- #1

JeffJo

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- TL;DR Summary
- Most versions of the Sleeping Beauty Problem that you may have seen introduce elements Adam Elga added to it, to make his solution eaasier. There is a better way.

First, a little history. I'm assuming you are familiar with how the details are governed, so I'll only list what the details were:

But now the solution is trivial. There are four possible states for (Quarter,Dime) in step 2, or at the start of steps 3 and 5. They are {(H,H),(H,T),(T,H),(T,T)}. Each has a credence (or probability), for a reasonable observer, of 1/4. Please note that this comes from the logic used by Halfers when they responded to Elga's modified problem. Elga himself worked backwards from the states where SB was given information that reduced the number of possible states from four to two.

And yes, the actual state changes in step 4, but the credence (or probability) distribution does not. By Halfer logic.

But whenever SB addresses the question, whether it is in step 3.2 or 5.2, she knows that the current state is not (H,H). This is new information. She can reasonably update her credence (or probability) of each remaining state from 1/4 to 1/3. That means her credence (or probability) that the Quarter is showing Heads is 1/3.

- The problem originated in an unpublished (until much later) work by Arnold Zuboff.
- The experiment was to last a trillion days.
- Depending on a fair coin toss (Heads or Tails was not specified), the subject (who I will call SB) would be wakened either on each of the trillion days, or on a single, randomly-chosen day in that period.
- The question was "How should [SB] answer the question of how many times [she is] being awakened?

- In 2000, Adam Elga recognized the hyperbole in the trillion-day version, and pared it down considerably.
- He proposed either one, or two, awakenings.
- He was ambiguous about the time period. He said the experiment was to occur over two days, but no awakenings were assigned to a specific day in that period.
- The one-awakening procedure would occur after a Heads result, and the two-awakening procedure after a Tails result.
- The question was "To what degree ought [SB to] believe that the outcome of the coin toss is Heads?"

- But in order to delineate the events in his solution, Elga:
- Re-introduced one potential awakening per day.
- Removed Zuboff's random-day specification for the single awakening, fixing it on the first day.
- Added a word that becomes important only in his solution.

Most of the arguments - including one recently closed in this forum - focus on points that, in my opinion, are irrelevant.Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

- There are various
*translations*of what Elga's "degree of belief" should mean. To be fair, Elga does use the term "credence" for it in the solution, but he also treats it as a straight-forward probability. - He added the order "Always awaken on Monday, but only on Tuesday after Tails."
- In his solution, it becomes clear that "when first awakened" means "before anybody might tell SB that it is Monday, or that the coin landed on Heads." He uses such revelations to reduce the probability problem to ones he thinks are obvious, and then to back-solve to get 1/3.

- Put SB to sleep.
- Flip
*two*coins, a Quarter and a Dime. - If
*either*of the coins is showing Tails, perform this procedure:- Wake SB.
- Ask SB for her degree of belief (or credence, knowledge based probability, or whatever else) that the Quarter
*is currently showing*Heads. - Put her back to sleep with amnesia.

- Turn the Dime over to show its other face.
- Repeat steps 3.1 to 3.3.
- End the experiment by waking SB.

But now the solution is trivial. There are four possible states for (Quarter,Dime) in step 2, or at the start of steps 3 and 5. They are {(H,H),(H,T),(T,H),(T,T)}. Each has a credence (or probability), for a reasonable observer, of 1/4. Please note that this comes from the logic used by Halfers when they responded to Elga's modified problem. Elga himself worked backwards from the states where SB was given information that reduced the number of possible states from four to two.

And yes, the actual state changes in step 4, but the credence (or probability) distribution does not. By Halfer logic.

But whenever SB addresses the question, whether it is in step 3.2 or 5.2, she knows that the current state is not (H,H). This is new information. She can reasonably update her credence (or probability) of each remaining state from 1/4 to 1/3. That means her credence (or probability) that the Quarter is showing Heads is 1/3.