https://github.com/woctezuma/consecutive-birthdays-probability

Probability that at least two people in a group are born on two consecutive days
https://github.com/woctezuma/consecutive-birthdays-probability

birthday birthdays consecutive probability stackoverflow-questions

Last synced: about 1 month ago
JSON representation

Probability that at least two people in a group are born on two consecutive days

• Host: GitHub
• URL: https://github.com/woctezuma/consecutive-birthdays-probability
• Owner: woctezuma
• License: mit
• Created: 2011-01-20T18:20:06.000Z (almost 14 years ago)
• Default Branch: master
• Last Pushed: 2024-12-02T21:26:00.000Z (about 1 month ago)
• Last Synced: 2024-12-02T22:27:24.258Z (about 1 month ago)
• Topics: birthday, birthdays, consecutive, probability, stackoverflow-questions
• Language: Python
• Homepage:
• Size: 26.4 KB
• Stars: 0
• Watchers: 3
• Forks: 0
• Open Issues: 1
• Metadata Files:
  • Readme: README.md
  • License: LICENSE

Awesome Lists containing this project

README

        
# Consecutive birthdays probability

[![Build status][Build image]][Build]

[![Code coverage][Codecov image]][Codecov]

  [Build]: 

  [Build image]: 

  [PyUp]: https://pyup.io/repos/github/woctezuma/Consecutive-birthdays-probability/

  [Dependency image]: https://pyup.io/repos/github/woctezuma/Consecutive-birthdays-probability/shield.svg

  [Python3 image]: https://pyup.io/repos/github/woctezuma/Consecutive-birthdays-probability/python-3-shield.svg

  [Codecov]: https://codecov.io/gh/woctezuma/Consecutive-birthdays-probability

  [Codecov image]: https://codecov.io/gh/woctezuma/Consecutive-birthdays-probability/branch/master/graph/badge.svg

Let n be a number of people. At least two of them may be born on the same day of the year with probability:

> 1 − ∏_{i=0}^{n−1} (365−i)/365

But what is the probability that at least two of them are born on two consecutive days of the year (considering December

31st and January 1st also consecutive)? Simulating pseudo-random integers with Python, we get a 99%-confidence interval

for any number of people. It seems a good approximation is:

> (1 − (n−1)(2×365+n−1)) × (1 − ∏_{i=0}^{n−1} (365−2*i)/365)

Here is the correct formula:

> 1 − ∑_{k=1}^{n} 1/((365^(n−k))*k) × (∏_{i=1}^{k−1} (365−(k+i))/(365×i)) × (∑_{j=0}^{k−1}(−1)^j C_j^k (k−j)^n)

An explanation can be found [here](http://math.stackexchange.com/questions/18268/consecutive-birthdays-probability/18363#18363).