JGWeissman 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: JGWeissman 14 April 2010 11:05:19PM *  1 point [-]

Now P(heads) = (1 + 1/3 + 1/3 + 0) / 4 = 5/12 = 0.417

This should be a weighted average, reflecting how many coin flips are observed in the four cases:

P(heads) = (2*1 + 3*1/3 + 3*1/3 + 4*0)/(2+3+3+4) = (2+1+1+0)/12 = 4/12 = 1/3
Comment author: Mallah 15 April 2010 06:00:50PM 0 points [-]

There are always 2 coin flips, and the results are not known to SB. I can't guess what you mean, but I think you need to reread Bostrom's paper.

Comment author: JGWeissman 15 April 2010 07:53:55PM 0 points [-]

It seems I was solving an equivalent problem. In the formulation you are using, the weighted average should reflect the number of wakeups.

What this results means is that SB should expect with probabilty 1/3, that if she were shown the results of the coin toss, she would observe that the result was heads.

Comment author: Mallah 15 April 2010 08:39:59PM *  0 points [-]

No, it shouldn't - that's the point. Why would you think it should?

Note that I am already taking observer-counting into account - among observers that actually exist in each coin-outcome-scenario. Hence the fact that P(heads) approaches 1/3 in the many-shot case.

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