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thomblake comments on Open Thread: April 2010 - Less Wrong
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Comments (524)
Maybe I'm misreading this, but it looks like you're missing a term...
You said: P(O) = P(B) * P(O|B)
Bayes's theorem: P(O) * P(B|O) = P(B) * P(O|B)
ne?
I agree that Jordan's equation needs to be adjusted (corrected), but I humbly suggest that in this context, it is better to adjust it to the product rule:
P(O and B) = P(B) * P(O|B).
ADDED. Yeah, minor point.
Yes, correct. I missed that. For the standard Doomsday Argument P(B|O) is probably 1, so it can be excluded, but for alternative classes of people this isn't so.