Sleeping Beauty as a decision problem (solved)

post by Bongo · 2010-10-10T03:15:08.755Z · LW · GW · Legacy · 3 comments

Contents

  Version 1
  Version 2 
  Version 3
  Version 4
None
3 comments

EDIT: User:Misha solved it

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First, here's the Sleeping Beauty problem, from Wikipedia:

The paradox imagines that Sleeping Beauty volunteers to undergo the following experiment. On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.

Each interview consists of one question, "What is your credence now for the proposition that our coin landed heads?"

I was looking at AlephNeil's old post about UDT and encountered this diagram depicting the Sleeping Beauty problem as a decision problem.

This diagram is underspecified, though. There are no specific payoffs in the boxes and it's not obvious what actions the arrows mean. So I tried to figure out some ways to transform the Sleeping Beauty problem into a concrete decision problem. I also made edited versions of AlephNeil's diagram for versions 1 and 2.

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The gamemaster puts Sleeping Beauty to sleep on Sunday. He uses a sleeping drug that causes mild amnesia such that upon waking she won't be able to remember any previous awakenings that may have taken place during the course of the game. The gamemaster flips a coin. If heads, he wakes her up on monday only. If tails, he wakes her up on monday and tuesday.

Version 1

Upon each awakening, the gamemaster asks Sleeping Beauty to guess which way the coin landed. For each correct guess, she's awarded $1000 at the end of the game. diagram

Version 2

Upon each awakening, the gamemaster asks Sleeping Beauty to guess which way the coin landed. If she all of her guesses are correct, she's awarded $1000 at the end of the game. diagram

Version 3

Upon each awakening, the gamemaster asks Sleeping Beauty for her credence as to whether the coin landed heads. For each awakening, if the coin landed x, and she declares a credence of p that it landed x, she's awarded p*$1000 at the end of the game.

Version 4

Upon each awakening, the gamemaster asks Sleeping Beauty for her credence as to whether the coin landed heads. At the end of the game, her answers are averaged to a single probability p, and she's awarded p*$1000.

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What's interesting is that while the suggested answers for the classic Sleeping Beauty problem are (1/2) and (1/3), for neither version 1 nor 2 is the correct answer to guess heads every second or third time, and for neither version 3 nor 4 is the correct answer to declare a credence of (1/2) or (1/3). The correct answers are (correct me if I'm wrong, I got these by looking at AlephNeil-style UDT diagrams and doing back-of-the-envelope calculations):

• Version 1: Always guess tails. Expected payoff $1000
• Version 2: Always guess heads, or always guess tails. Expected payoff $500.
• Version 3: Answer with a 0% credence of heads. Expected payoff $1000.
• Version 4: All answers seem to have an expected payoff of $500.

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Is there any way to transform Sleeping Beauty into a decision problem such that the correct answer in some sense is either (1/2) or (1/3)?

Is there a general procedure for transforming problems about credence into decision problems?

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