Continuity problem is that the 1⁄2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment, unless the ratio is 0 (or infinite) at which point special case logic is invoked to prevent the trivially absurd claim that credence of Heads is 1⁄2 when you are never woken under Heads.
If you are put through multiple sub-experiments in series, or probabilistically through some element of a set of sub-experiments, then the Expected number of times you are woken is linearly dependent on the distribution of sub-experiments. The probability that you are woken ever is not.
So the problem is that it’s not immediately clear what D should be. If D is split by total probability to be Heads or Tails, and the numbers worked separately in both cases, then to get 1⁄2 requires odd conditional probabilities, but 1⁄3 does not. If you don’t split D, and calculate back from 1⁄3, you get 3⁄2 as the “probability” of D. It isn’t. What’s happened is closer to E(H|D) = E(D|H) E(H) / E(D), over one run of the experiment, and this yields 1⁄3 immediately.
The issue is that certain values of “D” occur multiple times in some branches, and allowing those scenarios to be double counted leads to oddness. I second the observation that caution is generally required.
Continuity problem is that the 1⁄2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment
Why is this a problem? I’m perfectly comfortable with that property. Since you really just have one random variable in each arm. You can call them different days of the week, but with no new information they are all just the same thing
By D do you mean W?
What’s happened is closer to E(H|D) = E(D|H) E(H) / E(D), over one run of the experiment, and this yields 1⁄3 immediately.
Is this how you came up with the 1⁄3 solution? If so, I think it requires more explanation. Such as what D is precisely.
Continuity problem is that the 1⁄2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment
Why is this a problem?
The next clause of the sentence is the problem
unless the ratio is 0 (or infinite) at which point special case logic is invoked to prevent the trivially absurd claim that credence of Heads is 1⁄2 when you are never woken under Heads.
The problem is special casing out the absurdity, and thus getting credences that are discontinuous in the ratio. On the other hand, you seem to take 1/21in PSB (ie you do let it depend on the ratio) but deviate from 1⁄21 when multiple runs of PSB aggregate, which is not what I had expected...
D was used in the comment I was replying to as an “event” that was studiously avoiding being W.
The problem as I see it with W is that it’s not a set of outcomes, it’s really a multiset. That’s fine in it’s way, but it gets confusing because it no longer bounds probabilities to [0,1]. Your approach is to quash multiple membership to get a set back.
Continuity problem is that the 1⁄2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment, unless the ratio is 0 (or infinite) at which point special case logic is invoked to prevent the trivially absurd claim that credence of Heads is 1⁄2 when you are never woken under Heads.
If you are put through multiple sub-experiments in series, or probabilistically through some element of a set of sub-experiments, then the Expected number of times you are woken is linearly dependent on the distribution of sub-experiments. The probability that you are woken ever is not.
So the problem is that it’s not immediately clear what D should be. If D is split by total probability to be Heads or Tails, and the numbers worked separately in both cases, then to get 1⁄2 requires odd conditional probabilities, but 1⁄3 does not. If you don’t split D, and calculate back from 1⁄3, you get 3⁄2 as the “probability” of D. It isn’t. What’s happened is closer to E(H|D) = E(D|H) E(H) / E(D), over one run of the experiment, and this yields 1⁄3 immediately.
The issue is that certain values of “D” occur multiple times in some branches, and allowing those scenarios to be double counted leads to oddness. I second the observation that caution is generally required.
Why is this a problem? I’m perfectly comfortable with that property. Since you really just have one random variable in each arm. You can call them different days of the week, but with no new information they are all just the same thing
By D do you mean W?
Is this how you came up with the 1⁄3 solution? If so, I think it requires more explanation. Such as what D is precisely.
The next clause of the sentence is the problem
The problem is special casing out the absurdity, and thus getting credences that are discontinuous in the ratio. On the other hand, you seem to take 1/21in PSB (ie you do let it depend on the ratio) but deviate from 1⁄21 when multiple runs of PSB aggregate, which is not what I had expected...
D was used in the comment I was replying to as an “event” that was studiously avoiding being W.
http://lesswrong.com/lw/28u/conditioning_on_observers/201l shows multiple ways I get the 1/3 solution; alternatively betting odds taken on awakening or the long run frequentist probability, they all cohere, and yield 1/3.
The problem as I see it with W is that it’s not a set of outcomes, it’s really a multiset. That’s fine in it’s way, but it gets confusing because it no longer bounds probabilities to [0,1]. Your approach is to quash multiple membership to get a set back.