[Request] What's the chance that any given day is one of your ancestors' birthday?
Is it just number of ancestors over 365 or more complicated than that.
Posted Mar 18, 2016, 5:28 PM UTC
16
Is it just number of ancestors over 365 or more complicated than that.
Posted Mar 18, 2016, 5:28 PM UTC
[removed]
If a given person has a 1 in 365 (.548%) chance of sharing your birthday, and 2 people would have a 2 in 365 (1.096%) chance, by extrapolating out by generations, you have two parents, they each have two parents, etc. By the ninth generation, you would have 512 ancestors, and 512 in 365 = 140.274%
There is a 100% chance that any given day is an ancestor's birthday.
Not quite because not every date HAS to be used. It can't get above 100% if I'm not mistaken
So the simple model says that, including leap years there's a 364.25/365.25 chance of any particular day not being someone's birthday. This assumes birthdays are evenly spread out - which is false, there is an increased chance of people being born 9 months after winter (when exactly depends on the hemisphere) and there are a couple of other large scale trends like that. Eg: Mid-November for Valentines day sex babies.
I'm going to be lazy and use the simple model...
Now the number of ancestors you have is an interesting question - because simple maths says in each generation you have a total of 2^n ancestors, where n is the number of generations back it goes. The problem with this model is that it overlooks inbreeding, which becomes quite rampant once you start looking more than a couple of centuries back (I'm going to ignore this fact). It also doesn't tell us the total number of ancestors you have going back n generations, just the number in that generation. The total is actually given by:
2^n+1 - 2
ie. Going one generation back you have 2 - your parents, 2 generations = 6, 2x parents + 4x grandparents etc.
So the chance of it being one of a group of people's birthday is given by: 1-(364.25/365.25)^x where x is the number of people
We can combine the two of them to give:
P(an ancestor in n generations birthday) = 1-(364.25/365.25)^(2^n+1 -2)
As I mentioned, you will be the result of some mild inbreeding (nothing serious - it doesn't really have an effect so long as there's about 2-4 generations of non-inbreeding in between) so you might have to actually go to 12 generations to hit that 99.999% confidence. But given that mankind has existed for approximately 10,000 generations - it's almost certain that one of your ancestors has had a birthday on every day of the year
✓
Thanks!
Confirmed: 1 request point awarded to u/hilburn. ^[History]
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