Two barrels problem from the Intuitive Explanation (answered)

post by MarkusRamikin · 2011-07-13T07:11:57.387Z · LW · GW · Legacy · 7 comments

Contents

7 comments

I'm not sure if I'm doing something wrong here. EDIT: Yup, I'm allowing myself to be tricked.

I've finally sat down to reading http://yudkowsky.net/rational/bayes carefully, and I solved all story problems so far with no trouble. However, now I'm at this one:

Q.  Suppose that there are two barrels, each containing a number of plastic eggs.  In both barrels, some eggs are painted blue and the rest are painted red.  In the first barrel, 90% of the eggs contain pearls and 20% of the pearl eggs are painted blue.  In the second barrel, 45% of the eggs contain pearls and 60% of the empty eggs are painted red.  Would you rather have a blue pearl egg from the first or second barrel?
A.  Actually, it doesn't matter which barrel you choose!  Can you see why?

This doesn't look right to me.

In the first barrel, we have 18% blue eggs that contain pearls, and an unknown number of blue eggs that do not contain pearls, anywhere between 10% (worst case) and 0%. Depending on that, the proportion of blue eggs with pearls among all blue eggs can only be between 18/(18+10) = 64ish% in the worst case, to 100% in the best.

In the second barrel, we don't know how many pearls eggs are blue. We do know there are 45% eggs with pearls altogether, therefore 55% without pearls, and out of the latter 60% are red therefore 40% are blue. That means we have 40%*55% = 22% empty blue eggs. Pearl blue eggs are anywhere between 0 and 45%, so from 0% to 45/(45+22) = 67ish%.

Were we just supposed to conclude that there isn't enough information to answer that problem? But I'd say "anywhere between 64% and 100%" is a better shot than "anywhere between 0% and 67%". If I actually had to choose, and there were valuable pearls at stake, I'd choose the first barrel. Am I making some sort of a mistake?

7 comments

Comments sorted by top scores.