

A343015


Decimal expansion of the probability that at least 2 of 23 randomly selected people share a birthday, considering leap years.


0



5, 0, 6, 8, 7, 6, 0, 9, 3, 1, 6, 5, 2, 7, 8, 4, 5, 5, 2, 2, 2, 4, 3, 9, 3, 1, 3, 1, 6, 0, 5, 1, 1, 2, 3, 7, 7, 7, 3, 5, 2, 6, 9, 9, 8, 2, 5, 4, 8, 5, 2, 6, 1, 0, 5, 6, 1, 9, 4, 1, 2, 1, 4, 3, 8, 1, 4, 1, 3, 7, 2, 5, 8, 4, 6, 7, 8, 6, 3, 3, 5, 4, 8, 4, 9, 5, 1
(list;
constant;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

The usual solution of the Birthday Problem, 1  ((365!)/((365  23)! * 365^23)) = 0.507297... (A333507), is based on the assumption that all the years have 365 days.
The solution given by Nandor (2004) includes leap years, i.e., 97 years of 366 days in each cycle of 400 years of the Gregorian calendar.
With the addition of leapyear days, i.e., the possibility of having a birthday on February 29, the probability is reduced to 0.506876...
This constant is a rational number: its numerator and denominator have 111 and 112 digits, respectively.
The sequence has a period of 7.983424...*10^108.


LINKS

Table of n, a(n) for n=0..86.
M. J. Nandor, Including Leap Year in the Canonical Birthday Problem, The Mathematics Teacher, Vol. 97, No. 2 (2004), pp. 8789.
Eric Weisstein's World of Mathematics, Birthday Problem.
Wikipedia, Birthday problem.
Wikipedia, Gregorian calendar.


FORMULA

Equals 1  (365!/((365  23)! * 365^23)) * (146000/146097)^23 * (1 + 97 * 365 * 23/146000/(366  23)).


EXAMPLE

0.50687609316527845522243931316051123777352699825485...


MATHEMATICA

RealDigits[1  (365!/((365  23)! * 365^23)) * (146000/146097)^23 * (1 + 97 * 365 * 23/146000/(366  23)), 10, 100][[1]]


CROSSREFS

Cf. A011763, A014088, A333507.
Sequence in context: A320375 A200419 A271522 * A069206 A291800 A091685
Adjacent sequences: A343012 A343013 A343014 * A343016 A343017 A343018


KEYWORD

nonn,cons


AUTHOR

Amiram Eldar, Apr 02 2021


STATUS

approved



