This is the seventh post in my series on Anthropics. The previous one is Why Two Valid Answers Approach is not Enough for Sleeping Beauty. The next one is The Solution to Sleeping Beauty.

Introduction

As the Sleeping Beauty problem has been investigated for a long time, there have been multiple different attempts to apply various mathematical models to it. And as we are still talking about this problem, these attempts were not quite successful.

Figuring out where these models failed is crucial for understanding the Sleeping Beauty problem and solving it. The last thing we want is to repeat someone else's mistakes yet another time and get attached to a wrong solution.

In this post I'll investigate several such models, point out the issues due to which none of them manages to correctly capture the setting of the problem and thus collect valuable insights about the properties of the correct model.

Updateless Models

Updateless Models are the ones, according to which no update on awakening is supposed to happen, because regardless of the coin toss outcome Beauty can be certain to be awake.

$P (A w a k e | H e a d s) = 1$

$P (H e a d s | A w a k e) = P (H e a d s)$

Intuitively it may look as if such models should necessarily be in favor of answering 1/2, however, this is not the case and historically the first of updateless models is a thirder one.

Elga's Model

This model was originally applied to the Sleeping Beauty problem in the Self-locating belief and the Sleeping Beauty problem paper by Adam Elga. According to it, there are three possible outcomes with equal probability:

$Ω = {H e a d s & M o n d a y, T a i l s & M o n d a y, T a i l s & T u e s d a y}$

$P (H e a d s & M o n d a y) = P (T a i l s & M o n d a y) = P (T a i l s & T u e s d a y) = 1 / 3$

Every outcome describes an awakened state of the Beauty:

$P (A w a k e) = P (Ω) = 1$

And most of the awakenings happen on Monday:

$P (M o n d a y) = P (H e a d s & M o n d a y) + P (T a i l s & M o n d a y) = 2 / 3$

If the Beauty is awakened on Monday and knows about it, she can't guess Heads or Tails better than chance:

$P (H e a d s | M o n d a y) = P (T a i l s | M o n d a y) = 1 / 2$

And likewise, if she knows that the coin is Tails, her credence for Monday and Tuesday is the same:

$P (M o n d a y | T a i l s) = P (T u e s d a y | T a i l s) = 1 / 2$

The model is elegant, simple and wrong. The issue is, of course, that according to it the coin isn't fair.

$P (H e a d s) = 1 / 3$

$P (T a i l s) = 2 / 3$

Therefore, it can't possibly be describing Sleeping Beauty problem.

No-Coin-Toss Problem

The thing is, Elga's model is describin...