Thanks for posting that. I did almost exactly the same thing so it was easy to compare mine to yours. I saw my mistake right away. I calculated the chances for any given number of unique birthdays by multiplying all of the previous numbers. I can’t remember what chain of reasoning led me to do that, but when I corrected it my results match yours exactly.
The loose reasoning is that for any two particular people, the probability that they have the same birthday (assuming uniform distribution) is 1/(number of possible dates). But the number of possible pairs of people is (number of people)*(number of people - 1) which is pretty close to (number of people)^2.
These two effects cancel each other out when (number of people)^2 ~= number of possible dates.
I am the oldest of five kids, all born by Caesarean. My mother, a brother, and I all share the same birthday. And then a sister, and another brother, they share the same birthday (on a different date). Now *that’s *family planning!
I was once at a lecture on improbable sounding mathematical facts and he talked about the birthday question (for which 23 is the right answer, incidentally) and noticing that there were well over 100 in the audience stated that it was highly improbable not to have coincidental birth dates. I pointed out that there was a pair of twins in the audience so it was certain (same year too). It turned out there was someone in the audience who shared a birthday with the twins.
I was once teaching a class of 56 students and did a quick calculation that it was likely that one of the 8 students in the front row had the same birthday as another member of the class. And so it proved.
To get anything like an accurate answer to the OP you really do have to know something about the likely age distributions. Probably knowing the number in each decade would get you close.
Just a side note on the “20-something days and 50%” calculation.
The last time I did this calculation, I approached it from the viewpoint of calculating the opposite; that no two people’s birthday fell on the same day. I.E. take person 1 and calculate the odds that all the others in the room land on other days (roughly 364 in 365 each) and then proceed through the other 20-something people, do some other footwork, and subtract the result from 1. While I don’t recall right know exactly how I did it, I recall it felt easier than a previous time when I had tried calculating for one or more matches.
I.E the odds of something happening are 1 minus the odds of it not happening.