large group of people ($>365$), what is probability that no one has a birthday today What is the probability of a large group of employees, say $2,728$, and no one having a birthday today? and, if there are $254$ actual hiring days in a year, what is the probability that also no one was hired on this date? This actually occurred at my friends company today and we are curious.
 A: The probability that the first employee does not have a birthday today is
$$
P_1 = \frac{364}{365}
$$
in a normal year (and $\frac{365}{366}$ in a leap year).
The probability that no one has a birthday is the product of all these. If $N$ is the total number of employees, then it is:
$$
P = \prod_{n=1}^N P_n = \left(\frac{364}{365}\right)^N
$$
With $N=2\,728$ you get $P = 0.0005618758$.
So if the employee count stays constant, then what you describe happens on one day in about $2\,000$ on average.
EDIT: The question about the hirings is a bit more complicated: Imagine that there is only one hiring day per year, and it's a different day each year. This means that in 2013, maybe hundreds of people are hired on October 23, in 2014 maybe dozens on March 23 etc. In this case, the probability that no one has his anniversary today is maybe zero, if today is a day that has never been a hiring day.
With 254 hiring days, the distribution is much more constant, but there are other difficulties, for example a special day like March 26 (among others), which is one of the most common Good Fridays. This day is a free day in many countries, so the probability that it is a working day is lesser than that of other days, so fewer people have been hired on that day than on other days.
So, the answer on the hirings question depends on how the hiring days are distributed among the year.
For simplicity, let's assume that the hirings are distributed in a regular pattern over the year. For example, Monday-Friday are hiring days (even on holidays), Saturdays and Sundays are not. Then the probability that no one has his hiring day anniversary is the same as above:
$$
P = \left(\frac{364}{365}\right)^N
$$
Another simple case is if the hirings are always on the same dates. For example, on the first 20 days of a month, people are hired, then for the rest of the month they're not. In this case the probability is what you expected, if the total number of hiring days is $254$:
$$
P = \left(\frac{253}{254}\right)^N
$$
On non-hiring days, the probability is $P=1$ always, as no one was hired that day anyway.
I have to add that in my calculation of birthday probability, I assumed that birthdays are evenly distributed. This is not the case in reality. As far as I know, there are more people with birthdays in Summer, due to the fact the people live closer to each other in Winter (at least in northern countries).
