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Jonathan_Graehl comments on Book Club Update, Chapter 3 of Probability Theory - Less Wrong
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Shouldn't it be for exchanging the number of Drawn and Good balls? (In the wikipedia example, black is good and white is defective.) If the number of bad balls is 1 and the number drawn is 2, then the probability of getting 2 good balls when drawing 2 is high, but if you exchange drawn and bad, then the probability of getting 2 good balls becomes zero.
Drawn and Good are symmetric. I recommend a Venn diagram. There are two binary conditions: good vs bad and drawn vs left, both of which are effectively random. We are interested in the intersection, which is preserved by the symmetry.
Yes, that's right. I misremembered.
I like your argument that all the balls in the urn are labeled good/bad and drawn/not, and that two processes are causally orthogonal, but it's not so simple as each ball being independently randomly labeled. It's more like: sample without replacement some number of balls and mark them Good. Then replace them all, and sample without replacement some number of balls and mark them Drawn. Naturally, I mean for a full random shuffle of the balls in the urn to occur before both samples are taken. And, as you observed, we're asking about the distribution over the number of balls with the labels (Good,Drawn). Looking at it that way, I'm absolutely convinced. Thanks.