So I'm watching a Khan Academy video where it shows you how to solve the birthday problem. He point out that an easy way to do it is to solve first how many people DO NOT have the same birthday. Once you get that you subtract from the 100% of people and get the prob of how many people do.
So the problem I was given was the prob that at least 3 people share the same birthday in a group of $100$.
Following the video I get $\frac{(365!)}{(365-100)! \cdot 365^{100}} = 3.07$ but this does not make any sense since $3.07$ is larger than 1 and if I subtract $3.07$ from 1 i get a negative number.
The probability that at least two people among 100 share birthday $$1 - \frac{365 \cdots 265}{{365}^{100}} = 1- \frac{365!}{(365-100)! \cdot 365^{100}}$$ the second term is $\approx 3\cdot 10^{-7}$ so we are almost sure that some two people share the same birthday. This is not very surprising as already at $\approx 23$ people the chance is greater than $50\%$ ( famous problem I had in school ).
If any two people share a birthday, to avoid that a third shares the same birthday as any of the previous people would be $$365 \cdot 365 \cdot 364 \cdot 364 \cdots 315 / 365^{100} = \frac{\left(\frac{365!}{315!}\right)^2}{365^{100}} \approx 0.09\%$$
The complement to this is $1 - 0.0009 \approx 99.9\%$, so if my way of calculating works it would be almost $99.9\%$ chance that at least some 3 people among 100 should have the same b-day.