cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A343015 Decimal expansion of the probability that at least 2 of 23 randomly selected people share a birthday, considering leap years.

Original entry on oeis.org

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
Offset: 0

Views

Author

Amiram Eldar, Apr 02 2021

Keywords

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 leap-year 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.

Examples

			0.50687609316527845522243931316051123777352699825485...

Links

Crossrefs

Cf. A011763, A014088, A333507.

Programs

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

Formula

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

A099600 Inventory application: Numerator of the probability (rounded to the nearest trillionth: (10^-12)-th) of at least one shortage per month when n customers each request one item on one independently-chosen random day of a 30-day month, but when only exactly one such item is available per day.

Original entry on oeis.org

0, 0, 33333333333, 97777777778, 188000000000, 296266666667, 413555555556, 530844444444, 640314074074, 736230320988, 815361224691, 876907483128, 922041405981, 953224843588, 973494078033, 985863508285, 992931754142, 996701485266, 998570643615, 999428257446, 999790361064, 999930120355, 999979036106, 999994409628, 999998695580, 999999739116, 999999956519, 999999994203, 999999999420, 999999999961, 999999999999, 1000000000000
Offset: 0

Views

Author

Rick L. Shepherd, Oct 24 2004

Keywords

Comments

A shortage occurs each time a customer cannot immediately be provided the item (that is, on the day it is requested). The denominator is always 10^12 here. Inspired by just-in-time inventory considerations for pharmacies, where new prescriptions and refills are usually for 30-day supplies, many medications are expensive and it is important to minimize cost of inventory and inconvenience of shortages to patients, pharmacists and supply clerks. Of course similar sequences can be computed for probabilities of at least (or exactly) m shortages per month under the same conditions and/or with higher item availabilities.

Examples

			a(2) = 33333333333, so if only two customers each request the item on random days during the month, there is only a .033(3333...) or ~3.3% probability that they would do so on the same day - causing a shortage for the second customer. However, "collisions" of needs/wants go up rapidly with the number of customers.
a(7) = 530844444444, so 7 = A033810(30) is the least number of customers for which the probability of at least one shortage during the month, 0.530844444444, exceeds 0.5 or 50%. This corresponds to the solution of the Birthday Problem, where A014088(2) = 23 = A033810(365), on a planet (or moon) with a 30-day "year".

Links

Crossrefs

Cf. A014088, A033810.

Programs

  • PARI
    a(n) = if(n>30, 10^12, round(10^12*(1.-30!/((30-n)!*30^n))))
for(k=0,31,print1(a(k),","))

Formula

a(n) = round(10^12 * (1. - 30!/ ((30-n)! * 30^n) )) for 0 <= n <= 30. This corresponds, with d=30, to P_2(n, d) at the MathWorld Birthday Problem link. Clearly the probability is 1 and so a(n) = 10^12 here for all n > 30.

A219330 Number of random selections (with replacement) needed from a normal population to assure a greater than one-half chance that the selected group contains the top 10th percentile individual, top 1st percentile individual, the 0.1 percentile, 0.01 percentile etc...

Original entry on oeis.org

7, 69, 693, 6932, 69315, 693147, 6931472, 69314718, 693147181, 6931471806, 69314718056, 693147180560, 6931471805600, 69314718055995, 693147180559945, 6931471805599453, 69314718055994531, 693147180559945310, 6931471805599453094, 69314718055994530942
Offset: 1

Views

Author

Zacariaz Martinez, Apr 11 2013

Keywords

Comments

a(n) is the least number of random selections (with replacement) required that assures a group having at least a 0.5 probability of containing the top 100/(10^n)th percentile individual. Converges toward log(2)*10^n with increasing n.

Examples

			To assure a greater than 1/2 chance of finding an individual with, say, an IQ in the top one out of a thousand or 1/(10^3), you must select at least (with replacement) a(3) = 693 individuals.

References

  • F. Reif, Fundamentals of Statistical and Thermal Physics, McGraw-Hill, 1965

Crossrefs

Cf. A002162, A014088, A033810, A050255.

Programs

  • PARI
    a(n)=ceil((log(2))/(log(10^n/(10^n-1)))); \\ Joerg Arndt, Apr 15 2013

Formula

a(n) = ceiling((log(2))/(log(10^n/(10^n-1)))).
Previous Showing 11-13 of 13 results.

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