Hi. It’s been a while. I still find the implications of your reasoning fascinating. I am seeking to explore whether I agree with it. I recognise there are several double halfer positions. I’ll attempt to work with your version.
For all double halfers, in the original Sleeping Beauty Problem, the probability of Heads is 1⁄2 before she’s told what day it is and 1⁄2 after she learn it’s Monday. Whereas for a pure halfer and thirder, the elimination of Tuesday within Tails is subject to Bayesian updating and treated just like the elimination of a random event that had probability. Thus a thirder starts with 1⁄3 and updates to 1⁄2 for Heads, while a halfer starts with 1⁄2 and updates to 2⁄3.
Your particular reasoning, as I understand it, is that eliminating Tuesday within Tails only ruled out a self-location inside an outcome without removing any evidence for that outcome. A Tuesday awakening was not a random event and had no probability. It was a pre-existing self-location inside the Tails world. Likewise, Monday was a self-location in both worlds and can’t be assigned a probability. Tuesday’s elimination therfore did not reduce the probability of Tails having occurred.
Before she was told the day, Heads and Tails were equally likely. The only difference was that if the coin landed Heads her self-location was known, while if the coin landed Tails her self-location was unknown. In either case, only one self-location was relevant to her and this had no probability value. Regardless of the coin, she knew she was going to self-locate somewhere and her last memory would be Sunday. This was equally likely to happen with Heads and Tails. Being told it was Monday only confirmed and removed ignorance of when she currently existed within the coin outcome. It gave her no evidence about that outcome. If this is not your reasoning, you can correct me.
Now let’s do a familiar tweak. Suppose that, regardless of the coin, she is woken on both days with amnesia between. On Tuesday, if the coin landed Tails, her room is painted red. Otherwise her room is painted blue. What is her credence for the coin before she sees the room colour and after she sees that it’s blue. For pure halfers (and also thirders in this case), it’s straightforward. The probability of Heads is 1⁄2 before she sees the room colour, and 2⁄3 after she sees it’s blue.
How would you apply your self-location reasoning? From what I can tell it must logically be this. Seeing that the room is blue eliminates that her self-location is Tuesday within Tails. However, as with the original version, Tuesday within Tails was not a random event and had no probability. It’s elimination did not rule out, or reduce the probability of, Tails. Nor has she learnt anything new about the probability of Heads. Therefore the coin outcome is 1⁄2 both before and after she knows the room colour is blue. The only difference afterwards is that, if the coin landed Heads, her self-location remains unknown, while if the coin landed Tails she knows it’s Monday. Again, you can correct me if this is not your reasoning.
Finally lets do a version that I think may confound you. As before, regardless of the coin, she is woken on both Monday and Tuesday with amnesia between. This time, regardless of the coin outcome, on one of the days the room will be painted blue while on the other it will be painted red. If the coin lands Heads, it will be randomly decided which day the room is painted blue and on which it is painted red. If the coin lands Tails, the room will be painted blue on Monday and red on Tuesday.
The above version exposes the self-location issue. We can agree there is only one coin flip (not two as we explored in our previous exchange). Everyone can also agree that before she sees the room colour, there is complete parity. Heads and Tails are equally likely.
Now suppose she observes that the colour of the room is blue. Seeing this colour would appear to give you different information if the coin landed Heads than if it landed Tails. If the coin landed Heads, the room being painted blue is identified as an independent random event from the coin flip that had equal probability; either blue or red could have been selected for this day. Therefore the event sequence of the coin landing Heads and the room being painted red must be eliminated for Bayesian updating. However if the coin landed Tails, the room being painted blue was linked to her self-location. Monday was not a random event and has no probability. Assuming the coin landed Tails, today is definitely Monday and blue was the only colour that could have been selected for this self-location. Tuesday within Tails is of course ruled out as being the alternative self-location inside that outcome but this can’t be used for Bayesian updating and it doesn’t reduce the probability of Tails.
Therefore, upon seeing blue, it is now 1⁄3 the coin landed Heads and 2⁄3 it landed Tails. The same is would be true if she’d seen the colour red. In this version, using your self-location reasoning, whichever colour she sees, Beauty must become a thirder. Therefore she should logically be a thirder before she even sees the colour.
I think we can both agree that this is wrong. If I’m missing something that supports your particular double halfer position and reasoning about self-location, I’d be curious to know it.
Because you’re a double halfer, I see a contradiction in your conclusion about Lotaria’s colour room example. You’ve previously made a distinction between self-locating events, which are guaranteed to happen, and random outcomes that have genuine probability. Your position has been that rules of conditionalisation apply only to random events, not to self location.
In the colour room example, the coin flips are random events. The subsequently experienced colour ‘blue’ is not a random event; it is a confirmation of one of the fixed self-locations within each possible outcome.
By double halfer reasoning, when she sees a blue room, no conditionalisation will be applied apart from the elimination of TT. Since self-locations can’t be conditioned on in the same way, she would say that HH, HT and TH have an equal credence of 1⁄3, even though red room awakenings in two of the outcomes have been eliminated. If the coins landed HH, she doesn’t know whether she’s in the first or second awakening. If the coins landed HT, she know she’s in the first. If the coins landed TH, she knows she’s in the second. But for a double halfer, this information about her location or lack of it can’t be conditioned on to increase or decrease the probability of how the coins landed.
So because you’re a double halfer who treats self-location differently, you would have to say the probability is 2⁄3 that the coin tosses were different in the coloured room example, just as you do in the wake/sleep version. Do you see the consistency of that?
Incidentally I don’t share your view about self-location. I would argue that in the coloured room example, information about her self-location should be conditioned no differently than random events. If so, the answer is indeed 1⁄2 that the coin tosses are different. However, in the first scenario where she’s awake or asleep, I would agree it’s 2⁄3 the coin tosses are different. This makes me a pure halfer rather than a double halfer for the original SBP. I’ve touched on some of the inconsistencies I see with double-halfing.