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: Mallah 14 April 2010 04:00:22PM *  0 points [-]

Under a frequentist interpretation

In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.

Reading Bostrom's explanation of the SB problem, and interpreting 'what should her credence be that the coin will fall heads?' as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1/2 however many times Sleeping Beauty's later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin's probability of coming up heads is 1/2.

I am surprised you think so because you seem stuck in many-shot thinking, which gives 1/3.

Maybe you are asking the wrong question. The question is, given that she wakes up on Monday or Tuesday and doesn't know which, what is her creedence that the coin actually fell heads? Obviously in the many-shot case, she will be woken up twice as often during experiments where it fell tails, so in 2/3 or her wakeups the coin will be tails.

In the 1-shot case that is not true, either she wakes up once (heads) or twice (tails) with 50% chance of either.

Consider the 2-shot case. Then we have 4 possibilities:

  • coins , days , fraction of actual wakeups where it's heads
  • HH , M M , 1
  • HT , M M T , 1/3
  • TH , M T M , 1/3
  • TT , M T M T , 0

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

Obviously as the number of trials increases, P(heads) will approach 1/3.

This is assuming that she is the only observer and that the experiments are her whole life, BTW.

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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