I'm so sorry! Please would you promise that my comments will be taken as another point of view, and that there is no intention to rain in anybody's parade ? That none will feel attacked in his/her beleifs and enjoyment ?? :))

Let's see the first problem. Dr.Math takes this FIRST PREMISE: "We'll start by figuring out the probability that two people have the same birthday.

The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%."

This premise is absolutely false. He is taking for sure that the first person has a birthday that MATCHES everyone's birthday, so its probability is 1. In fact, the probability of any given couple of persons to have the same birthday is 1/365 multiplied by 1/365 and nothing else. Thus, the whole reasoning is false.

Problem of the cups: sorry, the probability of winning when there are only two cups left, is 1/2, exactly 50%, no matter what have been the previous results.It is the same apparent paradox of the probability of 50% in a coin toss, after 6, or after 200 tails in a row: still 50%. A different thing is to calculate the probability of the whole sequence (not one toss): two tails in a row is 1/4, (will probably happen in 1 of every four sequences of two tosses), three in a raw is 1/2 * 1/2 * 1/2 = 1/8 = 12.5%,.. and so on.

Interesting !

Un abrazo - Andrés (no Math degree..snif)