Variations on the Sleeping Beauty

post by casebash · 2016-01-10T13:00:52.115Z · LW · GW · Legacy · 6 comments

Contents

  Half*-Sleeping Beauty Problem
  Double-Half*-Sleeping Beauty problem
  Double-Half*-Sleeping Beauty problem with Known Day Variation
  Sleeping Couples Problem
  Sleeping Clones Problem
None
6 comments

This post won't directly address the Sleeping Beauty problem so you may want to read the above link to understand what the sleeping beauty problem is first.

Half*-Sleeping Beauty Problem

The asterisk is because it is only very similar to half of the sleeping beauty problem, not exactly half.

A coin is flipped. If it is heads, you are woken up with 50% chance and interrogated about the probability of the coin having come up heads. The other 50% of the time you are killed. If it is tails you are woken up and similarly interrogated. Given that you are being interrogated, what is the probability that the coin came up heads? And have you received any new information?

Double-Half*-Sleeping Beauty problem

A coin is flipped. If it is heads, a coin is flipped again. If this second coin is heads you are woken up and interrogated on Monday, if it is tails you are woken up and interrogated on Tuesday. If it is tails, then you are woken up on Monday and Tuesday and interrogated both days (having no memory of your previous interrogation). If you are being interrogated, what is the chance the coin came up heads? And have you received any new information?

Double-Half*-Sleeping Beauty problem with Known Day Variation

EDIT: This problem should have said: As above, but whenever you are being interrogated you are told the day. You may wish to consider this problem before the above one.

Sleeping Couples Problem

A man and his identical-valued wife have lived together for so many years that they have reached Aumann agreement on all of their beliefs, including core premises, so that they always make the same decision in every situation.

A coin is flipped. If it is heads, one of the couple is randomly woken up and interrogated about the probability of the coin having come up heads. The other is killed. If it is tales, both are woken up separately and similarly interrogated. If you are being interrogated, what is the probability that the coin came up heads? And have you received any new information?

Sleeping Clones Problem

A coin is flipped. If it is heads, you are woken up and interrogated about the probability of the coin having come up heads. If it is tails, then you are cloned and both copies are interrogated separately without knowing whether they are the clone or not. If you are being interrogated, what is the probability that the coin came up heads? And have you received any new information?

My expectation is that the Double-Half Sleeping Beauty and Sleeping Clones will be controversial, but I am optimistic that there will be a consensus on the other three.

Solutions (or at least what I believe to be the solutions) will be forthcoming soon.

6 comments

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comment by Anders_H · 2016-01-11T09:33:43.744Z · LW(p) · GW(p)

This post caused me to type up some old, unrelated thoughts about Sleeping Beauty. I posted it as a comment to the stupid questions thread at http://lesswrong.com/lw/n3v/stupid_questions_2nd_half_of_december/d14z . I'd very much appreciate any feedback on this idea. This comment is just to catch the attention of readers interested in Sleeping Beauty who may not see the comment in the stupid questions thread.

comment by Dagon · 2016-01-10T20:23:32.808Z · LW(p) · GW(p)

The first one is straightforward. Yes, you get evidence that not (coin heads && unlucky 50%), so it's definitely 33% heads. Would be the same to an outside observer (replace "you" with "the victim" and simply observe whether the victim is killed or not. The problem with SB isn't the probability, but the multiple indistinguishable observations.

The second one re-adds the memory wipe, making it pretty much sleeping beauty. You have received no information - all paths lead to the same observation (you wake up). So the prior holds: probability is 50/50, but interrogation-instance-weighed opportunities for prediction is 33/66 in favor of tails.

With known days, you do have information if woken on Monday. You know it wasn't heads-tails. On Tuesday, if you remember Monday's wake-up, you know it was heads-heads, and if you don't remember monday, you know it wasn't.

Couples: if partner1 is being interrogated, they know it wasn't heads-partner1kill. This is pretty much the same as problem 1 for each of them. Without communication, their existence is irrelevant.

Clones: no new information - outside prior holds. Same as sleeping beauty.

Replies from: casebash
comment by casebash · 2016-01-10T22:10:53.117Z · LW(p) · GW(p)

"Tuesday, if you remember Monday's wake-up" - You have no memory of Monday's interrogation because you are memory wiped.

Replies from: Dagon
comment by Dagon · 2016-01-11T04:25:19.963Z · LW(p) · GW(p)

Memory-wiped in the tails case, but not in the heads-heads case, right? If it's heads-heads, you're woken on Monday and nothing happens on Tuesday, so you presumably remember that you were woken and interrogated yesterday.

Replies from: casebash
comment by casebash · 2016-01-11T04:53:03.317Z · LW(p) · GW(p)

You are interrogated on Monday, so what happens on Tuesday is irrelevant to the problem for heads-heads. All that matters is what you say when you are being interogated

Replies from: Dagon
comment by Dagon · 2016-01-11T23:54:45.438Z · LW(p) · GW(p)

Probably doesn't matter how you phrase the handling of Monday night if there's no Tuesday interrogation. If you're interrogated on Tuesday, you know it wasn't heads-heads.