: A091712 A091713 A091714 this_sequence A091716 A091717 A091718
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KEYWORD
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frac,nonn
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AUTHOR
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Hugo Pfoertner (hugo(AT)pfoertner.org), Feb 04 2004
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EXTENSIONS
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Broken links corrected by S. R. Finch (Steven.Finch(AT)inria.fr), Jan 27 2009
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