Alejandro1 comments on LINK: Infinity, probability and disagreement - Less Wrong

2 Post author: Alejandro1 05 March 2013 04:36AM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (51)

You are viewing a single comment's thread. Show more comments above.

Comment author: Alejandro1 05 March 2013 05:27:38PM 0 points [-]

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.

Comment author: falenas108 06 March 2013 01:02:29AM 0 points [-]

Really? Whoops, didn't know that.

Comment author: Thomas 05 March 2013 05:34:29PM *  -1 points [-]

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?

Comment author: DanielLC 06 March 2013 02:12:27AM 0 points [-]

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.