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.

Apparently I misunderstood the question.  Thanks to Hagen's comment, this is how I approached the question:
First, find the probability that $r$ people will each have been born on different days of the week, and then subtract this from 1 to get probability that at least two from $r$ people will have been born on the same day of the week:
$Pr(\text{At least 2 of r were born on the same day of the week.)}=1-\cfrac{1-7(7-1)...(7-r+1)}{7^r}$
Substituting: 
r=3, Pr=39%.
r=4, Pr=65%!
