As I understand it, your proposal is essentially the principle of indifference applied to a sample space that isn't the same as the stated assumptions of the SB problem, namely your sample space is based on the triple

{Coin,Day,Wakefulness}

upon which you assign the distribution Pr(Heads,Monday,Awake) = Pr(Tails,Monday,Awake) = Pr(Heads,Tuesday,Asleep) = Pr(Tails,Tuesday,Awake) = 1/4. — sime

No.

The reason I keep asking for specific answers to specific questions, is that I find that nobody addresses "my sample space." Even though I keep repeating it. They change it, as you did here, to include the parts I am very intentionally trying to eliminate.

There are two, not three, random elements. They are COIN and DAY. WAKE and SLEEP are not random elements, they are the consequences of certain combinations, the consequences that SB can observe.

There are two sampling opportunities during the experiment, not two paths. The random experiment, as it is seen by SB's "inside" the experiment, is just one sample. It is not one day on a fixed path as seen by someone not going through the experiment, but one day only. Due to amnesia, each sample is not related, in any way SB can use, to any other.

Each of the four combinations of COIN+DAY is equally likely (this is the only application of the PoI), in the prior (this means "before observation") probability distribution. Since there are four combinations, each has a prior ("before observation") probability of 1/4.

In the popular problem, SB's observation, when she is awake, is that this sample could be H+Mon, T+Mon, or T+Tue; but not H+Tue. She knows this because she is awake. One specific question I ask, is what happens if we replace SLEEP with DISNEYWORLD. Because the point that I feel derails halfers is the sleep.

Halfers seem to think SLEEP means H+Tue cannot be sampled. So, I change the problem so it can be sampled. SB's observation is now that she is in an interview. So this sample could be H+Mon, T+Mon, or T+Tue; but not H+Tue. She knows this because she is in an interview, not at DisneyWorld. There is no difference it her utilization of the mechanics of the experiment, nor of what her observation means.

By contrast, the probability space for the classical SB problem is that of a single coinflip C = {H,T}, namely (C,{0,H,T,{H,T}},P) where P (C = H) = 0.5 .

This probability space includes a distribution says that a single sampling is happening on two days at the same time. Halfers convince themselves that there is no contradiction; after TAILS the "other" awakening is identical, and after HEADS it is not observed. But that doesn't work if we change to DISNEYWORLD.

But what makes your argument incorrect [is] the use of a non-permitted sample space.

What is non-permitted? It is a functionally equivalent one. Consequence #1 occurs with three of the four combinations, and Consequence #2 occurs with the fourth. The question is only asked with Consequence #2. I'm sorry, but this is a rationalization.

But its use is why I asked the specific question about my Camp Sleeping Beauty version. There is a 6x6 calendar for the six days of camp, and a six-sided die is used to pick one row (the columns are days of the week). If campers partake of, say, activity C then how do we deduce the conditional probability that the die roll was, say, 5?

I claim that the 36 cells each have a prior probability of 1/36 (PoI again). And that the conditional probability of a 5, given activity C, is the number of times C appears in row 5, divided by the number of times C appears in the calendar.

AND, this does not, and cannot, change if one of the "activities" is "sleep through the entire day." This is what addresses you concern here, and I'd love to hear why D="DARTS" produces a different result than D="DOZE."

There is nothing "non-permitted" about including either D in the sample space. In fact, since either D is part of the camp counselors' planning, they must be functionally equivalent. And it applies to any NxN isomoprhic experiment, even if N=2.

We could discuss this conclusion, but that discussion will need to include answers.

The reason I keep asking for specific answers to specific questions, is that I find that nobody addresses "my sample space." Even though I keep repeating it. They change it, as you did here, to include the parts I am very intentionally trying to eliminate. — JeffJo

I think you misunderstand me. I am simply interpreting the thrux of your position in terms of an extended sample space. This isn't miscontruing your position but articulating it in terms of Bayesian probabilities. This step is methodological and not about smuggling in new premises, except those that you need to state your intuitive arguments, which do constitute additional but reasonable premises. [/quote]

There are two, not three, random elements. They are COIN and DAY. WAKE and SLEEP are not random elements, they are the consequences of certain combinations, the consequences that SB can observe. — JeffJo

Look at this way: It is certainly is the case that according to the Bayesian interpretation of probabilities, one can speak of a joint probability distribution over (Coin State, Day State, Sleep State), regardless of one's position on the topic. But in the case of the frequentist halfer, the sleep-state can be marginalised out and in effect ignored, due to their insistence upon only using the coin information and rejecting counterfactual outcomes that go over and above the stated information.

There are two sampling opportunities during the experiment, not two paths. The random experiment, as it is seen by SB's "inside" the experiment, is just one sample. It is not one day on a fixed path as seen by someone not going through the experiment, but one day only. Due to amnesia, each sample is not related, in any way SB can use, to any other. — JeffJo

You have to be careful here, because you are in danger of arguing for the halfers position on their behalf. Counterfactual intuitions, which you are appealing to below, are in effect a form of path analysis, even if you don't see it that way.

Each of the four combinations of COIN+DAY is equally likely (this is the only application of the PoI), in the prior (this means "before observation") probability distribution. Since there are four combinations, each has a prior ("before observation") probability of 1/4.

In the popular problem, SB's observation, when she is awake, is that this sample could be H+Mon, T+Mon, or T+Tue; but not H+Tue. She knows this because she is awake. One specific question I ask, is what happens if we replace SLEEP with DISNEYWORLD. Because the point that I feel derails halfers is the sleep. — JeffJo

But the Sleeping Beauty Problem per-se does not assume that the Sleeping Beauty exists on tuesday if the coin lands heads, because it does not include an outcome that measures that possibility. Hence you need an additional variable if you wish to make your counterfactual argument that SB would continue to exist on tueday in the event the coin lands heads. Otherwise you cannot formalise your argument.

Just to clarify, I'm not confusing you for a naive thirder, as I mistook you for initially, where i just assumed that you were blindly assigning a naive prior over three possible outcomes. I think your counterfactual arguments are reasonable, and I verified that they numerally check out; but they do require the introduction of a third variable to the sample space in order to express your counterfactual intuition that I called "sleep state" (which you could equally call "the time independent state of SB").