Yet another Sleeping Beauty

The modification you're proposing is analogous to a modification (I think originally due to Michael Titelbaum) called "Technicolor Beauty", in which Beauty sees either a red or a blue piece of paper in her room when she awakens on Monday (determined by a fair coin toss, independently of the "main" coin toss that decides if she's woken once or twice), and on Tuesday (if she's awakened) sees a piece of paper of whichever color she didn't see on Monday. I'll use this example rather than yours because it requires less specification about which coin toss we're talking about. Let "RB" be the hypothesis that the "main" coin toss landed Tails and Red was shown on Monday and Blue was shown on Tuesday. Let BR be the same, except Blue on Monday and Red on Tuesday.

Titelbaum used this to generate the "thirder" (SIA) answer to the problem, but SSA doesn't actually give the same answer, as you suggest it does. Even though Beauty is twice as likely to observe red paper at some point in the experiment, at no point do her conditional probabilities (e.g. for observing red, conditional on Heads or Tails) differ. Briefly: conditional on Heads, she expects to see red with probability 0.5 (because the coin toss was fair). Conditional on Tails, suppose Beauty has her eyes shut while calculating her conditional probabilities for observing red upon opening them, and evenly splits her (conditional) credence between it being Monday and it being Tuesday (SSA requires this). Now, if it's Monday, Beauty's credence in RB and BR is 0.5 for both, so she expects to see red with probability 0.5. Same goes for Tuesday.

Alright, now it's time for my comment about why saying "I'd like to use the SSA" (or, for that matter, "I'd like to use the SIA") is misguided.

Suppose every time Beauty wakes up, she is asked to guess whether the coin landed Heads or Tails. She receives $3 for correctly saying Heads, and $2 for correctly saying Tails.

The SIA says Pr[Heads] = 1/3 and Pr[Tails] = 2/3, so saying Heads has an expected value of $1, and Tails an expected value of $1.33. On the other hand, the SSA says Pr[Heads]=Pr[Tails]=1/2, so saying Heads is expected to win $1.50, while saying Tails only wins $1.

These indicate different correct actions, and clearly only one of them can be right. Which one? Well, suppose Beauty decides to guess Heads. Then she wins $3 when Heads comes up. On the other hand if Beauty decides to guess Tails, she wins $4. So the SIA gives the "correct probability" in this case.

On the other hand, suppose the rewards are different. Now, suppose Beauty receives money on Wednesday -- $3 if she ever correctly said Heads, and $2 if she ever correctly said Tails. In this case, the optimal strategy for Beauty is to act as though Pr[Heads]=Pr[Tails]=1/2, as suggested by the SSA.

Of course, proponents of either assumption, assuming a working knowledge of probability, are going to make the correct guess in both cases, if they know how the game works; suddenly there is no more disagreement. I therefore argue that the things that these assumptions call Pr[Heads] and Pr[Tails] are not the same things. The SIA calculates the probability that the current instance of Beauty is waking up in a Heads-world or a Tails-world. The SSA calculates the probability that some instance of Beauty will wake up in a Heads-world or a Tails-world.

The way I phrase it makes them sound more different than they are, because this latter event is also the event that every instance of Beauty will wake up in a Heads-world or a Tails-world. Since it's certain that the current instance of Beauty wakes up in the same world that every instance of Beauty wakes up in, it's unclear why these probabilities are different.

This ambiguity disappears once you start talking some hand-wavy notion of probability that feels like it's perfectly okay to disagree about, and fix a concrete situation in which you need the correct probability in order to win, as illustrated in the example above.

(One final comment: by using payoffs of $2 and $3, I am technically only determining whether the probability in question is above or below 2/5. Since this separates 1/2 and 1/3, it is all that is necessary here, but in principle you could also use log-based payoffs to make Beauty give an actual probability as an answer.)

Okay, let's start over. Your math is incorrect. When assuming SSA:

1. In the HH outcome, you are certain to be shown heads. Pr[-H|HH] = 1.
2. In the HT outcome, you will never be shown heads. Pr[-H|HT] = 0.
3. In the TH outcome, you have a 50% chance of being the observer who is shown heads. Pr[-H|TH] = 1/2.
4. In the TT outcome, you have a 50% chance of being the observer who is shown heads. Pr[-H|TT] = 1/2.

Thus the probability of H* is (1+0)/(1 + 0 + 1/2 + 1/2) = 1/2. The extra coin toss cancels out.

Edit: this is essentially the same argument as in AlexSchell's comment.

Because there's no information flow between the coins, to stay self-consistent the "always assign 50% to heads" method has to not change its probabilities under irrelevant information. So this isn't so much a reconciliation as a demonstration that always assigning 50% to heads violates an axiom.

Leaving aside the SSA vs. SIA question, which I believe is fundamentally misguided but what do I know, you are incorrect here:

Thanks to this modification, she got the same answer as if she had used SSA.

When we make this modification, SIA gives the answer 1/3 (same as without it). SSA gives the answer 1/2 (same as without it). The modification does nothing. In the case of the unlikely observation (which doesn't have to be unlikely, actually -- we can compute the answer in terms of the unknown probability and it cancels out anyway), there is also no difference.

I'd like to use the SSA

Cthulhu's woundings, why?

a simple modification...Thanks to this modification, she got the same answer as if she had used SIA.

Assume it's -40 degrees Celsius outside. Thanks to this assumption, it's also -40 degrees Fahrenheit, we got the same answer as if we were using Fahrenheit.

In any case, if it's an unlikely observation, it probably won't happen twice, so she's about twice as likely to make it if she wakes up twice.

If it's a likely observation, such as noting that she is awake, then she is twice as likely to make it if she wakes up twice.