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AlephNeil comments on Beauty quips, "I'd shut up and multiply!" - Less Wrong

6 Post author: neq1 07 May 2010 02:34PM

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Comment author: AlephNeil 12 May 2010 04:02:13PM 2 points [-]

Variation Alpha:

10 people. If heads, one of the ten is randomly selected to be revived. If tails, all ten are revived. (If you like, suppose that the ten are revived one at a time on consecutive days - but it doesn't make any difference.)

Variation Beta:

Same as Alpha except the 10 people are clones of yours, with mental state identical to your own.

Variation Gamma:

Same as Beta except the cloning is done after you fall asleep.

Variation Delta:

Same as Gamma except that the way the clones are not created all at once. Rather, successive clones are created on subsequent days by erasing one days' worth of memory of the previous clone.

It seems clear to me that in variation Alpha, 1/11 is the answer and not 1/2. And clearly variation Delta is isomorphic to the Sleeping Beauty problem (except with 10 days rather than 2). And clearly each step from Alpha to Delta doesn't change anything essential.


Comment author: Jack 12 May 2010 11:52:35PM *  0 points [-]

Nice way of formulating the problem.

In variation Alpha we know beforehand of a particular event that will happen with P=1 if tails and P=1/10 if heads. Call this event "Jack wakes up and thinks a thought". So when we see that event we can conclude 1/11.

But in Beta and the remaining variations there is no such event. A clone can't tell which clone it is, going into the experiment my anticipated experience does not differ based on whether or not heads comes up. Either one of me will be woken up or 10 identical copies who don't know about each other will be woken up. "Jack wakes up and thinks a thought" happens at the same probability for heads and tails. At no point does any copy of me get new information to revise from 1/2.

Comment author: Morendil 12 May 2010 04:52:04PM 0 points [-]

It seems clear to me that in variation Alpha, 1/11 is the answer and not 1/2.

What is it that makes that clear to you?

Your variation Alpha strikes me as somewhat under-specified. Here is how I'm tempted to fill in:

We have 10 cryonics patients on hand and a revival procedure. Each patient, upon revival, awakens in a featureless room, alone, and is given either questionnaire Q then set free, or just set free without further ado.

We flip a coin. If it comes up heads, questionnaire Q is given to one patient; tails, it is given to all ten. Questionnaire Q consists of this very narrative plus the question, "What is your credence now that the coin came up heads?"

It seems to me that if the patient has no other relevant information (such as how many patients were revived), their answer ought to be 1/2, no matter how many revivals occur on tails. This looks a lot more like Stuart Armstrong's "proof of the SIA" than like SB, though, so I might have to reread that post.

The background information X'=(coin flip, revival with questionnaire) is different from the background information X=(coin flip), but not necessarily enough to alter the answer to the question - unless for some reason each patient is interested in maximizing the number of patients who would get the right answer if they were asked straight out how the coin came up. (Which is how some participants in the discussion have interpreted "credence", I now believe. Under some assumptions, such as having a payout involved, e.g. getting a candy bar for calling the coin correctly, this is even a legitimate interpretation.)

If you take "credence" to mean "your prior, updated with whatever information you've gained that has bearing on how the coin might have come up", and your prior for the coin is the 50/50 distribution, then it seems to me that you have nothing to update on, and that the answer is still 1/2.

Comment author: AlephNeil 12 May 2010 05:14:20PM 0 points [-]

Your filling in is not quite what I had in mind: When I said "one is randomly selected to be revived" I meant to imply "none of the others are revived".

Also, you may suppose that before entering hibernation, each patient knows that there's going to be a coin flip and what will happen in each case.

Deducing 1/11 is now just a matter of applying Bayes' theorem. This may be easier to comprehend if we introduce:

Variation Alpha':

Same as Variation Alpha except that one of the 10 people is (secretly) designated beforehand to be revived in the event of heads.

Comment author: Morendil 12 May 2010 05:22:18PM 0 points [-]

Your filling in is not quite what I had in mind: When I said "one is randomly selected to be revived" I meant to imply "none of the others are revived".

How do the variations you suggest make a difference? Do you agree with my conclusions in my own variant?

Comment author: AlephNeil 12 May 2010 05:36:35PM 0 points [-]

Well, as I'm sure you've guessed my aim is to present the "1/2"-er with a 'smooth spectrum' of scenarios beginning with something that's obviously 1/3 (or in this case 1/11) and ending with something isomorphic to the Sleeping Beauty puzzle, and challenging them to say where along this spectrum the "1/3"-er's argument breaks down.

In the case of Variation Morendil... hmm, I think the Bayesian reasoning for Variation Alpha goes through just the same, and the answer is 1/11. Doesn't it? (Does it make a difference if the patients know about the scenario beforehand, rather than being told about it only in the questionnaire? I don't think so. So pretend they are told beforehand...)

Comment author: Morendil 12 May 2010 05:52:59PM 0 points [-]

Effectively, either variant comes down to being told: "A fair coin has been flipped, and depending on the result of that flip you are either one of a group of 10 people or a lone subject, what credence do you have in being on the small-group branch?"

It doesn't seem obvious to me why, in such a situation, I should answer other than 1/2, so I'm still interested in what makes it obvious to you.

Comment author: AlephNeil 12 May 2010 06:00:12PM 0 points [-]

OK, well let's start with Variation Alpha'. Consider that there are 20 equally likely possibilities, which we can label (x, y) where x belongs to {heads, tails} and y belongs to {1, ..., 10}. Being in possibility (x, y) means "x is the result of the coin toss and y denotes the person we selected beforehand to be revived in the event of heads."

Suppose that (like Patrick McGoohan) you are number 6. Then out of the 20 possibilities, there are 11 in which you are revived, namely (heads, 6) and (tails, 1) to (tails, 10). Therefore, applying Bayes' theorem, given that you are revived, the probability of heads is 1/11.

Comment author: Morendil 12 May 2010 10:21:02PM *  1 point [-]

OK. I have a quibble with your formalization but I get a similar result when working it out formally: if my background information consists of the Alpha procedure, then updating on being revived does give me 1/11.

The quibble is that I only know, algebrically, to condition on something that is a variable, so to work out the joint probability distribution at issue I had to introduce the variable z, with values {revived, not revived}. The triplet (H,3,NR) codes for "the coin comes up heads, person 3 gets picked to be revived in the event of heads, and I don't get revived". (Clearly this entails that I'm not person 3.)

The joint probability distribution P(x,y,z) factors out, per the product rule, into P(x)P(y)P(z|x,y) since x and y are independent.

Let's use N=3 for the number of subjects involved, as I want to write out the full joint distribution (in case someone disagrees with that step) and N=10 makes it tedious. Arbitrarily I consider things from the perspective of Two.

  • (H,1,R)=0
  • (H,2,R)=1/6
  • (H,3,R)=0
  • (H,1,NR)=1/6
  • (H,2,NR)=0
  • (H,3,NR)=1/6
  • (T,1,R)=1/6
  • (T,2,R)=1/6
  • (T,3,R)=1/6
  • (T,1,NR)=0
  • (T,2,NR)=0
  • (T,3,NR)=0

This seems to check out: the marginal distribution for x is the expected 50/50, the marginal distribution for y is uniform, it all sums up to 1, it reproduces the setup as described. The conditional distribution P(x,y|z=R) is then:

  • (H,1)=0
  • (H,2)=1/4
  • (H,3)=0
  • (T,1)=1/4
  • (T,2)=1/4
  • (T,3)=1/4

Resulting in P(H|z=R)=1/4.

So I agree here that "I have been revived" is proper to update on, and yields 1/(N+1) credence for the coin having come up heads. (It wasn't obvious to me to start out, and I still don't rule out having made a mistake somewhere.)

I can see how this works out as equivalent to the variant I described, with z meaning "got the questionnaire" and y meaning "the label of the person picked to receive the questionnaire in the event of heads". It shouldn't matter, either, when we learn about the procedure.

Variations Beta and Gamma don't seem to introduce anything that should matter, because nothing in the original formulation hinges crucially on particular differences in the memories of the N people involved.

I'm not quite sure what Delta means. My interpretation of Delta would be:

We give you a handout describing the procedure, and some time to absorb it, then put you to sleep. We flip a coin; if it comes up head we wake you, if tails we make an atom-level scan of you, and create and wake N-1 copies from the original scan on successive days, inserting the original on the y-th day.

The triplet (H,3,NR) codes for... um... "the coin came up heads, day 3 was picked to awaken the original me in the even of tails, I (someone other than the person to be awakened in the case of heads) was not revived". Best I can do.

Something seems to have gone awry somewhere: Delta is not formally equivalent to the previous formulations.

Also, any interpretation of Delta has a big difference with Sleeping Beauty: it ends up with N distinct clones of me, whereas SB ends up with a single Beauty.

Comment author: AlephNeil 13 May 2010 03:34:18AM 0 points [-]

My description of Delta wasn't great, to be fair. So I'll clarify (and change it slightly) like this:

If (x, y) where x is in {H, T} and y is in {1,2,3} then:

If H then you are not cloned and wake up on day y. If T then a clone of you is created just before the beginning of day 1. Either you or the clone (doesn't matter which) is woken for day 1 while the other is kept in storage. Then the one that was kept in storage is cloned just before the beginning of day 2. Etc.

The idea of moving from Gamma to (my new) Delta is "it shouldn't matter whether the clones are created right away (and possibly never used) or 'just in time'".

Anyway, the following idea has occurred to me, for defending 1/3 as the answer to the original Sleeping Beauty problem: Imagine that there is a clock on the wall and that on any day when SB is woken, the time of day of her awakening is chosen randomly (from a uniform distribution). Then the information that SB gets on awakening is not simply "I was awakened at least once" but "I was awakened at least once at time x"...

...and I'll leave you guys to do the calculation, but you get 1/3, not 1/2.

Comment author: Morendil 13 May 2010 08:05:03AM 0 points [-]

We still have the same problem: there is no value of z that corresponds to "I am a non-special member of the initial set of N people, and I happen to get unlucky and not be revived". That makes Delta not equivalent to the other variants. It does very much matter whether "not revived" is subjectively possible!

It feels as if this might be the same point that neq1 made earlier in answer to one of the defenses of 1/3, so I'd urge you to press on with the formalization and calculation.

My take-away from the discussion (and the two occasions where I changed my mind so far) is that it confirms intuitions aren't reliable and need to be backed by detailed formalization.

Comment author: neq1 12 May 2010 10:28:20PM 0 points [-]

If you are the person that was selected beforehand to be revived in the event of heads, then I agree with 1/11. Unfortunately, in variation beta we lose the ability to label someone ahead of time. This changes things.

Comment author: AlephNeil 13 May 2010 03:11:13AM 1 point [-]

No it doesn't. Your clones are subjectively indistinguishable from you, but they're all in different places at least. Perhaps they're in rooms labelled 1-10, but not allowed to go outside and look at the number. So the experimenters can toss a D10 and randomly choose a subject without breaking the 'clone condition'.

Comment author: neq1 12 May 2010 04:34:13PM 0 points [-]

Variation Alpha is unclear, as worded. Let's say one of the 10 people is Sleeping Beauty, and the other people have different names. Sleeping Beauty was identified ahead of time, and she knows it. If she is not selected, then no one is interviewed. Then, if she is revived, she should think it was heads with probability 10/11.

But... if we will interview everyone who is revived, and no one was labeled as special ahead of time, then all each person that was interviewed knows is that at least one person was revived, which was a probability 1 event under heads and tails.

This is just the self-indication assumption situation.

Consider an example. Suppose we want to know if it's common for people to get struck by lightening. We could choose one person ahead of time to monitor. If they get struck by lightening in the next, say, year, then it's likely that getting struck by lightening is common. But... if instead everyone is monitored, but we are only told about one person who was struck by lightening (there could be others, we don't know), then we have no information about whether getting struck by lightening is common or not.

Comment author: AlephNeil 12 May 2010 06:24:22PM 0 points [-]

Variation Alpha is intended in such a way that, from the perspective of the experimenters, none of the ten subjects is 'special'.

See <a href="http://lesswrong.com/lw/286/beauty_quips_id_shut_up_and_multiply/1zzh">here</a> for why 1/11 is the correct posterior probability for heads.