Pigeonhole Principle: birthdays on same day of week

Question

How many people must be in a room so that at least 10 have a birthday on a Friday?

edit: Assume that no two people share the same birthday

I'm somewhat confused and see two different ways to solve the problem. First, since all birthdays fall on one of seven days, if there are 64 people in the room one would at least one day of the week to hold 10 people. This is a similar problem: picking certain number of days.

However, there are 366 possible birthdays and only 52 or 53 of those are Fridays, giving $\displaystyle366 - 52= 314$ days of the year that are not Friday. So to be absolutely certain there are 10 people with Friday birthdays, would one need $314+10=324$ people?

Answer 1

The latter reasoning of the OP is correct. Assuming that we are in a normal year with 365 days in it and that this year contains 52 Fridays and that no two people share the same birthdate, then there are $365-52=313$ ways to assign people so that no one has a birthday on a Friday. After that, you are forced to fill up the empty Fridays, thus you need exactly 10 more people to do that.

Thus any collection of $313+10=323$ people will have at least 10 Friday birthdays.

Answer 2

You can never be certain that at least 10 people have birthdays on Friday; even with a billion people they might all have birthdays on Thursday.