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neq1 comments on Conditioning on Observers - Less Wrong
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"Of course P(W) isn't bound within [0,1]"
Of course! (?) You derived P(W) using probability laws, i.e., solving for it in this equation: P(H)=P(H|W)P(W), where P(H)=1/2 and P(H|W)=1/3. These are probabilities. And your 1/3 solution proves there is an error.
If two variables have correlation of 1, I think you could argue that they are the same (they contain the same quantitative information, at least).
No. You will wake on Monday with probability one. But, on a randomly selected awakening, it is more likely that it's Monday&Heads than Monday&Tails, because you are on the Heads path on 50% of experiments
What is this random selection procedure you use in the last para?
("I select an awakening, but I can't tell which" is the same statement as "Each awakening has probability 1/3" and describes SB's epistemic situation.)
Random doesn't necessarily mean uniform. When Beauty wakes up, she knows she is somewhere on the tails path with probability .5, and somewhere on the tails path with probability .5. If tails, she also knows it's either monday or tuesday, and from her persepctive, she should treat those days as equally likely (since she has no way of distinguishing). Thus, the distribution from which we would select an awakening at random has probabilities 0.5, 0.25 and 0.25.
This appears to be where you are getting confused. Your probability tree in your post was incorrect. It should look like this:
If you think about writing a program to simulate the experiment this should be obvious.
No, because my probability tree was meant to reflect how beauty should view the probabilities at the time of an awakening. From that perspective, your tree would be incorrect (as two awakenings cannot happen at one time)
After the 1000 experiments, you divided 500 by 2 - getting 250. You should have multiplied 500 by 2 - getting 1000 tails observations in total. It seems like a simple-enough math mistake.
No, that's not what I did. I'll assume that you are smart enough to understand what I did, and I just did a poor job of explaining it. So I don't know if it's worth trying again. But basically, my probability tree was meant to reflect how Beauty should view the state of the world on an awakening. It was not meant to reflect how data would be generated if we saw the experiment through to the end. I thought it would be useful. But you can scrap that whole thing and my other arguments hold.
Well you did divide 500 by 2 - getting 250. And you should have multiplied the 500 tails events by 2 (the number of interviews that were conducted after each "tails" event) - getting 1000 "tails" interviews in total. 250 has nothing to do with this problem.
No, P(H)=P(H|W)P(W) is incorrect because the W in P(H|W) is different than the W in P(W): the former is a probability distribution over a set of three events, while the latter is a boolean. Using the former definition, as a probability distribution, P(W) isn't meaningful at all, it's just a type error.
It isn't a probability; the only use of it was to note the method leading to a 1/2 solution and where I consider it to fail, specifically because the number of times you are woken is not bound in [0,1] and thus "P(W)" as used in the 1/2 conditioning is malformed, as it doesn't keep track of when you're actually woken up. In as much as it is anything, using the 1/2 argumentation, "P(W)" is the expected number of wakings.
Sorry, but if we're randomly selecting a waking then it is not true that you're on the heads path 50% of the time. In a pair of runs, one head, one tail, you are woken 3 times, twice on the tails path.
On a randomly selected run of the experiment, there is a 1/2 chance of being in either branch, but: Choose a uniformly random waking in a uniformly chosen random run is not the same as Choose a uniformly random waking.
Why are you using the notation P(W) when you mean E(W)? And if you can get an expectation for it, you must know the probability of it.
Randomly selecting a waking does not imply a uniform distribution. On the contrary, we know the distribution is not uniform.