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Beauty quips, "I'd shut up and multiply!"
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:
P+(HEADS) = 1,000,001/1,000,002
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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Comments (335)
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.
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.
...yeah, I think you're right.
In the few minutes before I read your comment, I was thinking about reformulating this as an Omega-style problem. (I know, I know... I do try not to be too gratuitous with my use of Omega, but what can I say — omnipotence and omniscience are surprisingly useful for clarifying and simplifying reasoning/decision problems.) So Omega tells you she's going to flip a fair coin, and if it lands on tails, she's going to make a million copies of you and put all of them in identical rooms, and if it lands on heads, she'll just put the one of you in such a room. She flips the coin, you blank out for a moment, and as expected, you're in an unfamiliar room. In this case, it doesn't appear that adding or subtracting copies of you should have anything to do with what you believe about the coin flip. You saw her flip the coin yourself, and you knew that you'd be seeing the same thing no matter what side came up. She could come back a few minutes later and say "Hey, if and only if it was tails, I just made another million copies of you and put them in rooms identical to this one, kbye" which clearly shouldn't change your belief about the coin, but seems to be a situation identical to if she had just said "two million" in the first place.
Okay, I think I'm more confidently on the 1/2 side now.
How about the following scenario? Say instead of Omega, it's just a company doing a weird promotional scheme. They announce that they'll secretly flip a coin in their headquarters, and if it's tails, they'll hand out prizes to a million random people from the phone directory tomorrow, whereas if it's heads, they'll award the same prize to only one lucky winner. The next day, you receive a phone call from them. Would you apply analogous reasoning in this case (and how, or why not)?
The probability for the head is still the same.
On the additional information, that you got the call, it becomes more likely that it was the head this time.
In the situation you state, then yes, of course I place high probability on the coin having come up tails. However, in order for your situation to be truly analogous to the Sleeping Beauty problem, you would have to be guaranteed to get the phone call either way, which destroys any information you gain in your version.
I think that's very different... in the original scenario, heads and tails both result in you experiencing the same thing. In this case, if it comes up heads, it is a million times more likely that you will receive the prize, so getting a phone call from them is very significant Bayesian evidence.
Yes, you're right (as are the other replies making similar points). I tried hard once more to come up with an accurate analogy of the above problem that would be realizable in the real world, but it seems like it's impossible to come up with anything that doesn't involve implanting false memories.
After giving this some more thought, it seems to me that the problem with the copying scenario is that once we eliminate the assumption that each agent has a unique continuous existence, all human intuitions completely break down, and we can compute only mathematically precise problems formulated within strictly defined probability spaces. Trouble is, since we've breaking one of the fundamental human common sense assumptions, the results may or may not make any intuitive sense, and as soon as we step outside formal, rigorous math, we can only latch onto subjectively preferable intuitions, which may differ between people.
OK, I think I have a definite reductio ad absurdum of your point. Suppose you wake up in a room, and the last thing you remember is Omega telling you: "I'm going to toss a coin now. Whatever comes up, I'll put you in the room. However, if it's tails, I'll also put a million other people each in an identical room and manipulate their neural tissue so as to implant them a false memory of having been told all this before the toss. So, when you find yourself in the room, you won't know if we've actually had this conversation, or you've been implanted the memory of it after the toss."
After you find yourself in the room under this scenario, you have the memory of these exact words spoken to you by Omega a few seconds ago. Then he shows up and asks you about the expected value of the coin toss. I'm curious if your 1/2 intuition still holds in this situation? (I'm definitely unable to summon any such intuition at all -- your brain states representing this memory are obviously more likely to have originated from their mass production in case of tails, just like finding a rare widget on the floor would be evidence for tails if Omega pledged to mass-manufacture them if tails come up.)
But if you wouldn't say 1/2, then you've just reached an awful paradox. Instead of just implanting the memories, Omega can also choose to change these other million people in some other small way to make them slightly more similar to you. Or a bit more, or even more -- and in the limit, he'd just use these people as the raw material for manufacturing the copies of you, getting us back to your copying scenario. At which step does the 1/2 intuition emerge?
(Of course, as I wrote in my other comment, all of this is just philosophizing that goes past the domain of validity of human intuitions, and these questions make sense only if tackled using rigorous math with more precisely defined assumptions and questions. But I do find it an interesting exploration of where our intuitions (mis)lead us.)
I'd still say 1/2 is the right answer, yes.
But I'm trying to avoid using intuition here; when I do, it tends to find the arguments on both sides equally persuasive (obvious, even). If there is a right answer at all, then this is truly a case where we have no choice but to shut up and do the math.
Hm.. let's try pushing it a bit further.
Suppose you're a member of a large exploratory team on an alien planet colonized by humans. As a part of the standard equipment, each team member has an intelligent reconnaissance drone that can be released to roam around and explore. You get separated from the rest of your team and find yourself alone in the wilderness. You send out your drone to explore the area, and after a few hours it comes back. When you examine its records, you find the following.
Apparently, a local super-smart creature with a weird sense of humor -- let's call it Omega -- has captured several drones and released (some of?) them back after playing with them a bit. Examining your drone's records, you find that Omega has done something similar to the above described false memory game with them. You play the drone's audio record, and you hear Omega saying: "I'll toss a coin now. Afterwards, I'll release your drone back in any case. If heads come up, I'll destroy the other ten drones I have captured. If it's tails, I'll release them all back to their respective owners, but I'll also insert this message into their audio records." Assume that you've already heard a lot about Omega, since he's already done many such strange experiments on the local folks -- and from what's known about his behavior, it's overwhelmingly likely that the message can be taken at face value.
What would you say about the expected coin toss result now? Would you take the fact that you got your drone back as evidence in favor of tails, or does your 1/2 intuition still hold? If not, what's the difference relative to the false memory case above? (Unless I'm missing something, the combined memories of yourself and the drone should be exactly equivalent to the false memory scenario.)
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.
Intuitively, this is very convincing. But it definitely doesn't prove anything by itself...
Note jimmy's comparison to Blue Eyes. It's not necessarily the case that she's getting no new information here.
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?
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.
So, in the extreme sleeping beauty problem, when she is told the experimental procedure, she decides that it will be tails with near certainty?
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.
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.
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).
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?
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.
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?
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?
Yes, one pay out is the relevant case. The reason is because we are asking about her credence at an awakening.
How does the former follow from the latter, exactly? I seem to need that spelled out.
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?
What do you think the word "credence" means? I am thinking that perhaps that is the cause of your problems.
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.
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.
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.
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.
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:
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:
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.
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
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.
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.
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.)
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.
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?
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.
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.)
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.
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.
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.)
By "can't update" I refer to the problem with marking Thursday "impossible", since you'll encounter Thursday later.
It's not a problem with the model of ontology and preference, it's merely specifics of what kinds of observation events are expected.
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.
I humbly apologize for my inability to read (may the Values of Less Wrong be merciful).
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.
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]
NOW. One bet.
Again, consider 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.
If her credence for heads on an awakening, on every awakening (she can't distinguish between awakenings), really was 1/3, she would agree to accept the bet. But we all know accepting the bet would be irrational. Thus, her credence for heads on an awakening is not 1/3.
So: you are debating what:
"What is your credence now for the proposition that our coin landed heads?"
...actually means. Personally, I think your position on that is indefensible.
This would make it clear exactly where the problem lies - if not for the fact that you also appear to be in a complete muddle about how many times Beauty awakens and is interviewed.
We both know what question is being asked. We both know how many times she awakens and is interviewed. I know what subjective probability is (I assume you do too). I showed you my math. I also explained why your ratio of expected frequencies does not correspond to the subjective probability that you think it does.
Does it not concern you even a little that the Wikipedia article you linked to quite clearly says you are wrong and explains why?
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.
http://lesswrong.com/lw/np/disputing_definitions/
http://lesswrong.com/lw/208/the_iless_eye/
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.
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.
That's the wrong way to look at it. A certain bet may be the "correct" action to perform, or even a certain ritual of cognition may pay its rent, but it won't be about the concept of probability. Circumstances may make it preferable to do or say anything, but that won't influence the meaning of fixed concepts. You can't argue that 2+2 is in fact 5 on the grounds that saying that saves puppies. You may say that 2+2 is 5, or think that "probability of Tuesday" is 1/3 or 1/4 in order to win, but that won't make it so, it will merely make you win.
Subjective probability is not a well-defined concept in the general case. Fractions are well-defined, but only after you've decided where you are getting the numerator and denominator from.
That fractions are well-defined doesn't make them probabilities.
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.
Um... why? There are the same number of heads&Monday as tails&Monday; why would heads&Monday be more likely?
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.
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.
Interesting. You want to replicate an awakening 1000 times, and you end up with 1500 awakenings. I'd be concerned about that if I were you.
In 1000 replications of the experiment, there will be an average of 1500 awakenings - 1000 on Monday, and 500 on Tuesday.
"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."
What is it about this that you are not getting?
Complete replications of the entire experiment is not the right approach, because the outcome of interest occurs at a single awakening. We need 1000 replications of the process that lead to an awakening.
What you said further up this branch of the thread was:
"if you want to think about doing 1000 replications of the experiment, it should go like this".
Now you seem to be trying to shift the context retrospectively - now that you have found out that all the answers you gave to this were wrong.
You know that's not true. I didn't just discover the '500 500 500' answer -- I quoted it from wikipedia and showed why it was wrong.
I should have made it clear what I meant by experiment, but you know what I meant now, so why take it as an opportunity to insult?
I don't think it is accurate to describe my post as "insulting".
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.
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.
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.
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.
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".
'Credence' is not probability.
It means: "subjective probabilty":
"In probability theory, credence means a subjective estimate of probability, as in Bayesian probability."
An estimate of a thing is not the same thing as that thing. And Bayesian probability is probability, not an estimate of probability.
Or - to put it another way - for a Bayesian their estimated probability is the same as their subjective probability.
The concept of "estimated probability" doesn't make sense (in the way you use it).
? You can certainly estimate a probability - just like Wikipedia says.
Say you have a coin. You might estimate the probabiltiy of it coming down heads after a good flip on a flat horizontal surface as being 0.5. If you had more knowledge about the coin, you might then revise your estimate to be 0.497. You can consider your subjective probability to be an estimate of the probability that an expert might use.
You don't seem to understand the concept of Bayesian probability. Subjective probability is not estimation of "real probability", there is no "real probability". When you revise subjective probability, it's not because you found out how to approximate "real probability" better, it's because you are following the logic of subjective probability.
Really? Someone who's been posting around these parts for years, and your best hypothesis is "doesn't understand Bayesian probability"? How would you rank it compared to "Someone hijacked your Lw account" or "I'm not understanding you" or "You said something that would have made sense except for a fairly improbable typo"?
This seems a reasonable hypothesis specifically because it's Tim Tyler. It would be much less probable for most other old-timers (another salient exception that comes to mind is Phil Goetz, though I don't remember what he understands about probability in particular).
You seem to have to misattribute the phrase "real probability" to me in order to make this claim. What I actually said was "the probability that an expert might use".
I recommend you exercise caution with those quote marks when attributing silly positions to me: some people might be misled into thinking you were actually quoting me - rather than attacking some nonsense of your own creation.
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.
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
Why?
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.
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.
I can understand that, but the fact that a wager has been offered distorts the probabilities under a lot of circumstances.
How do you mean?
I just flipped a coin. Are you willing to offer me a wager on the outcome I have already seen? Yet tradition would suggest you have a degree of belief in the most probable possibilities.
The offering of the wager itself can act as useful information. Some people wager to win.
I see what you mean. Yes, actual, literal, wagers are messier than beliefs. Another example is a bet that the world is going to end: which you should obviously always bet against at any odds even if you believe the last days are upon us. The equivalence between degree of belief and fair betting odds is a more abstract equivalence with an idealized bookie who offers bets on everything, doesn't take a cut for himself and pays out even if you're dead.
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?
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.)
On the other hand...
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.
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.
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?
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.
By the way, you may have noticed that the wiki has an article on the Sleeping Beauty problem. Also, it's been referenced before Sleeping beauty gets counterfacutally mugged in a top-level post, and it was mentioned in the context of a general solution in The I-Less Eye. And the comments on How many LHC failures is too many are relevant to the problem too.
Did you even search to see if someone had done a post on this topic before?
Yes, and the posts you referenced don't cover the topic in the same way.
update: I did link to the wiki article in the original post (and quoted from it extensively). I'm surprised you didn't notice that
SB would start out with P(tails) = 1,000,001/1,000,002 and on being informed that it is monday would update:
The initial strong belief in tails is cancelled by the strong evidence of being told that it is monday, which only happens in one of many wakings if the coin landed tails.
The initial strong belief in tails is canceled by the strong evidence of being told what day it is, and then updated further to strong belief in heads by the strong evidence of it being Monday.
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?
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.
If the coin is tails, you would ask two random passerbies.
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?
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.
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?
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.
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.
I'm not sure I understand your "really extreme" formulation fully. Is the amnesia supposed to make the wins in chocolate bars non-cumulative?
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.
Surely, 1/3 is the correct answer - and is backed up by a perfectly reasonable model..
"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.
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.
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:
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.
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.
"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?
IMO, there's no problem with the form of this question. It is not ambiguous. The only way to make it so is with some pretty torturous misinterpretations.
"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.
I disagree; but I've already given my reasons.
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.
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.
Yes, this is very alarming, considering this is a forum for aspiring rationalists.
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.
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.
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.
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".
Actually I think it would make more sense to reply to my own comment in response to this. link
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.
I think that we've established the following:
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.
Which of your down-voted statements were correct?
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.
For example, on an awakening p(monday)=p(monday|tails)p(tails) + p(monday|heads) p(heads)=1/4+1/2=3/4
p(monday and heads)=p(heads)*p(monday|heads)=1/2
etc.
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.
Or they all should be 1/2.
Impossible - if they are to add up to 1.
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.
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.
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.
So you are not willing to tell me where I made a mistake?
P(heads)=1/2, p(monday | heads)=1. Which one of these is wrong?
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.
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.
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..."
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.
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)
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.
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.
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.
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.
If she thinks she will be asked what the coin shows more times if it is tails.
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:
In other words, neq1's probability-tree picture turned out to most clearly match my own reasoning on the problem. Does this make sense?
Yes, this is correct.
This was also my understanding of the problem. Are we missing something?
On awakening, I would give:
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).
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).
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!
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.
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.
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.
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.
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?
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.
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:
(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.)
"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.
Yes: her estimate of the chances of her being awakened is indeed 1.
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).
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).
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.
And I think this is roughly equivalent to the suggestion that the payoff matters.
The coverage on http://en.wikipedia.org/wiki/Sleeping_Beauty_problem seems much less confused than this post.
I disagree. That's why I quoted from that site and explained where I think the errors are.
Alas, that site is correct - and your whole post is totally wrong.
Except I showed why it's wrong. I understand both the 1/3 and 1/2 solutions. I showed where 1/3 reasoning fails.
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).
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.
I updated the post one more time. I think this time I more effectively explain where the thirder logic fails. Correct me if I'm wrong...