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JGWeissman comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong
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Comments (228)
Cupholder:
That is an excellent illustration ... of the many-worlds (or many-trials) case. Frequentist counting works fine for repeated situations.
The one-shot case requires Bayesian thinking, not frequentist. The answer I gave is the correct one, because observers do not gain any information about whether the coin was heads or tails. The number of observers that see each result is not the same, but the only observers that actually see any result afterwards are the ones in either heads-world or tails-world; you can't count them all as if they all exist.
It would probably be easier for you to understand an equivalent situation: instead of a coin flip, we will use the 1 millionth digit of pi in binary notation. There is only one actual answer, but assume we don't have the math skills and resources to calculate it, so we use Bayesian subjective probability.
Cupholder managed to find an analogous problem in which the Bayesian subjective probabilities mapped to the same values as frequentist probabilities, so that the frequentist approach really gives the same answer. Yes, it would be nice to just accept subjective probabilities so you don't have to do that, but the answer Cupholder gave is correct.
The analysis you label "Bayesian", on the other hand, is incorrect. After you notice that you have survived the killing you should update your probability that coin showed tails to
so you can then calculate
Or, as Academian suggested, you could have just updated to directly find