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PhilGoetz comments on Open Thread: April 2010 - Less Wrong
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Both your example are still taking from your current position to the end of all humans. What I said was that you get different results if you take one decile from your position, not all the way to the end. There's no reason to do one rather than the other.
P(Observing doomsday) = P(Being in some class of people) * P(Observing doomsday | you belong to the class of people)
You get a different probability for belonging to those classes, but the conditional probabilities of observing doomsday given that you belong to those classes are different. I'm not convinced that these differences don't balance out when you multiply the two probabilities together. Can you show me a calculation where you actually get two different values for your likelihood of seeing doomsday?
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.