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.
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.
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.
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.