For number 18, did you just assume that the probability for each day during the week is 1/7? Because when I was looking at it, I noticed that 366 doesn't divide evenly leaving two days with 53 occurrences and 5 days with 52 occurrences making the probability different. --Podarcze 14:55, 8 October 2008 (UTC)
while it is true that in any one year, the distribution of days is not even, given enough years, the distribution becomes even, and since we are not limiting ourselves to requiring the b-days happen in the same yar, but only on the same day of the week, we can safely assume that all days of the week are infact equally likely--mnoah 16:12, 8 October 2008 (UTC)
I just assumed each of the 366 days was equally likely. I calculated the probability none of them had the same birthday and then subtracted that from 1. --Norlow 15:08, 8 October 2008 (UTC)
You don't need to consider each day of the year, we are asked about the probability that randomly picked two people are born on the same day of the week. --Asuleime 02:46, 9 October 2008 (UTC)
Another interesting note: I was having some trouble with the final part of this problem, mainly due to the wording of the question. But it's very important to note that after n=8 (8 people are chosen at random), at least 2 people have to have the same birthday. This is very obvious after you read it, or think about it, but I overlooked it and thought I might be able to help anyone else who had done the same.
