# Beauty quips, "I'd shut up and multiply!"

5 07 May 2010 02:34PM

When it comes to probability, you should trust probability laws over your intuition.  Many people got the Monty Hall problem wrong because their intuition was bad.  You can get the solution to that problem using probability laws that you learned in Stats 101 -- it's not a hard problem.  Similarly, there has been a lot of debate about the Sleeping Beauty problem.  Again, though, that's because people are starting with their intuition instead of letting probability laws lead them to understanding.

The Sleeping Beauty Problem

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?"

Two popular solutions have been proposed: 1/3 and 1/2

The 1/3 solution

From wikipedia:

Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.

Yes, it's true that only in a third of cases would heads precede her awakening.

Radford Neal (a statistician!) argues that 1/3 is the correct solution.

This [the 1/3] view can be reinforced by supposing that on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads. (We suppose that Beauty knows such a bet will always be offered.) Beauty would not accept this bet if she assigns probability 1/2 to Heads. If she assigns a probability of 1/3 to Heads, however, her expected gain is 2 × (2/3) − 3 × (1/3) = 1/3, so she will accept, and if the experiment is repeated many times, she will come out ahead.

Neal is correct (about the gambling problem).

These two arguments for the 1/3 solution appeal to intuition and make no obvious mathematical errors.   So why are they wrong?

Let's first start with probability laws and show why the 1/2 solution is correct. Just like with the Monty Hall problem, once you understand the solution, the wrong answer will no longer appeal to your intuition.

The 1/2 solution

P(Beauty woken up at least once| heads)=P(Beauty woken up at least once | tails)=1.  Because of the amnesia, all Beauty knows when she is woken up is that she has woken up at least once.  That event had the same probability of occurring under either coin outcome.  Thus, P(heads | Beauty woken up at least once)=1/2.  You can use Bayes' rule to see this if it's unclear.

Here's another way to look at it:

If it landed heads then Beauty is woken up on Monday with probability 1.

If it landed tails then Beauty is woken up on Monday and Tuesday.  From her perspective, these days are indistinguishable.  She doesn't know if she was woken up the day before, and she doesn't know if she'll be woken up the next day.  Thus, we can view Monday and Tuesday as exchangeable here.

A probability tree can help with the intuition (this is a probability tree corresponding to an arbitrary wake up day):

If Beauty was told the coin came up heads, then she'd know it was Monday.  If she was told the coin came up tails, then she'd think there is a 50% chance it's Monday and a 50% chance it's Tuesday.  Of course, when Beauty is woken up she is not told the result of the flip, but she can calculate the probability of each.

When she is woken up, she's somewhere on the second set of branches.  We have the following joint probabilities: P(heads, Monday)=1/2; P(heads, not Monday)=0; P(tails, Monday)=1/4; P(tails, Tuesday)=1/4; P(tails, not Monday or Tuesday)=0.  Thus, P(heads)=1/2.

Where the 1/3 arguments fail

The 1/3 argument says with heads there is 1 interview, with tails there are 2 interviews, and therefore the probability of heads is 1/3.  However, the argument would only hold if all 3 interview days were equally likely.  That's not the case here. (on a wake up day, heads&Monday is more likely than tails&Monday, for example).

Neal's argument fails because he changed the problem. "on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads."  In this scenario, she would make the bet twice if tails came up and once if heads came up.  That has nothing to do with probability about the event at a particular awakening.  The fact that she should take the bet doesn't imply that heads is less likely.  Beauty just knows that she'll win the bet twice if tails landed.  We double count for tails.

Imagine I said "if you guess heads and you're wrong nothing will happen, but if you guess tails and you're wrong I'll punch you in the stomach."  In that case, you will probably guess heads.  That doesn't mean your credence for heads is 1 -- it just means I added a greater penalty to the other option.

Consider changing the problem to something more extreme.  Here, we start with heads having probability 0.99 and tails having probability 0.01.  If heads comes up we wake Beauty up once.  If tails, we wake her up 100 times.  Thirder logic would go like this:  if we repeated the experiment 1000 times, we'd expect her woken up 990 after heads on Monday, 10 times after tails on Monday (day 1), 10 times after tails on Tues (day 2),...., 10 times after tails on day 100.  In other words, ~50% of the cases would heads precede her awakening. So the right answer for her to give is 1/2.

Of course, this would be absurd reasoning.  Beauty knows heads has a 99% chance initially.  But when she wakes up (which she was guaranteed to do regardless of whether heads or tails came up), she suddenly thinks they're equally likely?  What if we made it even more extreme and woke her up even more times on tails?

Implausible consequence of 1/2 solution?

Nick Bostrom presents the Extreme Sleeping Beauty problem:

This is like the original problem, except that here, if the coin falls tails, Beauty will be awakened on a million subsequent days. As before, she will be given an amnesia drug each time she is put to sleep that makes her forget any previous awakenings. When she awakes on Monday, what should be her credence in HEADS?

He argues:

The adherent of the 1/2 view will maintain that Beauty, upon awakening, should retain her credence of 1/2 in HEADS, but also that, upon being informed that it is Monday, she should become extremely confident in HEADS:

This consequence is itself quite implausible. It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads.

It's correct that, upon awakening on Monday (and not knowing it's Monday), she should retain her credence of 1/2 in heads.

However, if she is informed it's Monday, it's unclear what she conclude.  Why was she informed it was Monday?  Consider two alternatives.

Disclosure process 1:  regardless of the result of the coin toss she will be informed it's Monday on Monday with probability 1

Under disclosure process 1, her credence of heads on Monday is still 1/2.

Disclosure process 2: if heads she'll be woken up and informed that it's Monday.  If tails, she'll be woken up on Monday and one million subsequent days, and only be told the specific day on one randomly selected day.

Under disclosure process 2, if she's informed it's Monday, her credence of heads is 1,000,001/1,000,002.  However, this is not implausible at all.  It's correct.  This statement is misleading: "It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads."  Beauty isn't predicting what will happen on the flip of a coin, she's predicting what did happen after receiving strong evidence that it's heads.

ETA (5/9/2010 5:38AM)

If we want to replicate the situation 1000 times, we shouldn't end up with 1500 observations.  The correct way to replicate the awakening decision is to use the probability tree I included above. You'd end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.

Suppose at each awakening, we offer Beauty the following wager:  she'd lose \$1.50 if heads but win \$1 if tails.  She is asked for a decision on that wager at every awakening, but we only accept her last decision. Thus, if tails we'll accept her Tuesday decision (but won't tell her it's Tuesday). If her credence of heads is 1/3 at each awakening, then she should take the bet. If her credence of heads is 1/2 at each awakening, she shouldn't take the bet.  If we repeat the experiment many times, she'd be expected to lose money if she accepts the bet every time.

The problem with the logic that leads to the 1/3 solution is it counts twice under tails, but the question was about her credence at an awakening (interview).

ETA (5/10/2010 10:18PM ET)

Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.

Another way to look at it:  the denominator is not a sum of mutually exclusive events.  Typically we use counts to estimate probabilities as follows:  the numerator is the number of times the event of interest occurred, and the denominator is the number of times that event could have occurred.

For example, suppose Y can take values 1, 2 or 3 and follows a multinomial distribution with probabilities p1, p2 and p3=1-p1-p2, respectively.   If we generate n values of Y, we could estimate p1 by taking the ratio of #{Y=1}/(#{Y=1}+#{Y=2}+#{Y=3}). As n goes to infinity, the ratio will converge to p1.   Notice the events in the denominator are mutually exclusive and exhaustive.  The denominator is determined by n.

The thirder solution to the Sleeping Beauty problem has as its denominator sums of events that are not mutually exclusive.  The denominator is not determined by n.  For example, if we repeat it 1000 times, and we get 400 heads, our denominator would be 400+600+600=1600 (even though it was not possible to get 1600 heads!).  If we instead got 550 heads, our denominator would be 550+450+450=1450.  Our denominator is outcome dependent, where here the outcome is the occurrence of heads.  What does this ratio converge to as n goes to infinity?  I surely don't know.  But I do know it's not the posterior probability of heads.

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Comment author: 07 May 2010 07:37:11PM *  5 points [-]

Beauty just knows that she'll win the bet twice if tails landed. We double count for tails...

That doesn't mean your credence for heads is 1 -- it just means I added a greater penalty to the other option.

You don't need a monetary reward for this reasoning to work. It's a funny ambiguity, I think, in what 'credence' means. Intuitively, a well-calibrated person A should assign a probability of P% to X iff X happens on P% of the occasions where A assigned a P% probability to X.

If we accept this, then clearly 1/3 is correct. If we run this experiment multiple times and Beauty guessed 1/3 for heads, then we'd find heads actually came up 1/3 of the times she said "1/3". Therefore, a well-calibrated Beauty guesses "1/3".

Comment author: 10 May 2010 06:23:47PM 2 points [-]

If we accept this, then clearly 1/3 is correct. If we run this experiment multiple times and Beauty guessed 1/3 for heads, then we'd find heads actually came up 1/3 of the times she said "1/3". Therefore, a well-calibrated Beauty guesses "1/3".

On the other hand...

Intuitively, a well-calibrated person A should assign a probability of P% to X iff X happens on P% of the occasions where A assigned a P% probability to X.

Here we're still left with "occasions". Should a well-calibrated person be right half of the times they are asked, or about half of the events? If (on many trials) Beauty guesses "tails" every time, then she's correct 2/3 of the times she's asked. However, she's correct 1/2 of the times that the coin is flipped.

If I ask you for the probability of 'heads' on a fair coin, you'll come up with something like '1/2'. If I ask you a million times before flipping, flip once, and it comes up tails, and then ask you once more before flipping, flip once, and it comes up heads, then you should not count that as a million cases of 'tails' being the correct answer and one of 'heads', even though a guess of 'tails' would have made you correct on a million occasions of being asked the question.

Comment author: 10 May 2010 07:49:11PM 0 points [-]

Well, the question was:

"What is your credence now for the proposition that our coin landed heads?"

No mention of "occasions". Your comment doesn't seem to be addressing that question, but some other ones, which are not mentioned in the problem description.

This explains why you can defend the "wrong" answer: you are not addressing the original question.

Comment author: 10 May 2010 08:03:03PM 0 points [-]

I did not claim that the problem statement used the word "occasions".

Beauty should answer whatever probability she would answer if she was well-calibrated. So does a well-calibrated Beauty answer '1/2' or 1/3'? Does Laplace let her into Heaven or not?

Comment author: 18 May 2010 01:39:28AM 0 points [-]

By the way, do you happen to remember the name or location of the article in which Eliezer proposed the idea of being graded for your beliefs (by Laplace or whoever), by something like cross-entropy or K-L divergence, such that if you ever said about something true that it had probability 0, you'd be infinitely wrong?

Comment author: 18 May 2010 02:05:09AM 2 points [-]
Comment author: 18 May 2010 02:31:03AM 0 points [-]

What Nick said. Laplace is also mentioned jokingly in a different context in An Intuitive Explanation of Bayes' Theorem.

Comment author: 10 May 2010 08:06:18PM *  0 points [-]

Well, 1/3. I thought you were supposed to be defending the plausibility of the "1/2" answer here - not asking others which answer is right.

Comment author: 09 May 2010 02:34:23AM 1 point [-]

We know she will have the same credence on monday as she does on tuesday (if awakened), because of the amnesia. There is no reason to double count those. Under the experiment, you should think of there being one occasion under heads and one occasion under tails. From that perspective, a well-calibrated person A will assign 1/2 for heads. I think that is the correct way to view this problem. If there was a way for her to distinguish the days, things would be different.

Comment author: 09 May 2010 11:22:41PM *  1 point [-]

We know she will have the same credence on monday as she does on tuesday (if awakened), because of the amnesia. There is no reason to double count those.

Well, she does say it twice. That seems like at least a potential reason to count it as two answers.

You could say that 1/3 of the times the question is asked, the coin came up heads. You could also say that 1/2 of the beauties are asked about a coin that came up heads.

To me, this reinforces my doubt that probabilities and beliefs are the same thing.

EDIT: reworded for clarity

Comment author: 10 May 2010 01:05:56AM 1 point [-]

To me, this reinforces my doubt that probabilities and beliefs are the same thing.

Why?

Comment author: 10 May 2010 02:28:24AM 1 point [-]

It illustrates fairly clearly how probabilities are defined in terms of the payoff structure (which things will have payoffs assigned to them and which things are considered "the same" for the purposes of assigning payoffs).

I've felt for a while that probabilities are more tied to the payoff structure than beliefs, and this discussion underlined that for me. I guess you could say that using beliefs (instead of probabilities) to make decisions is a heuristic that ignores, or at least downplays, the payoff structure.

Comment author: 10 May 2010 02:44:16AM 1 point [-]

I agree that probabilities are defined through wagers. I also think beliefs (or really, degrees of belief) are defined through wagers. That's the way Bayesian epistemologists usually define degree of belief. So I believe X will occur with P = .5 iff a wager on X and a wager on a fair coin flip are equally preferable to me.

Comment author: 10 May 2010 03:44:58AM 0 points [-]

That's fine. I guess I'm just not a Bayesian epistemologist.

If Sleeping Beauty is a Bayesian epistemologist, does that mean she refuses to answer the question as asked?

Comment author: 10 May 2010 04:03:43AM *  0 points [-]

I'm not sure I have an official position of Bayesian epistemology but I find the problem very confusing until you tell me what the payoff is. One might make an educated guess at the kind of payoff system the experiment designers would have had in mind-- as many in the this thread have done. (ETA: actually, you probably have to weigh your answer according to your degree of belief in the interpretation you've chosen. Which is of course ridiculous. Lets just include the payoff scheme in the experiment.)

Comment author: 12 May 2010 10:09:25PM 0 points [-]

I agree that more information would help the beauty, but I'm more interested in the issue of whether or not the question, as stated, is ill-posed.

One of the Bayesian vs. frequentist examples that I found most interesting was the case of the coin with unknown bias -- a Bayesian would say it has 50% chance of coming up heads, but a frequentist would refuse to assign a probability. I was wondering if perhaps this is an analogous case for Bayesians.

That wouldn't necessarily mean anything is wrong with Bayesianism. Everyone has to draw the line somewhere, and it's good to know where.

Comment author: 10 May 2010 02:47:57AM 0 points [-]

I can understand that, but the fact that a wager has been offered distorts the probabilities under a lot of circumstances.

Comment author: 10 May 2010 04:00:18AM 0 points [-]

How do you mean?

Comment deleted 07 May 2010 04:39:06PM *  [-]
Comment author: 07 May 2010 08:52:12PM 2 points [-]

In the sleeping beauty problem, whether the 2/3 or 1/2 is "right" is just a debate about words. The real issue is what kind of many-instance decision algorithm you are running.

Not quite. The question of what do we mean by probability in this case is valid, but probability shouldn't be just about bets. Probability is bound to a specific model of the situation, with sample space, probability measure, and events. The concept of "probability" doesn't just mean "the password you use to win bets to your satisfaction". Of course this depends on your ontological assumptions, but usually we are safe with a "possible worlds" model.

Comment deleted 07 May 2010 10:46:45PM [-]
Comment author: 07 May 2010 10:56:33PM *  0 points [-]

It is for making decisions, specifically for expressing preference under the expected utility axioms and where uniform distribution is suggested by indifference to moral value of a set of outcomes and absence of prior knowledge about the outcomes. Preference is usually expressed about sets of possible worlds, and I don't see how you can construct a natural sample space out of possible worlds for the answer of 2/3.

Comment deleted 08 May 2010 12:14:48AM [-]
Comment author: 08 May 2010 12:21:05AM *  0 points [-]

Of course that's the obvious answer, but it also has some problems that don't seem easily redeemable. The sample space has to reflect the outcome of one's actions in the world on which preference is defined, which usually means the set of possible worlds. "Experience-moments" are not carved the right way (not mutually exclusive, can't update on observations, etc.)

Comment deleted 08 May 2010 03:22:00AM [-]
Comment author: 08 May 2010 10:53:52AM *  0 points [-]

By "can't update" I refer to the problem with marking Thursday "impossible", since you'll encounter Thursday later.

However there is the possibility of subjectively indistinguishable experiences.

It's not a problem with the model of ontology and preference, it's merely specifics of what kinds of observation events are expected.

Experience moments are "mutually exclusive", in the sense that every experience moment can be uniquely identified in theory, and any given agent at any given time is only having one specific observer moment.

If the goal is to identify an event corresponding to observations in the form of a set of possible worlds, and there are different-looking observations that could correspond to the same event (e.g. observed at different time in the same possible world), their difference is pure logical uncertainty. They differ, but only in the same sense as 2+2 and (7-5)*(9-7) differ, where you need but to compute denotation: the agent running on the described model doesn't care about the difference, indeed wants to factor it out.

Comment deleted 08 May 2010 12:42:41PM [-]
Comment author: 08 May 2010 12:57:19PM 1 point [-]

I humbly apologize for my inability to read (may the Values of Less Wrong be merciful).

Comment author: 08 May 2010 05:22:04PM 3 points [-]

A bet where she can immediately win, be paid, and consumer her winnings seems to me far more directly connected to the probability of "what state am I in" than a bet where whether the bet is consummated and the bet paid depends on what else happens in other situations that may exist later. It seems crazy to treat both of those as equally valid bets about what state she is in at the moment.

Comment author: 07 May 2010 06:57:03PM *  0 points [-]

This has nothing to do with semantics. If smart people are saying "2+2=5" and I point out it's 4, would you say "what matters is why you want to know what 2+2 is"?

The question here is very well defined. There is only one right answer. The fact that even very smart people come up with the wrong answer has all kinds of implications about the type of errors we might make on a regular basis (and lead to bad theories, decisions, etc).

Comment deleted 08 May 2010 12:51:31PM [-]
Comment author: 08 May 2010 03:29:18PM 2 points [-]

So the difficult question here is which probability space to set up, not how to compute conditional probabilities given that probability space.

(Posted as an antidote to misinterpretation of your comment I committed a moment before.)

Comment author: 07 May 2010 07:06:02PM *  5 points [-]

If you mean something else by probability than "at what odds would you be indifferent to accepting a bet on this proposition" then you need to explain what you mean. You are just coming across as confused. You've already acknowledged that sleeping beauty would be wrong to turn down a 50:50 bet on tails. What proposition is being bet on when you would be correct to be indifferent at 50:50 odds?

Comment author: 07 May 2010 08:44:27PM 0 points [-]

There is a mismatch between the betting question and the original question about probability.

At an awakening, she has no more information about heads or tails than she had originally, but we're forcing her to bet twice under tails. So, even if her credence for heads was a half, she still wouldn't make the bet.

Suppose I am going to flip a coin and I tell you you win \$1 if heads and lose \$2 if tails. You could calculate that the p(H) would have to be 2/3 in order for this to be a fair bet (have 0 expectation). That doesn't imply that the p(H) is actually 2/3. It's a different question. This is a really important point, a point that I think has caused much confusion.

Comment author: 07 May 2010 09:02:00PM 1 point [-]

You seem to agree she should take a 50:50 bet on tails. What would be the form of the bet where she should be indifferent to 50:50 odds? If you can answer this question and explain why you think it is a more relevant probability then you may be able to resolve the confusion.

Roko has already given an example of such a bet: where she only gets one pay out in the tails case. Is this what you are claiming is the more relevant probability? If so, why is this probability more relevant in your estimation?

Comment author: 10 May 2010 02:12:06PM 0 points [-]

Yes, one pay out is the relevant case. The reason is because we are asking about her credence at an awakening.

Comment author: 10 May 2010 05:11:38PM 1 point [-]

How does the former follow from the latter, exactly? I seem to need that spelled out.

Comment author: 10 May 2010 06:56:48PM 0 points [-]

The interviewer asks about her credence 'right now' (at an awakening). If we want to set up a betting problem based around that decision, why would it involve placing bets on possibly two different days?

If, at an awakening, Beauty really believes that it's tails with credence 0.67, then she would gladly take a single bet of win \$1 if tails and lose \$1.50 if heads. If she wouldn't take that bet, why should we believe that her credence for heads at an awakening is 1/3?

Comment author: 07 May 2010 08:48:24PM 1 point [-]

Do you think this analysis works for the fact that a well-calibrated Beauty answers "1/3"? Do you think there's a problem with our methods of judging calibration?

Comment author: 10 May 2010 06:39:56AM 0 points [-]

What do you think the word "credence" means? I am thinking that perhaps that is the cause of your problems.

Comment author: 10 May 2010 02:17:48PM 0 points [-]

I'm treating credence for heads as her confidence in heads, as expressed as a number between 0 and 1 (inclusive), given everything she knows at the time. I see it as the same things as a posterior probability.

I don't think disagreement is due to different uses of the word credence. It appears to me that we are all talking about the same thing.

Comment author: 07 May 2010 07:11:13PM 1 point [-]

If smart people are saying "2+2=5" and I point out it's 4, would you say "what matters is why you want to know what 2+2 is"?

Yes. For example, let's take a clearer mathematical statement, "3 is prime". It seems that's true whatever people say. However, if you come across some mathematicians who are having a discussion that assumes 3 is not prime, then you should think you're missing some context rather than that they are bad at math.

I chose this example because I once constructed an integer-like system based on half-steps (the successor function adds .5). The system has a notion of primality, and 3 is not prime.

Comment author: 07 May 2010 08:57:42PM *  1 point [-]

What you should say when asked "What is 2+2?" is a separate question from what is 2+2. 2+2 is 4, but you should probably say something else if the situation calls to that. The circumstances that could force you to say something in response to a given question are unrelated to what the answer to that question really is. The truth of the answer to a question is implicit in the question, not in the question-answering situation, unless the question is about the question-answering situation.

Comment author: 07 May 2010 07:30:52PM 1 point [-]

If you want a standard system where 3 is not prime consider Z[omega] where omega^3=1 and omega is not 1. That is, the set of numbers formed by taking all sums, differences, and products of 1 and omega.

Comment author: 07 May 2010 07:37:32PM *  -2 points [-]

Given that Beauty is being asked the question, the probability that heads had come up is 1/3. This doesn't mean the probability of heads itself is 1/3. So I think this is a confusion about what the question is asking. Is the question asking what is the probability of heads, or what is the probability of heads given an awakening?

Bayes theorem:

• x = # of times awakened after heads
• y = # of times awakened after tails
• p(heads/awakened) = n(heads and awakened) / n(awakened) = x / (x+y)
• Yields 1/3 when x=1 and y=2.

Where is the probability of heads? Actually we already assumed in the calculation above that p(heads) = 0.5. For a general biased coin, the calculation is slightly more complex:

• p(T) = probability of tails
• x = # of times awakened after heads
• y = # of times awakened after tails
• p(heads/awakened) = n(heads and awakened) / n(awakened) = p(H)x / (p(H)x + p(T)y)
• Yields 1/3 when x=1 and y=2 and p(H)=p(T)=0.5.

I'm leaving this comment because I think the equations help explain how the probability-of-heads and the probability-of-heads-given-awakening are inter-related but, obviously -- I know you know this already -- not the same thing.

Comment author: 07 May 2010 08:49:49PM *  3 points [-]

To clarify, since the probability-of-heads and the probability-of-heads-given-single-awakening-event are different things, it is indeed a matter of semantics: if Beauty is asked about the probability of heads per event ... what is the event? Is the event the flip of the coin (p=1/2) or an awakening (p=1/3)? In the post narrative, this remains unclear.

Which event is meant would become clear if it was a wager (and, generally, if anything whatsoever rested on the question). For example: if she is paid per coin flip for being correct (event=coin flip) then she should bet heads to be correct 1 out of 2 times; if she is paid per awakening for being correct (event=awakening) then she should bet tails to be correct 2 out of 3 times.

Comment author: 07 May 2010 09:18:01PM *  1 point [-]

Actually .. arguing with myself now .. Beauty wasn't asked about a probability, she was asked if she thought heads had been flipped, in the past. So this is clear after all -- did she think heads was flipped, or not?

Viewing it this way, I see the isomorphism with the class of anthropic arguments that ask if you can deduce something about the longevity of humans given that you are an early human. (Being a human in a certain century is like awakening on a certain day.) I suppose then my solution should be the same. Waking up is not evidence either way that heads or tails was flipped. Since her subjective experience is the same however the coin is flipped (she wakes up) she cannot update upon awakening that it is more likely that tails was flipped. Not even if flipping tails means she wakes up 10 billion times more than if heads was flipped.

However, I will think longer if there are any significant differences between the two problems. Thoughts?

Comment author: 07 May 2010 08:20:49PM -1 points [-]

This is incorrect.

Given that Beauty is being asked the question, the probability that heads had come up is 1/2.

This is bayes' theorem:

p(H)=1/2

p(awakened|H)=p(awakened|T)=1

P(H|awakened)=p(awakened|H)P(H)/(p(awakened|H)p(H)+p(awakened|T)p(T))

which equals 1/2

Comment author: 07 May 2010 08:28:01PM 1 point [-]

By "awakened" here you mean "awakened at all". I think you've shown already that the probability that heads was flipped given that she was awakened at all is 1/2, since in both cases she's awakened at all and the probability of heads is 1/2. I think your dispute is with people who don't think "I was awakened at all" is all that Beauty knows when she wakes up.

Comment author: 10 May 2010 06:43:15AM 1 point [-]

Beauty also knows how many times she it likely to have been woken up when the coin lands heads - and the same for tails. She knew that from the start of the experiment.

Comment author: 07 May 2010 10:11:47PM *  0 points [-]

OK, I see now why you are emphasizing being awoken at all. That is the relevant event, because that is exactly what she experiences and all that she has to base her decision upon.

(But keep in mind that people are just busy answering different questions, they're not necessarily incorrect for answering a different question.)

Comment author: 08 May 2010 01:13:32PM 1 point [-]

Why was this comment down-voted so low? (I rarely ask, but this time I can't guess.) Is it too basic math? If people are going to argue whether 1/3 or 1/2, I think it is useful to know their debating about two different probabilities: the probability of heads or the probability of heads given an awakening.

Comment author: 09 May 2010 01:56:02PM *  1 point [-]

Re: "But if we specify that the money will be put into an account (and she will only be paid one winning) that she can spend after the experiment is over, which is next week, then she will find that 1/2 is the "right" answer"

That seems like a rather bizarre way to interpret: "What is your credence NOW for the proposition that our coin landed heads?" [emphasis added]

Comment author: 09 May 2010 01:39:53PM *  4 points [-]

A reasonable an idea for this and other problems that don't' seem to suffer from ugly asymptotics would simply to mechanically test it.

That is to say that it may be more efficient, requiring less brain power, to believe the results of repeated simulations. After going through the Monty Hall tree and statistics with people who can't really understand either, then end up believing the results of a simulation whose code is straightforward to read, I advocate this method--empirical verification over intuition or mathematics that are fallible (because you yourself are fallible in your understanding, not because they contain a contradiction).

Comment author: 09 May 2010 03:25:53PM *  2 points [-]

This is an interesting idea, that appeals to me owing to my earlier angle of attack on intuitions about "subjective anticipation".

The question then becomes, how would we program a robot to answer the kind of question that was asked of Sleeping Beauty?

This comment suggests one concrete way of operationalizing the term "credence". It could be a wrong way, but at least it is a concrete suggestion, something I think is lacking in other parts of this discussion. What is our criterion for judging either answer a "wrong" answer? More specifically still, how do we distinguish between a robot correctly programmed to answer this kind of question, and one that is buggy?

As in the robot-and-copying example, I suspect that which of 1/2 or 1/3 is the "correct" answer in fact depends on what (heretofore implicit) goals, epistemic or instrumental, we decide to program the robot to have.

Comment author: 10 May 2010 02:21:26PM 1 point [-]

As in the robot-and-copying example, I suspect that which of 1/2 or 1/3 is the "correct" answer in fact depends on what (heretofore implicit) goals, epistemic or instrumental, we decide to program the robot to have.

And I think this is roughly equivalent to the suggestion that the payoff matters.

Comment author: 07 May 2010 06:02:09PM 3 points [-]

The 1/3 argument says with heads there is 1 interview, with tails there are 2 interviews, and therefore the probability of heads is 1/3. However, the argument would only hold if all 3 interview days were equally likely. That's not the case here. (on a wake up day, heads&Monday is more likely than tails&Monday, for example).

Um... why? There are the same number of heads&Monday as tails&Monday; why would heads&Monday be more likely?

Comment author: 08 May 2010 06:37:30PM 0 points [-]

The smoke and mirrors with that solution is that the hypothetical repeated sampling is done in the wrong way. Think about one single awakening, which is when the question is asked. If you want to think about doing 1000 replications of the experiment, it should go like this: coin is flipped. if heads, it's monday. if tails, it's monday with prob .5 and tails with prob .5. repeat 1000 times. We'd expect 500 heads&monday, 250 tails&monday, 250 tails&tuesday. It should add up to 1000, which it does. If you do 1000 repeated trials and get more than 1000 outcomes, something is wrong. It's a very subtle issue here. (see my probability tree)

Another way to look at it: Beauty knows there's a 50% chance she's somewhere along the heads awakening sequence (which happens to be a sequence of 1 day) and a 50% chances she's somewhere along the tail awakening sequence (which is 2 days in the sleeping beauty problem or 1,000,000 days in the extreme problem). Once she's along one of these paths, she can't distinguish. So prior=posterior here.

Comment author: 09 May 2010 02:28:14PM *  1 point [-]

I make it: 500 heads & Monday ... 500 tails & Monday ... 500 tails & Tuesday.

You are arguing with http://en.wikipedia.org/wiki/Sleeping_Beauty_problem about the problem - and are making math errors in the process.

Comment author: 09 May 2010 02:00:04PM 5 points [-]

The coverage on http://en.wikipedia.org/wiki/Sleeping_Beauty_problem seems much less confused than this post.

Comment author: 07 May 2010 04:39:35PM 7 points [-]

This is one of those cases where we need to disentangle the dispute over definitions (1), forget about the notion of subjective anticipation (2), list the well-defined questions and ask which we mean.

If by the probability we mean the fraction of waking moments, the answer is 1/3.

If by the probability we mean the fraction of branches, the answer is 1/2.

Comment author: 07 May 2010 09:16:27PM *  2 points [-]

It's hard to make a sensible notion of probability out of "fraction of waking moments". Two subsequent states of a given dynamical system make for poor distinct elements of a sample space: when we've observed that the first moment of a given dynamical trajectory is not the second, what are we going to do when we encounter the second one? It's already ruled "impossible"! Thus, Monday and Tuesday under the same circumstances shouldn't be modeled as two different elements of a sample space.

Comment author: 08 May 2010 12:18:14AM 6 points [-]

As Wei Dai and Roko have observed, that depends on why you're asking in the first place. Probability estimates should pay rent in correct decisions. If you're making a bet that will pay off once at the end of the experiment, you should count the fraction of branches. If you're making a bet that will pay off once per wake-up call, you should count the fraction of wake-up calls.

Comment deleted 08 May 2010 12:56:17PM [-]
Comment author: 08 May 2010 01:01:04PM 0 points [-]

If reality gives you problems where you win by reasoning anthropically, but ordinary probability theory is not up to the job of facilitating, then invent UDT and use that instead.

The winning thing might be better than the probability thing, but it won't be a probability thing just because it's winning. Also, UDT weakly relies on the same framework of expected utility and probability spaces, defined exactly as I discuss them in the comments to this post.

Comment author: 13 May 2010 11:33:11PM 2 points [-]

Please insert a section break near the start of this post, so the whole thing doesn't show up on "NEW".

Comment author: 13 May 2010 10:06:27AM *  2 points [-]

So, I'm still working on this in my plodding, newbie-at-probability-math fashion.

What I took away from my exchanges with AlephNeil is that I get the clearest picture if I think in terms of a joint probability distribution, and attempt to justify mathematically each step of my building the table, as well as the operations of conditioning and marginalizing.

In the original Sleeping Beauty problem, we have three variables: x is how the coin came up {heads, tails}, y is the day of the week {monday, tuesday}, and z is whether I am asked for my credence (i.e. woken) {wake, sleep}.

P(x,y,z)=P(x)P(y|x)P(z|x,y) and unlike in the "revival" case x and y aren't clearly independent. In fact the answer very much seems to hinge on what we take the probability of it being tuesday, given that the coin came up heads.

The relevant possible outcomes are: (H,M,W) (H,T,W) (T,M,W) (T,T,W) (H,M,S) (H,T,S) (T,M,S) (T,T,S) - eight in all.

Conditioning on z=W consists of deleting the part of the table that has z=S, summing up all the remaining values, and renormalizing by dividing every cell in cell in the table by the total.

The rules for filling the table are: the values must add up to 1; the "heads" and "tails" branches must receive equal probability mass from P(x); and P(z|x,y) must reflect the experimental rules. So we must have the following:

• P(H,M,W) - see below
• P(H,T,W)=0
• P(T,M,W)=1/4
• P(T,T,W)=1/4
• P(H,M,S)=0
• P(H,T,S) - see below
• P(T,M,S)=0
• P(T,T,S)=0

The ambiguity seems to arise in allocating probability mass to the outcomes: "the coin comes up heads; it is Monday; I get woken up", and "the coin comes up heads; it is Tuesday; I do no get woken up". That is, I'm not sure what the correct conditional distribution P(y|x) should be.

The 1/2 answer corresponds to allocating all of the available 1/2 probability mass to the first of these outcomes in the joint table, saying P(y=M|x=H)=1 and P(y=T|x=H)=0. Or verbally, "it's certain that I get woken up on Monday if the coin comes up heads, and after that the experiment is over". The "not woken up" half of the table receives no probability mass at all.

The 1/3 answer corresponds to distributing that probability mass among the two outcomes, saying P(y=M|x)=P(y=T|x)=1/2. Verbally: "however the coin comes up, it could be either Monday or Tuesday". Here 1/4 of the total probability mass is in the "not woken up" half of the table and gets deleted when we condition on being woken.

(ETA: Where does the amnesia appear in this formalization? It doesn't, but neither does it need to. Its only practical consequence is to outlaw conditioning on the day, so working out the distribution P(x|z) conforms to the amnesia.)

Comment author: 13 May 2010 09:12:37PM 0 points [-]

I think the question we now have to ask to resolve the remaining confusion is - what, exactly, is it that Beauty is uncertain about, and at what time?

The variables we are considering only seem to make sense if Beauty is having woken up as part of the experiment. That is, assuming x means "the coin came up heads or tails", y means "it is Monday or Tuesday", and z means "I am awake or asleep" - i.e., we're dealing with uncertainty about facts that are already fixed, just unknown. Then these do not make sense outside that context.

Using that interpretation, then, and sticking to that context, we get the answer of 1/2, as if Beauty has just been woken up, she cannot allocate any probability mass to the possibility that she is asleep.

What other interpretations could there be? Perhaps the coin has not yet been flipped, and x is "the coin will come up heads (tails)", y is "it will be Monday (Tuesday) when I wake up", z is "I will be awake (asleep) when I wake up" (!). Of course, if the coin has not yet been flipped, I think we can agree 1/2 has to be the right answer. (Which then leads to the argument that it has to be 1/2 as she hasn't gained any information, but I guess that's been gone over before.) But the problem is that this y doesn't seem well-defined, as she might be woken up more than once. (Hm, this is sounding familiar as well...) We could perhaps introduce separate variables for being woken up on each day; from the pre-flip point of view, that makes more sense. But it still gets you an answer of 1/2.

This is all I can come up with; I'm not seeing what other interpretations there could be. Could someone explain just what 'x', 'y', and 'z' correspond to - if they do correspond to anything well-defined rather than having to be thrown out - in the interpretations that get you 1/3? I don't see any way for the probabilities to represent her uncertainty at the time of waking, while still having her assign nonzero probability to the possibility that she's asleep.

Comment author: 13 May 2010 09:55:35PM *  2 points [-]

I think the question we now have to ask to resolve the remaining confusion is - what, exactly, is it that Beauty is uncertain about, and at what time?

"At what time" doesn't matter in this formalism. You can be uncertain about future events or about past events, all that matters is how you update your uncertainty upon receiving new information.

So a triplet (x,y,z) represents, in the abstract, a conceivable configuration of the component uncertainties in the experimental setup. The coin could have come up heads or tails; it could be Monday or Tuesday; Beauty can be woken up on that day, or left asleep.

The joint probability P(x,y,z) is the plausibility we assign - in a timeless manner - to the corresponding propositions. Strictly speaking, it should be P(x,y,z|B) where B is our background information about the experiment: the rules, the fact that the coin is unbiased (or not known to be biased), and so on.

Our background information directs how we allocate probability mass to the various points in the sample space: P(T,T,S) corresponds to "the coin comes up tails, the day is Tuesday, Beauty is asleep". The rules of the experiment require that this be zero.

On the other hand, P(H,T,S) corresponds to "the coin comes up heads, the day is Tuesday, Beauty is asleep", and this can be non-zero.

When you learn ("condition on") some new information, the probability distribution is altered: you only keep the points which correspond to this particular variable having the value(s) you learned, and you renormalize so that the total probability is 1. So, on learning "heads" you keep only the points having x=H. On learning what day it is you keep only the points having that value for y.

When Beauty wakes up, she learns the value of z, so she can condition on z. That means she throws away the part of the joint distribution where she was supposed to be asleep. If that part of the joint distribution did contain some probability mass (as I've argued above it can), then that can make P(x|z=W) something other than 1/2.

Comment author: 13 May 2010 02:44:53PM 0 points [-]

Hm. Should "S" be representing "Beauty is asleep or the experiment is over"? Seeing as how the experiment ends after one day if heads comes up. But then, we can just modify the problem to say she's put back to sleep for the rest of Tuesday in the case of heads; that shouldn't change anything.

Comment author: 13 May 2010 03:33:14PM *  1 point [-]

It seems to me that if we make the experiment last three days instead of two, that ambiguity goes away: then it becomes clear that Beauty must assign non-zero probability mass to (H,T,S). (Or does it?)

However, that means I'd have to change my mind once again, and decide that the correct answer is in fact 1/3.

Comment author: 14 May 2010 12:45:23PM 0 points [-]

Can you explain what the three day version means in English, I'm having a little trouble parsing the spreadsheet.

Comment author: 14 May 2010 12:54:12PM 0 points [-]

See here and its grandparent.

The three day version goes: "Beauty is explained the rules on Sunday and put to sleep, then a coin is flipped. If it comes up heads, Beauty is awakened on Monday and sleeps through Tuesday and Wednesday. If it comes up tails, Beauty is awakened on Monday, Tuesday and Wednesday. On all awakenings (with the previous day's memories erased by the sleeping drug) she is asked for her credence in Heads."

This differs from the original which says "the experiment ends on Monday is the coin comes up heads". But Beauty would have the same uncertainty if you decided, in the original version, to wake Beauty on Tuesday in the event of heads, rather than Monday.

BTW the Google spreadsheet has a chat area, if you'd like to discuss this live.

Comment author: 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.

Right?

Comment author: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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.

Comment author: 07 May 2010 07:18:13PM *  2 points [-]

I agree with the others about worrying about the decision theory before talking about probability theory that includes indexical uncertainty, but separately I think there's an issue with your calculation.

"P(Beauty woken up at least once| heads)=P(Beauty woken up at least once | tails)=1"

Consider the case where a biased quantum coin is flipped and the people in 'heads' branches are awoken in green rooms while the 'tails' branches are awoken in red rooms.

Upon awakening, you should figure that the coin was probably biased to put you there. However, P(at least one version of you seeing this color room |heads) = P(at least one version of you seeing this color room |tails) = 1. The problem is that "at least 1" throws away information. p(I see this color|heads) != p(I see this color tails). The fact that you're there can be evidence that the 'measure' is bigger. The problem lies with this 'measure' thing, and seeing what counts for what kinds of decision problems.

The blue eyes problem is similar. Everyone knows that someone has blue eyes, and everyone knows that everyone knows that someone has blue eyes, yet "they gained no knew information because he only told them that at least one person has blue eyes!" doesn't hold.

Comment author: 07 May 2010 09:31:48PM 0 points [-]

The fact that you're there can be evidence that the 'measure' is bigger.

That "you are there" is evidence that the set of possible worlds consistent with your observations doesn't include the worlds that don't contain you, under the standard possible worlds sample space. Probabilistic measure is fixed in a model from the start and doesn't depend on which events you've observed, only used to determine the measure of events. Also, you might care about what happens in the possible worlds that don't contain you at all.

Comment author: 08 May 2010 06:03:42PM 0 points [-]

Probabilistic measure is fixed in a model from the start and doesn't depend on which events you've observed...

But the amount of quantum measure in each color room depends on which biased coin was flipped, and your knowledge of the quantum measure can change based on the outcome.

Comment author: 08 May 2010 12:52:47AM 6 points [-]

Add a payoff and the answer becomes clear, and it also becomes clear that the answer depends entirely on how the payoff works.

Without a payoff, this is a semantics problem revolving around the ill-defined concept of expectation and will continue to circle it endlessly.

Comment author: 08 May 2010 05:37:04AM *  5 points [-]

The problem posed is, p(heads | Sleeping Beauty is awake). There is no payoff involved. Introducing a payoff only confuses matters. For instance, Roko wrote:

But if we specify that the money will be put into an account (and she will only be paid one winning) that she can spend after the experiment is over, which is next week, then she will find that 1/2 is the "right" answer.

This is true; but that would be the answer to "What is the probability that the coin was heads, given that Sleeping Beauty was woken up at least once after being put to sleep?" That isn't the problem posed. If that were the problem posed, we could eliminate her forgetfulness from the problem statement.

If you agree that the forgetfulness is necessary to the story, then 1/2 is the wrong answer, and 1/3 is the right answer. If you don't agree it's necessary, then its presence suggests that the speaker intended a different semantics than you're using to interpret it.

ADDED: This is depressing. Here we have a collection of people who have studied probability problems and anthropic reasoning and all the relevant issues for years. And we have a question that is, on the scale of questions in the project of preparing for AGI, a small, simple one. It isn't a tricky semantic or philosophical issue; it actually has an answer. And the LW community is doing worse than random at it.

In fact, this isn't the first time. My brief survey of recent posts indicates that the LessWrong community's track record when tackling controversial problems that actually have an answer is random at best.

Comment author: 09 May 2010 03:30:34AM *  7 points [-]

There is no payoff involved. Introducing a payoff only confuses matters.

I define subjective probability in terms of what wagers I would be willing to make. I think a good rule of thumb is that if you can't figure out how to turn the problem into a wager you don't know what you're asking. And, in fact, when we introduce payoffs to this problem it becomes extremely clear why we get two answers. The debate then becomes a definition debate over what wager we mean by the sentence "what credence should the patient assign..."

Comment author: 09 May 2010 04:05:08AM 1 point [-]

As I just explained, the fact that the original author of the story wrote amnesia into it tells you which definition the author of the story was using.

Comment author: 09 May 2010 04:20:05AM *  3 points [-]

And that's a good argument you've got there, but I don't think that is totally obvious on the first read of the problem. It's a weird feature of a probability problem for the relevant wager to be offered once under some circumstances and twice under others. So people get confused. It is a little tricky. But, far from confusing things, that entire issue can be avoided if we specify exactly how the payoff works when we state the problem! So I don't know why you're freaking out about Less Wrong's ability to answer these problems when it seems pretty clear that people interpret the question differently, not that they can't think through the issues.

(Not my downvote, btw)

Comment author: 10 May 2010 07:18:37AM 3 points [-]

Re: "Introducing a payoff only confuses matters."

Personally, I think it clarifies things - though at the expense of introducing complication. People disagree over which bet the problem represents. Describing those bets highlights this area of difference.

Comment author: 10 May 2010 10:37:48PM *  1 point [-]

I see what you mean. But some comments have said, "I can set up a payoff scheme that gives this answer; therefore, this is an equally-valid answer." The correct response is to state the payoff scheme that gives your answer, and then admit your answer is not addressing the problem if you can't find justification for that payoff scheme in the problem statement.

Comment author: 11 May 2010 05:47:10AM *  1 point [-]

Indeed - that would be bad - and confusing.

It is both bad and confusing that people are defending the idea that this problem is not clearly-stated enough to answer.

I suspect this happens because, people don't like criticising the views of others. They would rather just say 'you are both right' - since then no egos get bruised, and a costly fight is avoided. So, nonsense goes uncriticised, and the innocent come to believe it - because nobody has the guts to knock it down.

Comment deleted 08 May 2010 06:08:18PM [-]
Comment author: 08 May 2010 06:46:32PM *  -1 points [-]

"ADDED: This is depressing. Here we have a collection of people who have studied probability problems and anthropic reasoning and all the relevant issues for years. And we have a question that is, on the scale of questions in the project of preparing for AGI, a small, simple one. It isn't a tricky semantic or philosophical issue; it actually has an answer. And the LW community is doing worse than random at it."

That's why I posted this to begin with. It is interesting that we can't come to an agreement on the solution to this problem, even though it involves very straightforward probability. Heck, I got heavily down voted after making statements that were correct. People are getting thrown off by doing the wrong kind of frequency counting.

--

However, I should note that the event 'sleeping beauty is awake' is equivalent to 'sleeping beauty has been woken up at least once' because of the amnesia. The forgetfulness aspect of the problem is why the solution is 1/2.

Comment author: 09 May 2010 02:21:20PM 1 point [-]

Which of your down-voted statements were correct?

Comment author: 10 May 2010 01:09:02PM 0 points [-]

Well, I got -6 for this statement: "P(monday and heads)=1/2. P(monday and tails)=1/4. P(tuesday and tails)=1/4. Remember, these have to add to 1."

Initially there is a 50% chance for heads and 50% chance for tails. Given heads, it's monday with certainty. So, P(heads)=1/2, p(monday | heads)=1.

Do you dispute either of those?

Similarly, p(tails)=1/2, p(monday | tails)=1/2. p(tuesday | tails)=1/2.

Do you dispute either of those?

The above are all of the probabilities you need to know. From them, you can derive anything that is of interest here.

etc.

Comment author: 10 May 2010 05:44:19PM 1 point [-]

Re: "P(monday and heads)=1/2. P(monday and tails)=1/4. P(tuesday and tails)=1/4. Remember, these have to add to 1."

Yes, but those Ps are wrong - they should all be 1/3.

Comment author: 10 May 2010 08:47:28PM 0 points [-]

My assumptions and use of probability laws are clearly stated above. Tell me where I made a mistake, otherwise just saying "you're wrong" is not going to move things forward.

Comment author: 10 May 2010 09:26:47PM 1 point [-]

Well, the correct sum is this one:

"Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3. This is the correct answer from Beauty's perspective."

That gives:

P(monday and heads)=500/1500. P(monday and tails)=500/1500. P(tuesday and tails)=500/1500.

You appear to have gone wrong by giving a different answer - based on a misinterpretation of the meaning of the interview question, it appears.

Comment author: 11 May 2010 02:06:45AM 0 points [-]

So you are not willing to tell me where I made a mistake?

You're using expected frequencies to estimate a probability, apparently. But you're counting the wrong thing. What you are calling P(monday and heads) is not that. There is a problem with your denominator. Think about it. Your numerator has a maximum value of 1000 (if the experiment was repeated 1000 times). Your denominator has a maximum value of 2000. If the maximum possible values of the numerator and denominator do not match, there is a problem. You have an outcome-dependent denominator. Try taking expectation of that. You won't get what you think you'll get.

Comment author: 11 May 2010 05:56:54AM 1 point [-]

Re: "If the maximum possible values of the numerator and denominator do not match, there is a problem.

The total possible number of awakenings is 2000.

That represents all tails - e.g.:

P(monday and heads) = 0/2000; P(monday and tails) = 1000/2000; P(tuesday and tails) = 1000/2000;

These values add up to 1 - i.e. the total numerators add up to the commonn denominator. That is the actual constraint. The maximum possible value of the numerator in each individual fraction is permitted to be smaller than the common denominator - that is not indicative of a problem.

Comment author: 10 May 2010 07:15:33PM 0 points [-]

Or they all should be 1/2.

Comment author: 10 May 2010 07:49:57PM 1 point [-]

Impossible - if they are to add up to 1.

Comment author: 10 May 2010 08:44:31PM 0 points [-]

For Jack's bookie, I agree, you have to use 1/3 – but if you want to calculate a distribution on how much cash Beauty has after the experiment given different betting behavior, it no longer works to treat Monday and Tuesday as mutually exclusive.

Comment author: 09 May 2010 04:06:18AM 1 point [-]

I'd like to see a model of how a group of people is supposed to improve their initial distribution of beliefs in a problem with a true/false answer.

Comment author: 10 May 2010 07:12:02AM *  0 points [-]

Distressingly few people have publicly changed their mind on this thread. Various people show great persistence in believing the wrong answer - even when the problem has been explained. Perhaps overconfidence is involved.

Comment author: 10 May 2010 05:55:40PM 3 points [-]

I changed my mind from "1/3 is the right answer" to "The answer is obviously 1/2 or 1/3 once you've gotten clear on what question is being asked". I'm not sure if I did so publicly. It seems to me that other folks have changed their minds similarly. I think I see an isomorphism to POAT here, as well as any classic Internet debate amongst intelligent people.

Comment author: 10 May 2010 06:07:32PM 1 point [-]

I'm not sure whether this is legitimate or a joke, but if the question is unclear about whether 1/2 or 1/3 is better, maybe 5/12 is a good answer.

Comment author: 10 May 2010 06:12:03PM 1 point [-]

I'm also not sure if you're serious, but if you assign a 50% probability to the relevant question being the one with the correct answer of '1/2' and a 50% probability to the relevant question being the one with the correct answer of '1/3' then '5/12' should maximize your payoff over multiple such cases if you're well-calibrated.

Comment author: 10 May 2010 06:08:59PM *  0 points [-]

Phil and I seem to think the problem is sufficiently clearly specified to give an answer to. If you think 1/2 is a defensible answer, how would you reply to Robin Hanson's comment?

FWIW, on POAT I am inclined towards "Whoever asked this question is an idiot".

Comment author: 10 May 2010 06:23:51PM 0 points [-]

Actually I think it would make more sense to reply to my own comment in response to this. link

Comment author: 10 May 2010 08:13:50PM *  0 points [-]

I am not sure that is going anywhere.

Personally, I think I pretty-much nailed what was wrong with the claim that the problem was ambiguous here.

Comment author: 10 May 2010 08:26:57PM 3 points [-]

I think that we've established the following:

1. there are some problems similar to this one for which the answer is 1/2
2. there are some problems similar to this one for which the answer is 1/3
3. people seem to be disagreeing which sort of problem this is
4. all debate has devolved to debate over the meanings of words (in the problem statement and elsewhere)

Given this, I think it's obvious that the problem is ambiguous, and arguing whether the problem is ambiguous is counterproductive as compared to just sorting out which sort of problem you're responding to and what the right answer is.

Comment author: 12 May 2010 06:20:03AM 3 points [-]

I'm coming around to the 1/2 point of view, from an initial intuition that 1/3 made most sense, but that it mostly depended on what you took "credence" to mean.

My main new insight is that the description of the set-up deliberately introduces confusion, it makes it seem as if there are two very different situations of "background knowledge", X being "a coin flip" and X' being "a coin flip plus drugs and amnesia". So that P(heads|X) may not equal P(heads|X').

This comment makes the strongest case I've seen that the difference is one that makes no difference. Yes, the setup description strongly steers us in the direction of taking "credence" to refer to the number of times my guess about the event is right. If Beauty got a candy bar each time she guessed right she'd want to guess tails. But on reflection what seems to matter in terms of being well-calibrated on the original question is how many distinct events I'm right about.

Take away the drug and amnesia, and suppose instead that Beauty is just absent-minded. On Tuesday when you ask her, she says: "Oh crap, you asked me that yesterday, and I said 1/2. But I totally forget if you were going to ask me twice on tails or on heads. You'd think with all they wrote about this setup I'd remember it. I've no idea really, I'll have to go with 1/2 again. Should be 1 for one or the other, but what can I say, I just forget."

I'm less than impressed with the signal-to-noise ratio in the recent discussion, in particular the back-and-forth between neq1 and timtyler. As a general observation backed by experience in other fora, the more people are responding in real time to a controversial topic, the less likely they are to be contributing useful insights.

I'm not ruling out changing my mind again. :)

Comment author: 12 May 2010 09:36:47AM *  1 point [-]

I've been thinking 1/2 as well (though I'm also definitely in the "problem is underdefined" camp).

Here is how describe the appropriate payoff scheme. Prior to the experiment (but after learning the details) Beauty makes a wager with the Prince. If the coin comes up heads the Prince will pay Beauty \$10. If it comes up tails Beauty will pay \$10. Even odds. This wager represents Beauty's prior belief that the coin is fair and head/tails have equal probability: her credence that heads will or did come up. At any point before Beauty learns what day of the week it is she is free alter the bet such that she takes tails but must pay \$10 more dollars to do so (making the odds 2:1).

Beauty should at no point (before learning what day of the week it is) alter the wager. Which means when she is asked what her credence is that the coin came up heads she should continue to say 1/2.

This seems at least as good an payoff interpretation as a new bet every time Beauty is asked about her credence.

Comment author: 12 May 2010 11:04:37AM *  0 points [-]

You don't measure an agent's subjective probability like that, though - not least because in many cases it would be bad experimental methodology. Bets made which are intended to represent the subject's probability at a particular moment should pay out - and not be totally ignored. Otherwise there may not be any motivation for the subject making the bet to give an answer that represents what they really think. If the subject knows that they won't get paid on a particular bet, that can easily defeat the purpose of offering them a bet in the first place.

Comment author: 12 May 2010 11:11:58AM *  0 points [-]

This doesn't make any sense to me. Or at least the sense it does make doesn't sound like sufficient reason to reject the interpretation.

Comment author: 12 May 2010 07:08:04AM *  1 point [-]

If Beauty forgets what is going on - or can't add up - her subjective probability could potentially be all over the shop.

However, the problem description states explicitly that: "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."

This seems to me to weigh pretty heavily against the hypothesis that she may have forgotten the details of the experiment.

Comment author: 12 May 2010 08:29:14AM 1 point [-]

In the case where she remembers what's going on, when you ask her on Tuesday what her credence is in Heads, she says "Well, since you asked me yesterday, the coin must have come up Tails; therefore I'm updating my credence in Heads to 0."

The setup makes her absent-minded (in a different way than I suggest above). It erases information she would normally have. If you told her "It's Monday", she'd say 1/2. If you told her "It's Tuesday", she'd say 0. The amnesia prevents Beauty from conditioning on what day it is when she's asked.

Prior to the experiment, Beauty has credence 1/2 in either Heads or Tails. To argue that she updates that credence to 1/3, she must be be taking into account some new information, but we've established that it can't be the day, as that gets erased. So what it is?

Jonathan_Lee's post suggests that Beauty is "conditioning on observers". I don't really understand what that means. The first analogy he makes is to an identical-copy experiment, but we've been over that already, and I've come to the conclusion that the answer in that case is "it depends".

Comment author: 12 May 2010 08:55:02AM *  0 points [-]

Re: "Prior to the experiment, Beauty has credence 1/2 in either Heads or Tails."

IMO, we've been over that adequately here. Your comment there seemed to indicate that you understood exactly when Beauty updates.

Comment author: 12 May 2010 09:42:41AM 0 points [-]

Yes. I noted then that the description of the setup could make a difference, in that it represents different background knowledge.

It does not follow that it does make a a difference.

When I say "prior to the experiment", I mean chronologically, i.e. if you ask Beauty on Sunday, what her credence is then in the proposition "the coin will come up heads", she will answer 1/2.

Once Beauty wakes up and is asked the question, she conditions on the fact that the experiment is now ongoing. But what information does that bring, exactly?

Comment author: 10 May 2010 12:56:04PM 0 points [-]

Yes, this is very alarming, considering this is a forum for aspiring rationalists.

Comment author: 09 May 2010 04:01:02AM 1 point [-]

However, I should not that the event 'sleeping beauty is awake' is equivalent to 'sleeping beauty has been woken up at least once' because of the amnesia.

I disagree; but I've already given my reasons.

Comment deleted 08 May 2010 12:39:54PM [-]
Comment author: 08 May 2010 01:54:20PM 2 points [-]

If you don't need to condition on it, why is it in the story?

The question asked in the story is "Sleeping Beauty, what is p(heads | you are awake now)?"

Someone is going to complain that you can't ask about p(heads) when it's already either true or false. Well, you can. That's how we use probabilities. If you are a determinist, you believe that everything is already either true or false; yet determinists still use probabilities.

Comment author: 08 May 2010 01:57:24PM 1 point [-]

If you don't need to condition on it, why is it in the story?

"On Sunday she is given a drug" is also in the story. Does it follow that it is imperative to explicitly condition on that as well?

Comment author: 08 May 2010 05:23:46PM *  3 points [-]

For my own benefit, i'll try to explain my thinking on this problem, in my own words, because the discussions here are making my head spin. Then the rest of you can tell me whether i understand. The following is what i reasoned out before looking at neq1's explanations.

Firstly, before the experiment begins, i'd expect a 50% chance of heads and a 50% chance of tails. Simple enough.

If it lands on heads, then i wake up only once, on Monday. If it lands on tails, then i wake up once on Monday, and a second time on Tuesday.

So, upon waking with amnesia, i'd expect a 50% chance of it being my first-and-only interview on Monday. I'd expect a 25% chance of it being my first-of-two interviews on Monday, and a 25% chance of it being my second-of-two interviews on Tuesday.

And due to the amnesia, and my having no indication of what day it is, i'd basically have no new information to act on after i wake up. So my probability estimates would remain the same after waking as they were before.

So, upon waking, i'd say:

• 50% chance that the coin landed on heads, and it's currently Monday.
• 25% chance that the coin landed on tails, and it's currently Monday.
• 25% chance that the coin landed on tails, and it's currently Tuesday.

In other words, neq1's probability-tree picture turned out to most clearly match my own reasoning on the problem. Does this make sense?

Comment author: 09 May 2010 01:56:53AM 1 point [-]

This was also my understanding of the problem. Are we missing something?

Comment author: 09 May 2010 02:08:45PM *  2 points [-]

On awakening, I would give:

``````* 33% chance that the coin landed on heads, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Tuesday.
``````

p(heads) and p(tails) on Monday should be equal (a fair coin was flipped). p(tails) on Monday and p(tails) on Tuesday should also be equal (nothing important changes in the interim).

Comment author: 10 May 2010 12:50:45PM 0 points [-]

Even though you knew ahead of time that there was a 50% chance you'd be on the heads path, and a 50% chance you'd be on the tails path, you'd shift those around without probability law justification?

I also think you are not careful with your wording. What does p(heads) on Monday mean? Is it a joint or conditional probability? p(heads | monday) = p(tails | monday), yes, but Beauty can't condition on Monday since she doesn't know the day. If you are talking about joint probabilities, p(heads and monday) does not equal p(tails and monday).

Comment author: 10 May 2010 05:53:39PM *  1 point [-]

Re: a 50% chance you'd be on the heads path, and a 50% chance you'd be on the tails path.

Those are not the probabilities in advance of the experiment being perfomed. Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails. These probabilities do not change during the experiment - since Beauty learns nothing.

Comment author: 12 May 2010 01:15:59PM 0 points [-]

"Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails."

Suppose 50% of the population has some asymptomatic form of cancer. We randomly select someone and do a diagnostic test. If they have cancer (we don't tell them), we wake them up 9 times and ask their credence for cancer (administering amnesia-inducing drug each time). If they don't have cancer, we wake them up once.

The person selected for this experiment knows there is a 50% chance they have cancer. And they decide ahead of time that, upon awakening, they'll be 90% sure they have cancer. And this makes sense to you.

Comment author: 10 May 2010 05:51:21PM 1 point [-]

Re: "but Beauty can't condition on Monday since she doesn't know the day."

She could make a bet. You do not have to know what day of the week it is in order to make a bet that it is Monday.

Comment author: 10 May 2010 05:47:53PM 1 point [-]

Re: "If you are talking about joint probabilities, p(heads and monday) does not equal p(tails and monday)."

Sure it does - if a fair coin was flipped!

Comment author: 10 May 2010 08:44:47PM 0 points [-]

Maybe instead of just saying it's true, you could look at my proof and show me where I made a mistake. I've done that with yours.

Comment author: 10 May 2010 09:06:32PM *  1 point [-]

I think you already clarified that here.

You interpreted:

"What is your credence now for the proposition that our coin landed heads?"

...as being equivalent a bet along these lines:

"the scenario where at each awakening we offer a bet where she'd lose \$1.50 if heads and win \$1 if tails, and we tell her that we will only accept whichever bet she made on the final interview."

...which is a tortured interpretation.

The question says "now". I think the correct corresponding wager is for Beauty to make a bet which is judged according to its truth value there and then - not for it to be interpreted later and the payout modified or cancelled as a result of other subsequent events.

Comment author: 09 May 2010 12:23:05PM 0 points [-]

My reasoning was a bit simpler.

Prior to the experiment, the probability of heads was 50%, tails 50%. Upon waking.. she learns no new information. She knew in advance she was going to wake up, and they tell her nothing.

So how could her beliefs possibly change?

Comment author: 09 May 2010 02:04:37PM 1 point [-]

She knew from the start that she is twice as likely to be asked when it is tails. So, her estimate of the chances of her being awakened facing tails should be bigger from the beginning.

Comment author: 09 May 2010 07:03:29PM 2 points [-]

Thank you, your explanation for the 1/3 answer makes sense to me. I'm still a bit confused about it, but i think i feel like i might be changing my mind.

I'll try to figure out what would happen if SB makes a bet on the coin flip at each interview. Suppose she guesses heads each time, then:

• Given that the result was heads, then she is interviewed once, and she is right once.
• Given that the result was tails, then she is interviewed twice, and she is wrong twice.
• ... meaning that if the experiment is repeated several times, the guess "heads" will be correct for one out of three guesses. Just like you said.

(Perhaps it's important to realize that, if the coin lands on tails, then she's guaranteed to wake up once on Monday, and also guaranteed to wake up once on Tuesday. Now that i read your other comment again, i see your meaning when you say that p(heads) and p(tails) for each day is the same.)

Comment author: 14 May 2010 08:02:04PM 0 points [-]

I couldn't decide exactly you meant by "twice as likely to be asked (woken) when it's tails" either. I'm going to guess that you're averaging evenly over Monday and Tuesday, in which case I agree. After marginalizing over M/T, P(wake|heads)=1/2 and P(wake|tails)=1.

Comment author: 10 May 2010 12:52:31PM 0 points [-]

"She knew from the start that she is twice as likely to be asked when it is tails. "

The probability that she would be asked is 1, regardless of the outcome of the coin. Her estimate of the chances of her being awakened should have been 1.

Comment author: 10 May 2010 05:45:56PM 1 point [-]

Yes: her estimate of the chances of her being awakened is indeed 1.

Comment author: 09 May 2010 03:25:49PM 1 point [-]

One of the major take-aways I got from actually reading Jaynes was how he is always careful to write probabilities as conditioned on all prior knowledge: P(A|X) where X is our "background knowledge".

This is useful in the present case since we can distinguish X, Beauty's background knowledge about which way a given coin might land, and X', which represents X plus the description of the experimental setup, including the number of awakenings in each case.

That - the difference between X and X' - is the new information that Beauty learns and which might make P(heads|X') different from P(heads|X).

Comment author: 07 May 2010 08:02:52PM 2 points [-]

I have a question for those more familiar with the discussions surrounding this problem: is there anything really relevant about the sleeping/waking/amnesia story here? What if instead the experimenter just went out and asked the next random passerby on the street each time?

It seems to me that the problem could be formulated less confusingly that way. Am I missing something?

Comment deleted 08 May 2010 03:15:06AM [-]
Comment author: 08 May 2010 07:32:35PM 0 points [-]

I'm not sure I understand your "really extreme" formulation fully. Is the amnesia supposed to make the wins in chocolate bars non-cumulative?

Comment author: 07 May 2010 08:06:32PM 1 point [-]

What if instead the experimenter just went out and asked the next random passerby on the street each time?

I'm confused about how that's supposed to have the same relevant features, so the answer to your question is probably "Yes".

Are you suggesting the following?: Flip a coin. Go out and ask a random passerby what the probability is that the coin came up heads.

If so, you've entirely eliminated Beauty's subjective uncertainty about whether she's been woken up once or more than once, which is putatively relevant to subjective probability.

Comment author: 07 May 2010 08:32:20PM *  3 points [-]

The exact equivalent of the original problem would be as follows. You announce that:

(1) You're about to flip a coin at some secret time during the next few days, and the result will be posted publicly in (say) a week.

(2) Before the flip, you'll approach a random person in the street and ask about their expectation about the result that's about to be posted. After the flip, if and only if it lands tails, you'll do the same with one additional person before the result is announced publicly. The persons are unaware of each other, and have no way to determine if they're being asked before or after the actual toss.

So, does anyone see relevant differences between this problem and the original one?

Comment author: 18 May 2010 01:23:45AM 0 points [-]

I'm guessing you already understood this, but as a person accosted and informed of this procedure, I know it's more likely that I heard about it because the result was tails (than I was to hear about it before the toss). Those experiments that resulted in heads, I (most likely) never got to hear about.

So in asking if there's any relevant thing that's different, you expect a halfer to come forth and explain himself. Unfortunately, I'm not one. But it does seem to me that the only possible important difference is that Beauty knows about the experiment before the coin is tossed; but perhaps the amnesia compensates exactly for that.

As far as your "an instrument sold by a (so far completely ignorant) third party that pays off \$100 if the announced result is tails", then of course Beauty would value it exactly as your interviewees, provided she knew that the offer was to be made at every interview.

Comment author: 07 May 2010 08:36:37PM 0 points [-]

Well you also have to note in the problem description that a particular person is asked, and ask what should their guess be (so far you just got as far as the announcement).

But I think that's equivalent.

Comment author: 07 May 2010 08:52:02PM *  3 points [-]

Well, yes, I should also specify that you'll actually act on the announcement.

But in any case, would anyone find anything strange or counterintuitive about this less exotic formulation, which could be readily tried in the real world? As soon as the somewhat vague "expectation about the result" is stated clearly, the answer should be clear. In particular, if we ignore risk aversion and discount rate, each interviewee should be willing to pay, on the spot, up to \$66.66 for an instrument sold by a (so far completely ignorant) third party that pays off \$100 if the announced result is tails.

Comment author: 07 May 2010 08:16:39PM 2 points [-]

If the coin is tails, you would ask two random passerbies.

Comment author: 07 May 2010 08:20:22PM 0 points [-]

Aha. In that case, I'd say it's analogous, but I might just be granting that since the correct answer there is 1/3 as well. Or are there folks that would answer 1/2 to this scenario?

Comment author: 08 May 2010 07:17:31AM 1 point [-]

Yes, the answer is 1/3, because I am more likely to be asked if it was tails. But in the original problem, I am not more likely to be asked, I am just asked more often, so there is no analogy.

Comment author: 07 May 2010 09:34:23PM *  1 point [-]

It doesn't make sense to assert that probability of Tuesday is 1/4 (in the sense that it'd take a really bad model to give this answer). Monday and Tuesday of the "tails" case shouldn't be distinct elements of the sample space. What happens when you've observed that "it's not Tuesday", and the next day it's Tuesday? Have you encountered an event of zero probability? This is exactly the same reason why the solution of 1/3 can't be backed up by a reasonable model.

In the classical possible worlds model, you've got two worlds for each outcome of the coin flip, with probabilities 1/2 apiece, and so (Tuesday, tails) is the same event as (Monday, tails), weighing probability of 1/2. Thus, for example, probability that we are in the possible world where Monday can be observed, given that Tuesday can be observed, is 1, but it doesn't make sense to ask "What is probability of it being Tuesday?", unless this question is interpreted as "What is probability of us being in the possible world where it's possible to observe Tuesday?", in which case the question "What is the probability of it being Monday, given that it's Tuesday?", interpreted the same way, has "100%" as the answer.

Comment author: 09 May 2010 07:35:36PM 0 points [-]

Surely, 1/3 is the correct answer - and is backed up by a perfectly reasonable model..

Comment author: 10 May 2010 03:02:35PM 0 points [-]

"It doesn't make sense to assert that probability of Tuesday is 1/4 (in the sense that it'd take a really bad model to give this answer)."

Suppose if heads we wake Beauty up on Monday, and if tails we wake her up either on Monday or Tuesday (each with probability 1/2). In that case, when Beauty is awakened, she should it's Monday with probability .75 and tails with probability .25.

"In the classical possible worlds model, you've got two worlds for each outcome of the coin flip, with probabilities 1/2 apiece, and so (Tuesday, tails) is the same event as (Monday, tails), weighing probability of 1/2."

I agree with this. I just thought it would be more intuitive if people thought of "(Tuesday, tails) is the same event as (Monday, tails), weighing probability of 1/2" from the perspective of the experiment that I describe above (where we imagine Beauty is awakened on a random day within the space of possible days for each coin result).

I have no problem imagining a probability distribution for Tuesday, just like I can imagine a probability distribution for the mean of some random variable.

Comment author: 08 May 2010 02:42:20AM *  0 points [-]

If Sleeping Beauty doesn't know what day it is, what could possibly motivate her to say that the probability of heads is something other than 50%? I mean, she knows nothing about the coin except that it's round and shiny, and the metal costs more than the coin does.

Unless I misunderstood, this problem is smoke and mirrors.

Comment author: 09 May 2010 07:41:23PM 2 points [-]

If she thinks she will be asked what the coin shows more times if it is tails.

Comment author: 05 August 2013 09:13:04PM 1 point [-]

I agree with the author of this article. After having done a lot of research on the Sleeping Beauty Problem as it was the topic of my bachelor's thesis (philosophy), I came to the conclusion that anthropic reasoning is wrong in the Sleeping Beauty Problem. I will explain my argument (shortly) below:

The principle that Elga uses in his first paper to validate his argument for 1/3 is an anthropic principle he calls the Principle of Indifference:

"Equal probabilities should be assigned to any collection of indistinguishable, mutually exclusive and exhaustive events."

The Principle of Indifference is in fact a more restricted version of the Self-Indication Assumption:

"All other things equal, an observer should reason as if they are randomly selected from the set of all possible observers."

Both principles are to be accepted a priori as they can not be attributed to empirical considerations. They are therefore vulnerable to counterarguments...

The counterargument:

Suppose that the original experiment is modified a little:

If the outcome of the coin flip is Heads, they wake Beauty up at exactly 8:00. If the outcome of the first coin flip is Tails, the reasearchers flip another coin. If it lands Heads they wake Beauty at 7:00, if Tails at 9:00. That means that when Beauty wakes up she can be in one of 5 situations:

Tails and Monday 7:00

Tails and Monday 9:00

Tails and Tuesday 7:00

Tails and Tuesday 9:00

Again, these situations are mutually exclusive, indistinguishable and exhaustive. Hence thirders are forced to conclude that P(Heads) = 1/5.

Thirders might object that Beauty's total credence in the Tails-world would still have to equal 2/3, as Beauty is awakened twice as many times in the Tails-world as in the Heads-world. They are then forced to explain why temporal uncertainty regarding an awakening (Monday or Tuesday) is different from temporal uncertainty regarding the time (7:00 or 9:00 o’clock). Both classify as temporal uncertainties within the same possible world, what could possibly set them apart?

An explanation could be that Beauty is only is asked for her credence in Heads during an awakening event, regardless of the time, and that such an event occurs twice in the Tails-world. That is, out of the 4 possible observer-moments in the Tails-world there are only two in which she is interviewed. That means that simply the fact that she is asked the same question twice is reason enough for thirders to distribute their credence, and it is no longer about the number of observer moments. So if she would be asked the same question a million times then her credence in Heads would drop to 1/1000001!

We can magnify the absurdity of this reasoning by imagining a modified version of the Sleeping Beauty Problem in which a coin is tossed that always lands on Tails. Again, she is awakened one million times and given an amnesia-inducing potion after each awakening. Thirder logic would lead to Beauty’s credence in Tails being 1/1000000, as there are one million observer-moments where she is asked for her credence within the only possible world; the Tails-world. To recapitulate: Beauty is certain that she lives in a world where a coin lands Tails, but due to the fact that she knows that she will answer the same question a million times her answer is 1/1000000. This would be tantamount to saying that Mt. Everest is only 1m high when knowing it will be asked 8848 times! It is very hard to see how amnesia could have such an effect on rationality.

Conclusion:

The thirder argument is false. The fact that there are multiple possible observer-moments within a possible world does not justify dividing your credences equally among these observer-moments, as this leads to absurd consequences. The anthropic reasoning exhibited by the Principle of Indifference and the Self-Indication Assumption cannot be applied to the Sleeping Beauty Problem and I seriously doubt if it can be applied to other cases...

Comment author: 29 October 2011 03:15:13PM 1 point [-]

After tinkering with a solution, and debating with myself how or whether to try it again here, I decided to post a definitive counter-argument to neq1's article as a comment. It starts with the correct probability tree, which has (at least) five outcomes, not three. But I'll use the unknown Q for one probability in it:

••••••• Monday---1---Waken; Pr(observe Heads and Monday)=Q/2 ••••••••••/ ••••••••Q •••••••/ ••• Heads •••••/••\••••••••••••1---Sleep; Pr(sleep thru Heads and Tuesday)=(1-Q)/2 ••••/•••1-Q•••••••/ ••1/2••••\••••••••/ ••/•••• Tuesday--0---Waken; Pr(observe Heads and Tuesday)=0 •/ + •\ ••\•••• Monday---1---Waken; Pr(observe Tails and Monday)=1/4 ••1/2••••/ ••••\••1/2 •••••\••/ ••• Tails •••••••\ •••••••1/2 ••••••••••\ ••••••• Tuesday--1---Waken; Pr(observe Tails and Tuesday)=1/4

What halfers refuse to recognize, is that whether Beauty is awakened in any specific circumstance is a decision that is part of the process. It is based on the other two random variables, after both – repeat, both – have been determined. The event “Heads and Tuesday” is an event that exists in the sample space, and the decision to not awaken her is made only after that event has occurred. Halfers think they have to force that event into non-existence by making Q=1, when all the experiment requires is that the probability Beauty will observe it is zero. This is the point one thirder argument utilizes, that of Radford Neal’s companion Prince who is always awakened but only asked if Beauty is awakened.

In fact, there is no reason why the probability that it is Monday, given Heads, should be any different than the probability it is Monday, given Tails. So, with Q=1/2, we get that Pr(observe heads)=1/4, Pr(observe anything)=3/4, so Pr(Heads|observe anything)=1/3. QED.

Neq1’s arguments that the thirder positions are wrong are all examples of circular reasoning. He makes some assumption equivalent to saying the answer is 1/2, and from that proves the answer is 1/2. For example, when he uses “Beauty woken up at least once” as a condition, all his terms are also conditioned on the fact that the rules of the experiment were followed. So when he inserts the completely unconditional “Pr(Heads)=1/2” on the right-hand side of the equation, he really should use Pr(heads|rules followed), which is the unknown we are trying to find. It is then unsurprising that he gets the number he inserted, especially if you consider what using a probability-one event as a condition in Bayes’ Rule means.

Where neq1 claims that Nick Bostrom’s argument is wrong in “Disclosure Process 1,” I suggest he go back and use the values from his probability tree. Her credence of heads is (1/2)/(1/2+1/2/1,000,000). In the second process, it is either (1/2)/(1/2+1/2/7,000,000) of (1/2)/(1/2+1/2/1,000,000,000,000), depending on what “specific day” means.

Comment author: 08 September 2011 02:11:19PM 1 point [-]

Sleeping Beauty does not sleep well. She has three dreams before awakening. The Ghost of Mathematicians Past warns her that there are two models of probability, and that adherents to each have little that is good to say about adherents to the other. The Ghost of Mathematicians Present shows her volumes of papers and articles where both 1/2 and 1/3 are "proven" to be the correct answer based on intuitive arguments. The Ghost of Mathematicians Future doesn't speak, but shows her how reliance on intuition alone leads to misery. Only strict adherence to theory can provide an answer.

Illuminated by these spirits, once she is fully awake she reasons: "I have no idea whether today is Monday or Tuesday; but it seems that if I did know, I would have no problem answering the question. For example, if I knew it was Monday, my credence that the coin landed heads could only be 1/2. On the other hand, if I knew it was Tuesday, my credence would have to be 0. But on the gripping hand, these two incontrovertible truths can help me answer as my night visitors suggested. There is a theorem in probability, called the Theorem of Total Probability, that says the probability for event A is equal to the probability of the sum of the events (A intersect B(i)), where B(i) partitions the entire event space.

"Today has to be either Monday or Tuesday, and it can't be both, so these two days represent such a partition. Since I want to avoid making any assumptions as long as I can, let me say that the probability that today is Monday is X, and the probability that it is Tuesday is (1-X). Now I can use this Theorem to state, unequivocally, that my credence that the coin landed heads is P(heads)=(1/2)X+0(1-X)=X/2.

"But I know that it is possible that today is Tuesday; even a Bayesian has to admit that X<1. So I know that 1/2 cannot be correct; the answer has to be less than that. A Frequentist would say that X=2/3 because, if this experiment were repeated many times, two out of every three interviews would take place on Monday. And while a Bayesian could, in theory, choose any value that is less than 1, it is a violation of Occam's Razor to assume there is a factor present that would make X different than 2/3. So, it seems my answer must be 1/3.

Comment author: 05 August 2013 11:45:01PM 0 points [-]

You can have a credence of 1/2 for heads in the absence of which-day knowledge, but for consistency you will also need P(Heads | Monday) = 2/3 and P(Monday) = 3/4. Neither of these match frequentist notions unless you count each awakening after a Tails result as half a result (in which case they both match frequentist notions).

Comment author: 11 April 2011 11:25:02PM 1 point [-]

The whole anthropics debate is over things that you have taken as assumptions e.g. whether waking up is identical evidence to merely knowing that you wake at least once, whether the three days are equally likely

Comment author: 17 May 2010 11:15:34PM *  1 point [-]

Proof that neq1 is wrong:

Let H be the event that heads was flipped in this experiment instance. We're going to let Beauty experience a waking now. Let M be the event that the waking is on Monday. Let B be the information that Beauty (knowing the experiment design) has upon waking. Let h=P(H|B), and let m=P(M|B).

We wish to discover the true values of h and m. Clearly in the context of someone being asked about the expected outcome of the experiment, P(H)=1/2, but h may (or may not) differ from 1/2.

• Fact 1: P(H|M,B)=P(H)=1/2

• Fact 2: P(H|~M,B)=0 (by ~M I mean the complement of M, i.e. that it's not Monday)

Given the above two facts, we know enough to solve for h and m.

lemma 1:

• P(~H|B)=P(M,B)P(~H|M,B)+P(~M)P(~H|~M,B) ; probability axiom
• (1-h)=m(1/2)+(1-m)(1) ; by facts 1-2 and above axiom
• 1-(h)=1-(m/2) ; above simplified
• h=m/2

lemma 2:

• P(H|B)=P(M)P(H|M,B)+P(~M)(P(H|~M,B) ; probability axiom
• h=m(1/2)+(1-m)(0) ; facts 1-2 and above
• h=m/2 ; simplified
• m=2h

(oops, that turned out to be redundant; not surprising since I'm using in lemma 2 the variants p(~X)=1-P(X) from the same facts 1+2).

P(H|B) is a weighted average of the probability for heads given Monday (1/2) and given Tuesday (0). It turns out that, according to thirders, it's more likely that it's Monday (m=2h=2/3).

The thirder argument is that m=2/3 (that is, 2 out of 3 wakings on average are on Monday). The halfer argument that h=1/2 implies that m=1; that is, that Beauty is certain that it's Monday (but this is obviously stupid of her).

I was originally sympathetic to neq1's argument that B is merely "1 or more wakings occur" and that P(H|1 or more wakings occur)=P(H)=1/2, since 1 or more wakings always occur, no matter whether H or ~H. But B is better characterized as "Beauty has just been woken, not knowing whether it's the first or second waking, but knowing the experiment design".

I would like to strengthen this argument to prove that m=2/3.

Comment author: 18 May 2010 12:03:29AM 2 points [-]

Lemma 1 is wrong. -h=(-1/2)m, m=2h. So your two lemmas are just saying the same thing.

Comment author: 18 May 2010 12:24:04AM *  1 point [-]

I agree. I should have used a computer algebra program ;) I've revised my post so that it's correct. It's funny to me that I let slip a computation error that happened to accidentally give me the result I expected.

Comment author: 13 May 2010 05:01:21PM 1 point [-]

Just an observation: I've mostly ignored this discussion, but it appears to have generated a lot of meaningful debate about the very fundamental epistemic issues at play (though a lot of unproductive debate as well). No consensus on which position is idiotic has apparently arisen.

With that in mind, surely this article should be rated above 1? Are the upvotes being canceled by downvotes, or are people just not voting it either way? Why isn't this rated higher?

Comment author: 13 May 2010 05:12:49PM 2 points [-]

I, for one, rate articles by the article text alone, not by the discussions generated in their comment threads.

Comment author: 13 May 2010 05:17:07PM *  1 point [-]

Okay. That's good -- I agree with that standard. So is the consensus that, however productive the debate might be that is going on in the comments, the article that prompted them wasn't very good? If so, the rating seems reasonable. (I felt the same way about the top-level article that was basically just the question, "What are you doing, and why are you doing it?")

Comment author: 13 May 2010 05:15:56PM 1 point [-]

I suspect that people don't like the tone/conclusions/analysis, and much of the debate was instigated by the article's author. If someone wrote a post that successfully managed to explain what people actually mean when they say the answer goes one way or the other, then I'd expect that one to be rated higher.

Frankly, I think the Wikipedia article on the sleeping beauty problem tells you everything you'd get out of this article and more, without the implication that 1/2 is the right answer and people who answer 1/3 are doing something basically stupid.

Comment author: 13 May 2010 05:18:58PM *  0 points [-]

Alright, sounds good to me. Rating seems reasonable then.

ETA: And let me add that such restraint in voting gives me renewed confidence LW's karma system.

Comment author: 10 May 2010 09:45:24PM 1 point [-]

Your update doesn't solve the problem. It's a semantic issue about what credence we are being asked. If we are being asked about the probability of our coin flip associated with this iteration of the experiment, then the answer is 1/2. If we are being asked about the probability of the coin flip associated with this particular awakening, then it must be 1/3.

You say that you must use cell counts of 500,250,250, but the fact is that if you repeat the experiment 1000 times, sleeping beauty will be awoken 1500 times, not 1000. So what are you doing with the other 500 awakenings? I would say you are implicitly ignoring them, as you do when you say "we only accept her last decision" in the bet scenario. The reformulations of this using different people, rather than the same person being awoken multiple times don't seem to cause as much trouble.

The semantic issue here is reminiscent of arguments I've seen over the Monty Hall problem when it is misstated so that Monty's algorithm is not clear. People who assume what he usually does on the show come up with 2/3, and people who don't make any assumptions come up with 1/2 (as do most of the people who simply don't understand restricted choice).

Comment author: 10 May 2010 09:54:24PM *  1 point [-]

Re: "If we are being asked about the probability of our coin flip associated with this iteration of the experiment, then the answer is 1/2. If we are being asked about the probability of the coin flip associated with this particular awakening, then it must be 1/3."

What is actually asked at each awakening is:

"What is your credence now for the proposition that our coin landed heads?"

I figure that makes the answer 1/3 - and not 1/2.

If the question had been: "What is your credence that this is the last time you awaken and our coin landed heads-up?"

...then the answer would have been 1/2.

...but that wasn't the question that was asked.

Comment author: 07 May 2010 04:10:03PM 1 point [-]

The OP is correct. There are actually all the same issues here as with the Self Indication Assumption; it is wrong for the same reasons as the 1/3 probability. I predict that a great majority of those who accept SIA will also favor the probability of 1/3.

Comment author: 26 May 2010 07:58:35AM 0 points [-]
Comment author: 15 May 2010 12:06:04AM 0 points [-]

this is a probability tree corresponding to an arbitrary wake up day

Huh? If tails, then Beauty is (always) woken on Monday. Why do you have probability=1/2 there?

(likewise for Tuesday)

Comment author: 15 May 2010 12:28:55AM 0 points [-]

The probability represents how she should see things when she wakes up.

She knows she's awake. She knows heads had probability 0.5. She knows that, if it landed heads, it's Monday with probability 1. She knows that, if it landed tails, it's either Monday or Tuesday. Since there is no way for her to distinguish between the two, she views them as equally likely. Thus, if tails, it's Monday with probability 0.5 and Tuesday with probability 0.5.

Comment author: 15 May 2010 05:47:59AM *  0 points [-]

Okay, I now understand what you mean by that tree.

Comment author: 15 May 2010 12:02:36AM 0 points [-]

If we want to replicate the situation 1000 times, we shouldn't end up with 1500 observations. The correct way to replicate the awakening decision is to use the probability tree I included above. You'd end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.

Beauty ends up with 1500 observations on average (maybe as few as 1000 or as many as 2000). Imagine a sequence of Beauty-observations in (H|TT)^1000 , where by r^1000 I mean 1000 repetitions of r. This string is from 1000-2000 letters long.

If you consider the scenario from a non-amnesiac perspective, then you can consider the TT - the two forgetful-Beauty observations in the tails case, as a single event, which is indeed equally likely to the alternative, H. In fact, the shortest possible coding to describe one of the beauty-observation strings is just a 1000-bit string where the nth bit indicates the result of the nth coin flip.

But what are you thinking when you say there are two "cells" each with p=1/4 (count 250 out of 1000)? What, exactly, would happen to 250 times on average? Certainly we expect Beauty waking on Mon. with it being tails 500 times (and also 500 times on Tue.).

Comment author: 13 May 2010 06:43:37PM 0 points [-]

PhilGoetz writes:

I had expected that people would read posts and comments by other people, and take special note of comments by people who had a prior history of being right, and thereby improve their own accuracy.

I would like to do this. However, it's time consuming to sort through people's posts and see what they think. (You have to read carefully, because they may be critiquing a particular argument rather than the value 1/2 or 1/3 presented in the parent.) Would people mind stating their position on the Sleeping Beauty problem with a single sentence explaining the core detail of the argument that persuades them?

Comment author: 11 May 2010 07:00:26AM *  0 points [-]

I don't follow your latest argument against thirders. You claim that the denominator

#(heads & monday) + #(tails & monday) + #(tails & tuesday)

counts events that are not mutually exclusive. I don't see this. They look mutually exclusive to me-- heads is exclusive of tails, and monday is exclusive of tuesday, Could you elaborate this argument? Where does exclusivity fail? Are you saying tails&monday is not distinct from tails&tuesday, or all three overlap, or something else?

You also assert that the denominator is not determined by n. (I assume by n you mean replications of the SB experiment, where each replication has a randomly varying number of awakenings. That's true in a way-- particular values that you will see in particular replications will vary, because the denominator is a random variable with a definite distribution (Bernoulli, in fact). But that's not a problem when computing expected values for random processes in general; they often have perfectly definite and easily computed expected values. Are you arguing that this makes that ratio undefined, or problematic in some way? I can tell easily what this ratio converges to, but you won't like it.

Comment author: 07 May 2010 07:31:43PM *  0 points [-]

The next program works well:

R=Random(0,1) If R=0 SAY "P(R=0)=1/2" Elseif SAY "P(R=0)=1/2": SAY "P(R=0)=1/2" Endif

The next doesn't:

R=Random(0,1) If R=0 SAY "P(R=0)=1/3" Elseif SAY "P(R=0)=1/3": SAY "P(R=0)=1/3" Endif

Run it many times and you will clearly see, that the first program will be right, since it will be about the same number of cases when R will be 0 and the other cases when R will be 1.

Just what the first program keep saying.

Comment author: 07 May 2010 03:10:26PM *  0 points [-]

I'm not convinced that 1/2 is the right answer. I actually started out thinking it was obviously 1/2, and then switched to 1/3 after thinking about it for a while (I had thought of Bostrom's variant (without the disclosure bit) before I got to that part).

Let's say we're doing the Extreme version, no disclosure. You're Sleeping Beauty, you just woke up, that's all the new information you have. You know that there are 1,000,001 different ways this could have happened. It seems clear that you should assign tails a probability of 1,000,000/1,000,001.

Now I'll go think about this some more and probably change my mind a few more times.

Comment author: 07 May 2010 03:34:54PM 6 points [-]

We can tweak the experiment a bit to clarify this. Suppose the coin is flipped before she goes to sleep, but the result is hidden. If she's interviewed immediately, she has no reason to answer other than 1/2 - at this point it's just "flip a fair coin and estimate P(heads)". What information does she get the next time she's asked that would cause her to update her estimate? She's woken up, yes, but she already knew that would happen before going under and still answered 1/2. With no new information she should still guess 1/2 when woken up.

Comment author: 10 May 2010 07:14:34AM *  4 points [-]

She knows in advance how many times she will be woken up (on each coin result). It says so in the problem description. So, she never answers 1/2 in the first place. She doesn't update on awakening. She updates when she is told the experimental procedure in the first place.

Comment author: 10 May 2010 01:24:21PM 1 point [-]

So, in the extreme sleeping beauty problem, when she is told the experimental procedure, she decides that it will be tails with near certainty?

Comment author: 07 May 2010 08:13:07PM 4 points [-]

Let's look at the ultimate extreme version. Assume she's woken up once (or arbitrarily many non-zero times) for tails, and not at all for heads. Now the fact that she's been woken up implies tails with certainty. So if the answer remains 1/2 in the extreme versions, then there must be a discontinuous jump, rather than convergence, when the ratio of the number of awakenings for heads vs. tails tends towards zero.

Comment author: 07 May 2010 08:18:45PM 0 points [-]

Intuitively, this is very convincing. But it definitely doesn't prove anything by itself...

Comment author: 07 May 2010 09:40:59PM *  1 point [-]

This was similar to my intuition also. She will wake up in the end no matter what regardless of how many times she has been woken up before, so how does her wakefulness add any new information? If there was a scenario where she would never wake up, then her being awake would actually mean something, but that isn't the question.

I can't see how this a problem of conditional probability. Isn't it just "what is P(heads)"? Am I missing something?

Comment author: 07 May 2010 08:17:42PM 1 point [-]

Note jimmy's comparison to Blue Eyes. It's not necessarily the case that she's getting no new information here.

Comment author: 07 May 2010 04:14:46PM 1 point [-]

As I said in the other comment, this argument is just like arguing that if I exist, then there are likely more people, since there would be more ways that it could happen that I existed, i.e. I could be any of them of the people who exist.

But in fact "I" am just one of the people who exist, no matter how many or few there are, so the inevitable fact of my existence cannot increase the probability of many people existing. Likewise, when Sleeping Beauty wakes up, that is just the one case or one of the million cases; either way it would still have happened. It would not have happened with greater likelihood in the million cases.