@Baldimo,
Quote:Are you? I should point out that you write just like Bobsal did and even had the same job in the military if I remember correctly. Funny how we could have 2 different people on this site who had the same MOS during the same time frame of service.
Is it. I live in a small town where another members sister and nephew live and none of us was born here AND the member and my wife both only had one job in their live and the retired from different divisions of the SAME company!!! NOT ONLY THAT - but they are both from the same town in an upper mid Atlantic state and to this date from when they were born - HAVE NEVER EVER MET!
I meet fellow boomers every where and WE thought we were fighting Soviet fascism the way our dads fought German and Italian and Japanese fascists.
I found this for you, whoever the hell you are:
Probability of Shared Birthdays
or, How to Win Money in Bar Bets
Copyright © 2001–2017 by Stan Brown
Summary: In a group of 30 people, would you be surprised if two of them have the same birthday? As it turns out, you should be more surprised if they don’t.
There are 365 possible birthdays. (To keep the numbers simpler, we’ll ignore leap years.) The key to assigning the probability is to think in terms of complements: “Two (or more) people share a birthday” is the complement of “All people in the group have different birthdays.” Each probability is 1 minus the other.
(a) What is the probability that any two people have different birthdays? The first person could have any birthday (p = 365÷365 = 1), and the second person could then have any of the other 364 birthdays (p = 364÷365). Multiply those two and you have about 0.9973 as the probability that any two people have different birthdays, or 1−0.9973 = 0.0027 as the probability that they have the same birthday.
(b) Now add a third person. What is the probability that her birthday is different from the other two? Since there are 363 days still “unused” out of 365, we have p = 363÷365 = about 0.9945. Multiply that by the 0.9973 for two people and you have about 0.9918, the probability that three randomly selected people will have different birthdays.
(c) Now add a fourth person, and a fifth, and so on until you have 22 people with different birthdays (p ≈ 52.4%). When you add the 23rd person, you should have p ≈ 49.3%.
(d) If the probability that 23 randomly selected people have different birthdays is 49.3%, what is the probability that two or more of them have the same birthday? 1−0.493 = 0.507 or 50.7%. In a randomly selected group of 23 people, it is slightly more likely than not that two or more of them share a birthday.
For the rest of it and a lot of formuli and equations and grapheseses
https://brownmath.com/stat/birthday.htm