I recall getting a D+ in the last high school math class I took. That said…
There are nine workers in my office. Three of them share the same birthday. What are the odds of that happening?
Please show your work 
mmm
The odds are slim. 
Rather than thinking about the odds of people sharing a birthday, you need to look at the odds of people NOT sharing a birthday.
Fortunately, someone has done the work for you.
According to the link - if I’m reading it correctly - the probability of 3 people out of a group of 9 sharing a birthday is about 0.0006.
How does that translate into ‘x out of x’?
6 out of 10,000 (or 1 out of 1666)?
mmm
I assume that none are siblings. In my daughter’s primary school class there were two sets of twins who were born on the same day/month/year. I imagine that the odds against four in a class of 20 or so sharing the exact same birth date are pretty high.
There are 84 groups of 3 people within 9 people. ABC, ABD, ABE etc.
The chances of each of the two “other” people in a group of three matching birthdays with the first person in the group are 1 in 365.
So the chances of all 3 people sharing the same birthday is 1 in 365 * 365 or 1/ 133,225.
So for all 84 groups it is 84/133,225 or about 1 chance in 1,586.
I get a slightly lower probability than don’t ask, whose calculation I think includes the few cases where there is more than one group of three people sharing the same birthday, and also cases where there are more than three people with the same birthday.
I interpreted the OP as “there are three people who have the same birthday, and nobody else shares a birthday with anybody.” I ignored the possibility of birthdays on February 29.
So, 84 ways of choosing three people, as said. Their common birthday has 365 possibilities. The other six people have 364 x 363 x … x 359 possibilities. And the total number of birthday permutations is 365^9.
84 x 365 x 364 x … x 359 / 365^9 ~= 0.000595, or about one chance in 1,681.
Are we talking about the same date, or the same date and year?
So I used don’t ask’s method but included February 29th … [adorable grin] … not worth mentioning the difference … so I won’t …
Ximean and don’t ask are clearly only looking at date, not year as well.
The bigger point, though, is that, starting from scratch, the odds that a particular nine-person group have three people who share a birthday are roughly 1 in 1,600. Not too likely. BUT there are a lot of existing nine-person groups in the U.S. – thousands if not millions. So we’d expect to find quite a few existing nine-person groups who happened to have a three-person shared birthday, so if someone finds one, it’s not really a sign of cosmically ordained fate; there’s no reason to think it’s not coincidence.
Date only, in my example.
mmm
Nitpick: Setting aside February 20, calculations assume that the 365 dates are equally likely for birth. This is not the case. In the U.S., births peak in summer and early autumn, and are lowest in springtime.
A study of several hundred thousand insurance applicants found dates in September to be 12% more common than dates in April. Perhaps it’s climatic differences nine months prior that explain this.
I used an online calculator that calculates the probability of at least 3 people sharing the same birthday, not exactly 3.
For 356 days in a year it calculated 1 in 1606 and for 365.25 days 1 in 1608.
I found a calculator that does both “at least” and “exactly”, but it’s not working right for me.
There are all kinds of different things we can and could calculate. I will calculate the probability that there is ANY group of 3 people with the same birthday (whether or not there is more than one such group, and whether or not such group is in fact part of a group of more than three people who share the same birthday), given 9 people whose birthdays are independently uniformly drawn from 365 possible days (ignoring leap years for convenience).
This is precisely 1 - 114911963532096447717000/365^9, which is a hair over 0.0006227. It’s 1 in a hair over 1605.8.
The calculation is as noted in the very first link: given 9 distinct people to be assigned one of 365 birthdays each in one of 365^9 equally probable ways overall, there’s the number of ways everyone could have distinct birthdays (104103381771080265974400), the number of ways precisely 1 pair could have a matching birthday and no other matches (10497820010529102451200), the number of ways precisely 2 pairs could have matching birthdays and no other matches (307896955616077027200), 3 pairs (2858838956509536000), 4 pairs (5955914492728200), and finally, the leftover is the number of ways to have what we’re interested in happen.
As for how to calculate the number of ways to have precisely P pairs match with no other matches: there are (9 choose 2P) ways to choose which people are in the pairs, and (2P - 1)!! ways to pair those 2P people off [!! is like factorial but only includes odd factors], and then 365 * 364 * 363 * …, with 9 - P factors, many ways to assign distinct birthdays to each of these pairs as well as each of the remaining people.
So, for example, the number of ways to have precisely 3 pairs match with no other matches is (9 choose 2 * 3) * (2 * 3 - 1)!! * (365 * 364 * 363 * …) with (9 - 3) factors. This yields the 2858838956509536000 above.
[On edit: Ah, just scooped by x-ray vision, but I’ve at least included all the relevant details that you can recreate and understand this calculation on your own]
For exactly 3 people within 356 days, the calculator spit out 0.0005950549892282834744463
Not sure what took you so long. I did it in seconds.
You keep saying 356 where you mean 365. But, yes.
The number of ways for exactly 3 people to share a birthday, and no one else to share any birthdays with anyone, is (9 choose 3) ways to choose the 3 people, * 365 ways to give those sharers a birthday, * 364 * 363 * 362 * 361 * 360 * 359 to give the other 6 people distinct birthdays. This comes out to 68421545692461561600. Dividing this by 365^9, we get 0.0005950549892…
[On edit, re: “Not sure what took you so long. I did it in seconds”: :D]
Doh! I probably typed it wrong once and copied and pasted from there. I can’t remember now, it’s been several minutes. Getting old.
For a real answer this is one of those things that could have a multitude of answer varying greatly the odds if all factors were considered. Including knowing details of the parents lives. You could even look at slightly less general info such as the areas birthrate for that month. Lots of things could and probably did change the odds of that happening.
These calculations are also inaccurate in assuming the the ‘9 people in my office’ are a random, independent group. But that’s unlikely to be true in msny offices.
In a small office, there might be several people from the same family. Birthdates tend to cluster together in families. (Especiaslly pre-birth control. There were several families in my grade school how had multiple children born in the same month of the year.)
Also, office groups are often about the same age. And similar race & ethnic groups. Those could also lead to non-random birthdates.
Yes. All computations so far have assumed that an equal number of people are born on every day of the year, and we know that’s not true. So to do a better job, a first pass would be to take the probability distribution of births over the year into account. That’s complicated enough to move this out of the Mathematical Recreations-style pen-and-paper math puzzle realm and into the realm of something you’d need an actual statistics package (such as R) to deal with reasonably.