CronoDAS comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong

18 Post author: Stuart_Armstrong 23 September 2009 02:54PM

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Comment author: CronoDAS 23 September 2009 04:41:35PM *  1 point [-]

I'm not sure about the transition from A to B; it implies that, given that you're alive, the probability of the coin having come up heads was 99%. (I'm not saying it's wrong, just that it's not immediately obvious to me.)

The rest of the steps seem fine, though.

Comment author: gjm 23 September 2009 06:19:05PM 1 point [-]

Pr(heads|alive) / Pr(tails|alive) = {by Bayes} Pr(alive|heads) / Pr(alive|tails) = {by counting} (99/100) / (1/100) = {by arithmetic} 99, so Pr(heads|alive) = 99/100. Seems reasonable enough to me.

Comment author: eirenicon 23 September 2009 08:49:22PM *  0 points [-]

It doesn't matter how many observers are in either set if all observers in a set experience the same consequences.

(I think. This is a tricky one.)