Intuition Pump
Suppose 50% of people in a population have an asymptomatic form of cancer. None of them know if they have it. One of them is randomly selected and a diagnostic test is carried out (the result is not disclosed to them). If they don't have cancer, they are woken up once. If they do have it, they are woken up 9 times (with amnesia-inducing drug administered each time, blah blah blah). Each time they are woken up, they are asked their credence (subjective probability) for cancer.
Imagine we do this repeatedly, randomly selecting people from a population that has 50% cancer prevalence.
World A: Everyone uses thirder logic
Someone without cancer will say: "I'm 90% sure I have cancer"
Someone with cancer will say: "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer." "I'm 90% sure I have cancer."
Notice, everyone says they are 90% sure they have cancer, even though only 50% of them ...
POLL
There will be six replies to this post. Two will be answers to the question "What is the best answer to the sleeping beauty problem?'. Two more will be answers to the question "Can the Sleeping Beauty problem, as it is written in this post, be reasonably interpreted as returning either 1/3 or 1/2. Two more comments will be karma dumps.
Somewhat off-topic, but it should broaden our view of the past to know that people were thinking like this long ago: An essay from 1872 by a French political radical arguing that in an infinite universe, each of our lives is infinitely duplicated on other worlds, and so we are all effectively immortal.
Indeed, using the 1/3 answer and working back to try to find P(W) yields P(W) = 3/2, which is a strong indication that it is not the probability that matters,
This is expected value: the expected number of wakenings per coin flip is 1.5.
Expected value, the probability of heads, the probability of heads given an awakening are all well-defined things with well-defined numbers for this problem. While I understand needing to develop novel methods for 'subjective observer mathematics' for these types of problems, I think it would be useful to depart from the...
There’s a lot of terminology in this article that I simply don’t understand: what are the “continuity problems” mentioned in the first paragraph? And this sentence means almost nothing to me: “The source of confusion appears to be based on the distinction between the probability of an observer and the expectation number of observers, with the former not being a linear function of problem definitions.”
It’s possible that some of the other commenters who are having trouble with this article are in the same position as me, but are too polite to say so. Without...
playing this game with a single die roll and all possible values of k recovers the Sleeping Beauty problem
I have no idea what this sentence is supposed to mean.
I've done the same work of formalization for PSB that I did on AlephNeil's revival question and the joint distribution table does have the same structure, in particular I agree that 1/21 is the right answer in PSB. So I agree that this could be a promising start - but it's unclear (and was also unclear with AlephNeil's question) how we get from there to SB.
Given that structure for the joint proba...
Maybe this will help:
Where does the 1/3 solution come from?
It comes from taking a ratio of expected counts. First, a motivating example.
Suppose people can fall into one of three categories. For example, we might create a categorical age variable, catage, where catage is 1 if age50.
Suppose we randomly select N people from the population. Let n1 be the number of people with catage=1, with n2 and n3 defined similarly. Given the sample size N, the random variables n1, n2 and n3 follow a multinomial distribution, with parameters (probabilities) p1, p2 and p...
What is supposed to happen, in "Probabilistic Sleeping Beauty", if the coin comes up heads and the die doesn't come up k?
Wait, I didn't catch this the first time:
"using the 1/3 answer and working back to try to find P(W) yields P(W) = 3/2, which is a strong indication that it is not the probability that matters"
No. It's proof that your solution is wrong.
And I know exactly why your solution is wrong. You came up with P(Monday|W) using a ratio of expected counts, but you relied on an assumption that trials are independent. Here, the coin flips are indpendent but the counts are not. Even though you are using three counts, there is just one degree of freedom. Vlad...
Nitpick:
Hence in 900 observations, there would be 300 heads and 300 tails, with 600 observations following a tail and 300 following a head.
Change the second 300 to 600.
EDIT: Oops. Brain malfunction. Never mind.
"Under these numbers, the 1000 observations made have required 500 heads and 250 tails, as each tail produces both an observation on Monday and Tuesday. "
I must have been unclear in explaining my probability tree. The tree represents how Beauty should view things on an awakening. I thought it would be helpful. Apparently it just created more confusion (although some people got it).
"P(Monday|W) = 2/3"
Why? I believe it is 0.75. How did you come up with 2/3?
Two ways to iterate the experiment:
- Replicate the entire experiment 1000 times. That is, there will be 1000 independent tosses of the coin. This will lead between 1000 and 2000 awakenings, with expected value of 1500 awakenings.
and
- Replicate her awakening-state 1000 times. Because her epistemic state is always the same on an awakening, from her perspective, it could be Monday or Tuesday, it could be heads or tails.
The distinction between 1 and 2 is that, in 2, we are trying to repeatedly sample from the joint probability distributions that she should have on an awakening. In 1, we are replicating the entire experiment, with the double counting on tails.
This seems a distinction without a difference. The longer the iterated SB process continues, the less important is the distinction between counting tosses versus counting awakenings. This distinction is only about a stopping criterion, not about the convergent behavior of observations or coin tosses to expected values as it's ongoing. Considered as an ongoing process of indefinite duration, the expected number of tosses and of observations of each type are well-defined, easily computed, and well-behaved with respect to each other. Over the long run, #awakenings accumulates 1.5 times more frequently than #tosses. Beauty is never more than two awakenings away from starting a new coin toss, so whether you choose to stop as soon as an awakening has completed or until you finish a coin-toss cycle, the relative perturbation in the statistics collected so far goes to zero. Briefly, there is no "natural" unit of replication independent of observer interest.
She knows that it was a fair coin. She knows that if she's awake it's definitely Monday if heads, and could be either Monday or Tuesday if tails. She knows that 50% of coin tosses would end up heads, so we assign 0.5 to Monday&heads.
This would be an error. You are assigning a 50% probability to an observation (that it is Heads&Monday) without taking into account the bias that's built in to the process for Beauty to make observations. Alternatively, if you are uncertain whether Monday is true or not--you know it might be Tuesday--then you should be uncertain that P(Heads)=P(Heads&Monday).
You the outside observer know the chance of observing that the coin lands Heads is 50%. You presumably know this because you have corroborated it through an unbiased observation process: look at the coin exactly once per toss. Once Beauty is put to sleep and awoken, she is no longer an outside observer, she is a particpant in a biased observation process, so she should update her expectation about what her observation process will show. Different observation process, different observations, different likelhoods of what she can expect to see.
Of course, as a card-carrying thirder, I'm assuming that the question about credence is about what Beauty is likely to see upon awakening. That's what the carefully constructed wording of the question suggests to me.
She knows that 50% of coin tosses would end up tails,
except that as we agreed, she's not observing coin tosses, she's observing biased samples of coin tosses. The connection between what she observes and the objective behavior of the coin is just what's at issue here, so you can't beg the question.
In 1, people are using these ratios of expected counts to get the 1/3 answer. 1/3 is the correct answer to the question about the long-run frequencies of awakenings preceded by heads to awakenings preceded by tails. But I do not think it is the answer to the question about her credence of heads on an awakening.
Agreed, but for this: it all depends on what you want credence to mean, and what it's good for; see discussion below.
In 2, the joint probabilities are determined ahead of time based on what we know about the experiment.
Let n2 and n3 are counts, in repeated trials, of tails&Monday and tails&Tuesday, respectively. You will of course see that n2=n3. They are the same random variable. tails&Monday and tails&Tuesday are the same.
Let me uphold a distinction that's continually skated over, but which is crucial point of disagreement here. I think you're confusing your evidence for the thing evidenced. And you are selectively filtering your evidence, which amounts to throwing away information. Tails&Monday and Tails&Tuesday are not the same; they are distinct observations of the same state of the coin, thus they are perfectly correlated in that regard. Aside from the coin, they observe distinct days of the week, and thus different states of affairs. By a state of affairs I mean the conjunction of all the observable properties of interest at the moment of observation.
It's like what Jack said about types and tokens. It's like Vladimir_Nesov said:
The distinction between types and tokens is only relevant when you want to interepret your tokens as being about something else, their types, rather than about themselves. But types are carved out of observers' interests in their significance, which are non-objective, observer-dependent if anything is. Their variety and fineness of distinction is potentially infinite. As I mentioned above, a state of affairs is a conjunction of observable properties of interest. This Boolean lattice has exactly one top: Everything, and unknown atoms if any at bottom. Where you choose to carve out a distinction between type and token is a matter of observer interest.
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.
I'll certainly agree it isn't desirable, but oughtn't isn't the same as isn't, and in the Sleeping Beauty problem we have no choice. Monday and Tuesday just are different elements in a sample space, by construction.
if she starts out believing that heads has probability 1/2, but learns something about the coin toss, her probability might go up a little if heads and down a little if tails.
What you seem to be talking about is using evidence that observations provide to corroborate or update Beauty's belief that the coin is in fact fair. Is that a reasonable take? But due to the epistemic reset between awakenings, there is never any usable input to this updating procedure. I've already stipulated this is impossible. This is precisely what the epistemic reset assumption is for. I thought we were getting off this merry-go-round.
Suppose, for example, she is informed of a variable X. If P(heads|X)=P(tails|X), then why is she updating at all? Meaning, why is P(heads)=/=P(heads|X)? This would be unusual. It seems to me that the only reason she changes is because she knows she'd be essentially 'betting' twice of tails, but that really is distinct from credence for tails.
Ok, I guess it depends on what you want the word "credence" to mean, and what you're going to use it for. If you're only interested in some updating process that digests incoming information-theoretic quanta, like you would get if you were trying to corroborate that the coin was inded a fair one to within a certain standard error, you don't have it here. That's not Sleeping Beauty, that's her faithful but silent, non-memory-impaired lab partner with the log book. If Beauty herself is to have any meaningful notion of credence in Heads, it's pointless for it be about whether the coin is indeed fair. That's a separate question, which in this context is a boring thing to ask her about, because it's trivially obvious: she's already accepted the information going in that it is fair and she will never get new information from anywhere regarding that belief. And, while she's undergoing the process of being awoken inside the experimental setup, a value of credence that's not connected to her observations is not useful for any purpose that I can see, other than perhaps to maintain her membership in good standing in the Guild of Rational Bayesian Epistomologists. It doesn't connect to her experience, it doesn't predict frequencies of anything she has any access to, it's gone completely metaphysical. Ok, what else is there to talk about? On my view, the only thing left is Sleeping Beauty's phenomenology when awakened. On Bishop Berkeley's view, that's all you ever have.
Beauty gets usable, useful information (I guess it depends on what you want "information" to mean, too) once, on Sunday evening, and she never forgets it thereafter. This information is separate from, in addition to the information that the coin itself is fair. This other information allows her to make a more accurate prediction about the likelihood that, each time she is awoken, the coin is showing heads. Or whether it's Monday or Tuesday. The information she receives is the details of the sampling process, which has been specifically constructed to give results that are biased with respect to the coin toss itself, and the day of the week. Directly after being informed of the structure of the sampling process, she knows it is biased and therefore ought to update her prediction about what relative frequencies per observation will be of each observable aspect of the possible state of affairs she's awoken into-- Heads vs. Tails, Monday vs. Tuesday.
I think I might understand the interpretation that a halfer puts on the question. I'm just doubtful of its interest or relevance. Do you see any validity (I mean logical coherence, as opposed to wrong-headedness) to this interpretation? Is this just a turf war over who gets to define a coveted word for their purposes?
Response to Beauty quips, "I'd shut up and multiply!"
Related to The Presumptuous Philosopher's Presumptuous Friend, The Absent-Minded Driver, Sleeping Beauty gets counterfactually mugged
This is somewhat introductory. Observers play a vital role in the classic anthropic thought experiments, most notably the Sleeping Beauty and Presumptuous Philosopher gedankens. Specifically, it is remarkably common to condition simply on the existence of an observer, in spite of the continuity problems this raises. The source of confusion appears to be based on the distinction between the probability of an observer and the expectation number of observers, with the former not being a linear function of problem definitions.
There is a related difference between the expected gain of a problem and the expected gain per decision, which has been exploited in more complex counterfactual mugging scenarios. As in the case of the 1/2 or 1/3 confusion, the issue is the number of decisions that are expected to be made, and recasting problems so that there is at most one decision provides a clear intuition pump.
Sleeping Beauty
In the classic sleeping beauty problem, experimenters flip a fair coin on Sunday, sedate you and induce amnesia, and wake you either on just the following Monday or both the following Monday and Tuesday. Each time you are woken, you are asked for your credence that the coin came up heads.
The standard answers to this question are that the answer should be 1/2 or 1/3. For convenience let us say that the event W is being woken, H is that the coin flip came up heads and T is that the coin flip came up tails. The basic logic for the 1/2 argument is that:
P(H)=P(T)=1/2, P(W|H) = P(W|T) = P(W) = 1 so by Bayes rule P(H|W) = 1/2
The obvious issue to be taken with this approach is one of continuity. The assessment is independent of the number of times you are woken in each branch, and this implies that all non zero observer branches have their posterior probability equal to their prior probability. Clearly the subjective probability of a zero observer branch is zero, so this implies discontinuity in the decision theory. Whilst not in and of itself fatal, it is surprising. There is apparent secondary confusion over the number of observations in the sleeping beauty problem, for example:
Under these numbers, the 1000 observations made have required 500 heads and 250 tails, as each tail produces both an observation on Monday and Tuesday. This is not the behaviour of a fair coin. Further consideration of the problem shows that the naive conditioning on W is the point where it would be expected that the number of observations comes in. Hence in 900 observations, there would be 300 heads and 300 tails, with 600 observations following a tail and 300 following a head. To make this rigorous, let Monday and Tuesday be the event of being woken on Monday and Tuesday respectively. Then:
P(H|Monday) = 1/2, P(Monday|W) = 2/3 (P(Monday|W) = 2*P(Tuesday|W) as Monday occurs regardless of coin flip)
P(H|W) = P(H ∩ Monday|W) + P(H ∩ Tuesday|W) (Total Probability)
= P(H|Monday ∩ W).P(Monday|W) + 0 (As P(Tuesday|H) = 0)
= P(H|Monday).P(Monday|W) = 1/3 (As Monday ∩ W = Monday)
Which would appear to support the view of updating on existence. The question of why this holds in the analysis is immediate to answer: The only day on which probability of heads occuring is non zero is Monday, and given an awakening it is not guaranteed that it is Monday. This should not be confused with the correct observation that there is always one awakening on Monday. This has caused problems because "Awakening" is not an event which occurs only once in each branch. Indeed, using the 1/3 answer and working back to try to find P(W) yields P(W) = 3/2, which is a strong indication that it is not the probability that matters, but the E(# of instances of W). As intuition pumps, we can consider some related problems.
Sleeping Twins
This experiment features Omega. It announces that it will place you and an identical copy of you in identical rooms, sedated. It will then flip a fair coin. If the coin comes up heads, it will wake one of you randomly. If it comes up tails, it will wake both of you. It will then ask what your credence for the coin coming up heads is.
You wake up in a nondescript room. What is your credence?
It is clear from the structure of this problem that it is almost identical to the sleeping beauty problem. It is also clear that your subjective probability of being woken is 1/2 if the coin comes up heads and 1 if it comes up tails, so conditioning on the fact that you have been woken the coin came up heads with probability 1/3. Why is this so different to the Sleeping Beauty problem? The fundamental difference is that in the Sleeping Twins problem, you are woken at most once, and possibly not, whereas in the Sleeping Beauty problem you are woken once or many times. On the other hand, the number of observer moments on each branch of the experiment is equal to that of the Sleeping Beauty problem, so it is odd that the manner in which these observations are achieved should matter. Clearly information flow is not possible, as provided for by amnesia in the original problem. Let us drive this further
Probabilistic Sleeping Beauty
We return to the experimenters and a new protocol. The experimenters fix a constant k in {1,2,..,20}, sedate you, roll a D20 and flip a coin. If the coin comes up tails, they will wake you on day k. If the coin comes up heads and the D20 comes up k, they will wake you on day 1. In either case they will ask you for your credence that the coin came up heads.
You wake up. What is your credence?
In this problem, the multiple distinct copies of you have been removed, at the cost of an explicit randomiser. It is clear that the structure of the problem is independent of the specific value of the constant k. It is also clear that updating on being woken, the probability that the coin came up heads is 1/21 regardless of k. This is troubling for the 1/2 answer, however, as playing this game with a single die roll and all possible values of k recovers the Sleeping Beauty problem (modulo induced amnesia). Again, having reduced the expected number of observations to be in [0,1], intuition and calculation seem to imply a reduced chance for the heads branch conditioned on being woken.
This further suggests that the misunderstanding in Sleeping Beauty is one of naively looking at P(W|H) and P(W|T), when the expected numbers of wakings are E(#W|H) = 1, E(#W|T) = 2.
The Apparent Solution
If we allow conditioning on the number of observers, we correctly calculate probabilities in the Sleeping Twins and Probabilistic Sleeping Beauty problems. It is correctly noted that a "single paying" bet is accepted in Sleeping Beauty with odds of 2; this follows naturally under the following decision schema: "If it is your last day awake the decision is binding, otherwise it is not". Let the event of being the last day awake be L. Then:
P(L|W ∩ T) = 1/2, P(L|W ∩ H) = 1, the bet pays k for a cost of 1
E(Gains|Taking the bet) = (k-1) P(L|W ∩ H)P(H|W) - P(L|W ∩ T) P(T|W) = (k-1) P(H|W) - P(T|W)/2
Clearly to accept a bet at payout of 2 implies that P(H|W) - P(T|W)/2 ≥ 0, so 2.P(H|W) ≥ P(T|W), which contraindicates the 1/2 solution. The 1/3 solution, on the other hand works as expected. Trivially the same result holds if the choice of important decision is randomised. In general, if a decision is made by a collective of additional observers in identical states to you, then the existence of the additional observers does not change anything the overall payoffs. This can be modelled either by splitting payoffs between all decision makers in a group making identical decisions, or equivalently calculating as if there is a 1/N chance that you dictate the decision for everyone given N identical instances of you ("Evenly distributed dictators"). To do otherwise leads to fallacious expected gains, as exploited in Sleeping Beauty gets counterfactually mugged. Of course, if the gains are linear in the number of observers, then this cancels with the division of responsibility and the observer count can be neglected, as in accepting 1/3 bets per observer in Sleeping Beauty.
The Absent Minded Driver
If we consider the problem of The Absent-Minded Driver, then we are faced with another scenario in which depending on decisions made there are varying numbers of observer moments in the problem. This allows an apparent time inconsistency to appear, much as in Sleeping Beauty. The problem is as follows:
You are an mildly amnesiac driver on a motorway. You notice approaching junctions but recall nothing. There are 2 junctions. If you turn off at the first, you gain nothing. If you turn off at the second, you gain 4. If you continue past the second, you gain 1.
Clearly analysis of the problem shows that if p is the probability of going forward (constant care of the amnesia), the payout is p[p+4(1-p)], maximised at p = 2/3. However once one the road and approaching a junction, let the probability that you are approaching the first be α. The expected gain is then claimed to be αp[p+4(1-p)]+(1-α)[p+4(1-p)] which is not maximised at 2/3 unless α = 1. It can be immediately noticed that given p, α = 1/(p+1). However, this is still not correct.
Instead, we can observe that all non zero payouts are the result of two decisions, at the first and second junctions. Let the state of being at the first junction be A, and the second be B. We observe that:
E(Gains due to one decision|A) = 1 . (1-p)*0 + 1/2 . p[p+4(1-p)]
E(Gains due to one decision|B) = 1/2 . [p+4(1-p)]
P(A|W) = 1/(p+1), P(B|W) = p/(p+1), E(#A) = 1, E(#B) = p, (#A, #B independent of everything else)
Hence the expected gain per decision:
E(Gains due to one decision|W) = [1 . (1-p)*0 + 1/2 . p[p+4(1-p)]]/(p+1) + 1/2 . [p+4(1-p)].p/(p+1) = [p+4(1-p)].p/(p+1)
But as has already been observed in this case the number of decisions made is dependent on p, and thus
E(Gains|W) = [p+4(1-p)].p , which is the correct metric. Observe also that E(Gains|A) = E(Gains|B) = p[p+4(1-p)]/2
As a result, there is no temporal inconsistency in this problem; the approach of counting up over all observer moments, and splitting outcomes due to a set of decisions across the relevant decisions is seemingly consistent.
Sleeping Beauty gets Counterfactually Mugged
In this problem, the Sleeping Beauty problem is combined with a counterfactual mugging. If Omega flips a head, it simulates you, and if you would give it $100 it will give you $260. If it flips a tail, it asks you for $100 and if you give it to Omega, it induces amnesia and asks again the next day. On the other hand if it flips a tail and you refuse to give it money, it gives you $50.
Hence precommitting to give the money nets $30 on the average, whilst precommiting not to nets $25 on the average. However since you make exactly 1 decision on either branch if you refuse, whilst you make 3 decisions every two plays if you give Omega money, per decision you make $25 from refusing and $20 from accepting (obtained via spreading gains over identical instances of you). Hence correct play depends on whether Omega will ensure you get a consistent number of decisions or plays of the whole scenario. Given a fixed number of plays of the complete scenario, we thus have to remember to account for the increased numbers of decisions made in one branch of possible play. In this sense it is identical to the Absent Minded Driver, in that the number of decisions is a function of your early decisions, and so must be brought in as a factor in expected gains.
Alternately, from a more timeless view we can note that your decisions in the system are perfectly correlated; it is thus the case that there is a single decision made by you, to give money or not to. A decision to give money nets $30 on average, whilst a decision not to nets only $25; the fact that they are split across multiple correlated decisions is irrelevant. Alternately conditional on choosing to give money you have a 1/2 chance of there being a second decision, so the expected gains are $30 rather than $20.
Conclusion
The approach of using the updating on the number observer moments is comparable to UDT and other timeless approaches to decision theory; it does not care how the observers come to be, be it a single amnesiac patient over a long period or a series of parallel copies or simulations. All that matters is that they are forced to make decisions.
In cases where a number of decisions are discarded, the splitting of payouts over the decisions, or equivalently remembering the need for your decision not to be ignored, yields sane answers. This can also be considered as spreading a single pertinent decision out over some larger number of irrelevant choices.
Correlated decisions are not so easy; care must be taken when the number of decisions is dependent on behaviour.
In short, the 1/3 answer to sleeping beauty would appear to be fundamentally correct. Defences of the 1/2 answer appear to have problems with the number of observer moments being outside [0,1] and thus not being probabilities. This is the underlying danger. Use of anthropic or self indication probabilities yields sane answers in the problems considered, and can cogently answer typical questions designed to yield a non anthropic intuition.