Another Birthday Problem (Probability/Combinatorics) What is the smallest number of people in a room to assure that the probability that at least two were born on the same day of the week is at least 40%?
I understand when approaching this type of problem, you simplify it so there's only 365 days.  Also, I thought you go about the question by finding the probability that no one is born on the same day of the week.  Then you subtract by 1 to get the solution:
Therefore, if the first person can have a birthday on any of 365 days, and the second is (365-8) because 1 week has to be removed (since question asks at least two born on the same day).  I thought the answer is:
$1-\cfrac{365(365-8)...(365-r+1)}{365^{r}}\tag{1}$
The solution is 4 people but when I enter r=4, I get 5.6% which is obviously wrong.  Any help is appreciated.  Thank you.

Here are the first several results for the same day of the week:
$$
\begin{align}
1-\frac77&=0&\text{$1$ person}&(0\%)\\
1-\frac77\frac67&=\frac17&\text{$2$ people}&(14.29\%)\\
1-\frac77\frac67\frac57&=\frac{19}{49}&\text{$3$ people}&(38.78\%)\\
1-\frac77\frac67\frac57\frac47&=\frac{223}{343}&\text{$4$ people}&(65.01\%)\\
\end{align}
$$
