Group of $r$ people at least three people have the same birthday? What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday?


*

*$\displaystyle 1- \frac{365\cdot364 \cdots(365-r+1)}{365^r}$  

*$\displaystyle \frac{365\cdot364
        \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots
        (364-(r-2) +1)}{364^{r-2}}$ 

*$\displaystyle 1- \frac{365\cdot364
            \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots (364-(r-2) +1)}{364^{r-2}}$ 

*$\displaystyle\frac{365\cdot364 \cdots(365-r+1)}{365^r}
                $



My attempt :
(May be typo in option $(3)$ of question !)
$$P(\text{at least 3 persons have same birthday})$$
$$= 1 - \{P\text{(no one has same birthday) + P(any 2 have same birthday)\}}$$
So, option $(3)$ is true.

Can you explain it, please?

It asked here before,  but I'm not satisfied by explanation.
 A: There are three possible situations: either 1. everyone has a different birthday, 2. there are 1 or more pairs of birthdays, but no triplets, 3. there is any triplet.
The three situations add up to all possible situations.
The calculations for each of the cases:


*

*You have the possibility to have all different birthdays: $365 \choose r$

*You have the possibility that there is any pair that has the same birthday, this can be calculated with inclusion/exclusion method, taking into account that there can be $1,2,3,..,(r/2)$ pairs.
Please invest the time to fully understand the inclusion/exclusion method. Counting the combinations with $(a,b)$ having the same birthday regardless of $c, d$, has to exclude counting twice $(c,d)$ having the same birthday (though different from $(a,b)$.

*Final solution: ${(365^r - 1. - 2.)} / {365^r}$


More elaborate discussion on:
Probability of 3 people in a room of 30 having the same birthday
Where you obviously have to generalize to $r$ iso 30. (And take the second answer, the first is only an estimation)
Note: Maybe it is more clear when you have 6 people throwing dice. What is the probability that no 3 throw the same number?
There can be 0..3 pairs of pairs. Extend this solution to 365 sided dice, and $r$ people.
A: Given $2k$ items, there are $(2k-1)!!$ ways to arrange them into pairs: the first item can be paired with $2k-1$ possibilities; the first unpaired item can be matched with $2k-3$ items; the new first unpaired item can be paired with $2k-5$ items; etc.
The number of functions from $n$ people to $365$ dates with $n-2k$ singles and $k$ pairs is
$$
\begin{array}{cc}
&\displaystyle\underbrace{
\overbrace{\binom{365}{n-k}}^{\substack{\text{ways to choose}\\\text{$n-k$ dates}\\\text{for birthdays}}}
\overbrace{\binom{n-k}{k}}^{\substack{\text{ways to choose}\\\text{$k$ dates}\\\text{for pairs}}}
}&\displaystyle\underbrace{
\overbrace{\ \ \binom{n}{2k}\ \ }^{\substack{\text{ways to choose}\\\text{$2k$ people}\\\text{for pairs}}}
\overbrace{(n-2k)!\vphantom{\binom{n}{k}}}^{\substack{\text{ways to arrange}\\\text{$n-2k$ singles}}}
\overbrace{(2k-1)!!\vphantom{\binom{n}{k}}}^{\substack{\text{ways to pair}\\\text{$2k$ people}}}
\overbrace{\ \ \ \ \ k!\ \ \ \ \ \vphantom{\binom{n}{k}}}^{\substack{\text{ways to arrange}\\\text{$k$ pairs}}}
}\\
\displaystyle=
&\displaystyle\frac{365!}{(365-n+k)!\,(n-2k)!\,k!}
&\displaystyle\frac{n!}{2^k}
\end{array}
$$
Thus, the probability of getting at least one triple is
$$
1-\frac{365!\,n!}{365^n}\sum_{k=0}^{365}\frac1{(365-n+k)!\,(n-2k)!\,k!\,2^k}
$$
where we take $\frac1{n!}=0$ for $n\lt0$.
Here is a plot from $n=0$ to $n=730$. For $n\lt3$, the probability of getting a triple is $0$, and for $n\gt730$, the probability is $1$.

