Birthday Paradox (Math Questions V)

To wrap up this string of math questions... the answers to Part II and III Jerry answered very well in the comments thread. The Monty Hall Paradox (Math Questions IV) is easily google-able... but the gist of it is that if you have a one in three chance of winning a prize (or, identically, a three-card monte winning card), and the host of the game shows you a non-winner of the remaining two options and offers to switch, you have a 2/3 chance of winning if you switch, and only a 1/3 chance of winning if you stay. That is because you will only choose correctly originally 1 out of 3 times, and if you have chosen incorrectly you should switch and will therefore win. This should happen two out of three times. Essentially the host is giving you the choice of "do you want to stay with the option you've chosen? Or, do you want BOTH of the other two?" to which one would obviously say, "Well, I'd rather have the other two." Very counterintuitive and a cool problem

The birthday paradox that Jerry referred to is not a question as much as an interesting fact: In a randomly chosen group of 23 people, the probability that two people have the same birthday is about 50%. So, if you are a teacher and have a class of 23 people and ask them all their birthdays, half the time two students will have the same birthday. More mathematical explanation that you probably care about is on the Wikipedia page, which is linked to above.