tag:blogger.com,1999:blog-5944248932558389199.post2041620283942049203..comments2024-02-15T19:40:29.872-08:00Comments on neopolitan's philosophical blog: Two balls, one urnneopolitanhttp://www.blogger.com/profile/02501854905476808648noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5944248932558389199.post-76457117318443699622015-02-09T22:50:49.157-08:002015-02-09T22:50:49.157-08:00I've tried to simplify things in a revision to...I've tried to simplify things in a revision to the scenario, B. Perhaps it won't be as confusing - or rather, only confusing because the answer is counterintuitive.neopolitanhttps://www.blogger.com/profile/02501854905476808648noreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-5405474614258674952015-02-08T18:28:00.258-08:002015-02-08T18:28:00.258-08:00From your friend's point of view, the logic yo...From your friend's point of view, the logic you provide in the above paragraph seems to be the best she has to work with, so a 50% chance that the remaining ball is white would be a safe bet. But from our point of view, we know that the chance of her drawing another white ball is 1 in a million or less. At best, there was a 2 in a million chance that either ball in the urn would be white (as you chose twice with a 1 in a million chance each time). There is a 1 in 10^11 chance of both balls being white (1/1M * 1/1M). But we know that despite all odds, one of the balls was in fact white, as this is the ball your friend drew. The chance that the remaining ball is white then could be seen as 1/1M (the chance of any single ball being white) or even less (closer to the 1/(10^11) chance of both balls being white). That's all I got.<br /><br />-BAnonymousnoreply@blogger.com