# Probability problem, birthdays **URL:** \ **Category:** Factual Questions\ **Created:** [April 23, 2001, 6:43pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026 "2001-04-23T18:43:58Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![aubries](https://avatars.discourse-cdn.com/v4/letter/a/53a042/32.png) [@aubries](https://boards.straightdope.com/u/aubries)\ **Post date:** [April 23, 2001, 6:43pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/1 "2001-04-23T18:43:58Z")
What is the probability that of eight randomly selected people, at least two have the same birthday? Assume no leap years. I tried to set this up as a binomial formula problem, with p=(1/365), but I’m sure that can’t be right. -Ed ---
**Author:** ![ftredeau](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@ftredeau](https://boards.straightdope.com/u/ftredeau)\ **Post date:** [April 23, 2001, 6:55pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/2 "2001-04-23T18:55:26Z")
The probability is 0.0743353… You’re welcome. ---
**Author:** ![aubries](https://avatars.discourse-cdn.com/v4/letter/a/53a042/32.png) [@aubries](https://boards.straightdope.com/u/aubries)\ **Post date:** [April 23, 2001, 7:38pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/3 "2001-04-23T19:38:19Z")
> [@](#): > > \*Originally posted by ftredeau \* > \*\*The probability is 0.0743353… > > You’re welcome. \*\* Thanks. How did you do that? -Ed ---
**Author:** ![JeffB](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jeffb/32/188_2.png) [@JeffB](https://boards.straightdope.com/u/JeffB)\ **Post date:** [April 23, 2001, 7:48pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/4 "2001-04-23T19:48:49Z")
Here are a couple of sites that explain it: [Ask Dr. Math FAQ](http://forum.swarthmore.edu/dr.math/faq/faq.birthdayprob.html) [Two People with the Same Birthday](http://www.people.virginia.edu/~rjh9u/birthday.html) The second one has a nice graph. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [April 23, 2001, 7:54pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/5 "2001-04-23T19:54:26Z")
With p=1/365 the answer is (7+6+5+4+3+2+1)/365 = 0.076712329 ---
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [April 23, 2001, 8:02pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/6 "2001-04-23T20:02:48Z")
Sailor is way off. By that logic, the same problem with 50 people would have well over 100% probability, clearly wrong. The correct way is to envision the following: First phrase the question the opposite way, what are the odds that none share a common birthday? The first person can have any ol’ day, 365/365. The second person can have any day but the 1st persons, so 364/365. The third person can have any day but the 1st and 2nd’s, so 363/365. and so on. The answer to that question is (364_363_362_361_360_359_358)/(365^7) The answer to the OP is that number subtracted from one. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [April 23, 2001, 8:14pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/7 "2001-04-23T20:14:05Z")
I take that back, … never mind ---
**Author:** ![ftredeau](https://avatars.discourse-cdn.com/v4/letter/f/bbe5ce/32.png) [@ftredeau](https://boards.straightdope.com/u/ftredeau)\ **Post date:** [April 23, 2001, 8:20pm UTC](https://boards.straightdope.com/t/probability-problem-birthdays/63026/8 "2001-04-23T20:20:16Z")
For n people: p = 1 - {[364!/(365-n)!]/365^(n-1)} I mean it. You’re welcome. Really.