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.
If you want, you can make it even more extreme. You could have one room with one person, one with two, one with three, etc., each of which contain exactly one loser. Now it seems like the probability of having lost is infinitesimal.
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.
Yes. And there is also a way, to join every looser who does not have the 6 with the 999 winners in a bigger room.
Say, that you find yourself in one of those. How likely it is, that you are the sole looser among nearly 1000 winners?
If you want, you can make it even more extreme. You could have one room with one person, one with two, one with three, etc., each of which contain exactly one loser. Now it seems like the probability of having lost is infinitesimal.