I think the logic is wrong since the probability of two persons having different birthdays is dependent on the fact that they need to have different birthdays than all the others. A simple example birthday paradox for A,B and C not having birthday on the same weekday. Each of these pairs are 1/7 in a vacuum. But given A had birthday an a Monday and B and C did not have birthday the same day. The probability of B and C not having birthday on the same day given they not having birthday on the same day as A is 1/6.
The logic you should apply is the following. Let the person enter one by one and stop the experiment if two has the same birthday.
- Person 1 enters, so cant have the same birthday as anyone else
- Person 2 enters, so there is 1/365 chance that she has the same birthday as person 1. If so the experiments stops otherwise the number of days taking goes up to 2.
- Person 3 enters, so there is 2/365 chance that he will have the same birthday as either person 1 or 2.
Now the pattern is clear. The probability of k (k < 366) persons all having different birthdays are: $$ P(k)=\prod^{k}_{i=1}\left[1-\frac{i-1}{365}\right]=\prod^{k}_{i=1}\frac{366-i}{365}=\frac{365!}{(366-k)!*365^k}$$$$ P(k)=\prod^{k}_{i=1}\left[1-\frac{i-1}{365}\right]=\prod^{k}_{i=1}\frac{366-i}{365}=\frac{365!}{(365-k)!*365^k}$$
