Interesting maths question

Imagine a speaker is in front of 300 random people in an auditorium. Starting at the first row, he asks that person their date of birth. Then he asks the next person, and so on. Assume that everyone will give an honest answer. We are only interested in the day and month, not the year.

Question 1 : On average, how many people will the speaker have to ask before a birthdate is repeated?

Question 2 : What is the chance that he will make it through all 300 people and not have any birthdate repeated?

I'm not sure how to do a, but b is

(364 factorial/365*300*65 factorial)

Well this isn't comedy really. It's just interesting that it is such a low number. It makes mathematical sense, but when he asked the audience what they thought the answer was, a lot of people said 182 (at which point there is about a 50% chance that the next person will say a date already mentioned) and others just called out random numbers. I think the lowest guess was 100.

The answer is 23.

When the second person calls out their birthdate, there is already 1 known birthdate, so there are 364 left, and therefore a .274% chance that it will be the same date.

When the third person calls out their birthdate, there are 2 known birthdates, leaving 363, and therefore a 0.5479% chance that it will share a date with one of them. HOWEVER, you need to multiply by the chance that there hasn't already been a match. The best way I can explain this is multiply the chance there has not been a cumulative match, with the current chance there will not be a match (The chance of there not being a match is 100% minus the chance of there being a match) So in this case it would be 99.726% (100 - 0.274) * 99.4521% (100 - 0.5479) which makes the cumulative total of there not being a match 99.1795%; therefore there is 0.8205% there will be a match.

You carry this cumulative total forward until you get to number 23, at which point there is a 50.792% chance that a match will have occurred by that time.

Before he explained the math behind it (which I don't think I have explained adequately, sorry), he went through the audience about half a dozen times and did just as I had suggested in the problem. He never got to 30.

For the second question, what's the chance that 300 random people will have no matching birthdates? I really shouldn't have chosen 300. I can't even get it to display in Excel. It's 6.2453E-80, which if my limited memory of maths serves, that means there is a 0.[insert 79 zeros]62453% chance that this will happen.

Makaveli, I tried putting that equation into Excel, and it turned up a #NUM! response. I know solving this problem can be expressed as a formula, but I took the long hand approach as I don't remember how to express it :)