Step 1, 2, and 3: 5/6 chance of not rolling six. Some infinities are bigger than others, by taking the limit you can show that there are 5 times as many people who didin't roll 6 than those who did.
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, and of those on that list, 1/2 of them rolled a 6. Until you have more knowledge, there is no reason to suspect you are on that list.
Step 5: Assuming you can know that being transported to the room means you are on that list, then by being transported you just gained new information: You are on the list, which as said before, means your probability of rolling and not rolling a six is 1/2.
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.
I saw this conundrum at Alexander Pruss's blog and I thought LWers might enjoy discussing it: