Step 4 For every person who rolled a six and is on that list, there is one person who did not: So, 2/6=1/3 of all the people will be on that list
This is wrong. The lists can be made (and in the problem, they are made) in such a way that each person will be in one of the lists. There is a one-one correspondence between people who rolled 6 and people who didn't, in the same way that, as Cantor showed, there is a one-one correspondence between even integers and all integers.
Really? Whoops, didn't know that.
I saw this conundrum at Alexander Pruss's blog and I thought LWers might enjoy discussing it: