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.
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:
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.
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.
I'd quibble about calling it an assumption. The 1/3 solution notes that this is the ratio of observations upon awakening of heads to the total number of observations, which is one of the problematic facts about the experimental setup. The 1/3 solution assumes that this is relevant to what we should mean by "credence", and makes an argument that this is a justification for the claim that Sleeping Beauty's credence should be 1/3.
Your argument is, I take it, that these counts of observations are irrelevant, or at best biased. Something else should be counted, or should be counted differently. The disagreement seems to center on the denominator; it should count not awakenings, but coin-tosses. Then there is a difference in the definition of the relevant events and the probabilities that get calculated from them.
Did I get that right? Is this a fair description?
I think a key difference between halfers and thirders is that for thirders, the occurrence of an awakening constitutes evidence of the current state of the system that's being asked about--whether the coin shows heads or tails, because the frequency with which the state of the system is asked about (or, equivalently, an observation is made) is influenced by the current state of the system. To ward off certain objections, it is of no consequence whether this influence is deterministic, probabilistic or mixed in nature, the mere fact that it exists can and should be exploited. I don't think there's disagreement that it exists, but there is over how it's relevant.
Halfers deny that any new evidence becomes available on awakening, because the operation of the process is completely known ahead of time. (Alternatively, if any new evidence could be said to become available, it cannot be exploited.) From what I can tell, and my understanding is surely imperfect, there is some kind of cognitive dissonance about what kinds of things can constitute evidence in some epistomological theory, such that drawing a distinction between the actual occurrence of an event and the knowledge that at least one such event will surely occur is illegitimate for halfers. Is this a fair description?
That's as may be, but it doesn't help Sleeping Beauty in her quandary. If you think this example helps to prove your point, I think it helps to prove the opposite. Although she knows, in this variation, that a randomly selected person will be tested, the random person selection process is not accessible to her, only the opportunity to know that one of three possible test results has been collected. She knows very well, given a randomly selected person (resp. a coin toss), what the probability they are male is (resp. the given coin toss came Heads). She isn't being asked about that conditional probability. (Or maybe you think she is? Please clarify.) To follow your analogy, upon being awakened, she's informed that a test result has been collected from an unknown person, and now, given that a test result has been collected, what are the chances it cames from a male?
Clearly the selection process for asking Sleeping Beauty questions is biased. If bias had not been introduced by an extra awakening on Tuesday, the problem would collapse into triviality. The puzzle asks how this sampling bias should affect Sleeping Beauty's calculations of what to answer on awakening, if at all. One of the reasons for doing statistical analysis of sampling schemes is to quantify how the mechanism that's introducing bias changes the expected values of observations. In the SB case, the biased selection process is a mixture of random and deterministic mechanisms. Untangling the random from the deterministic parts is difficult enough for the participants in this discussion-- they can't even agree on a forking path diagram! Untangling it for Sleeping Beauty while she's in the experiment is epistemically impossible. She has no basis whatsoever inside the game for saying, "this one is randomly different from the last one" versus "this one is deterministically identical to the last one, therefore this one doesn't count."
The same considerations apply to the case of the cancer test. Let me elaborate on your scenario to see if I understand it, and let me know if I'm mischaracterizing the test protocol in any material way. There is a test for a disease condition. Every person knows they have a 50% chance going in of testing positive for the disease. We'll stipulate that the repeatability of the test is perfect, though in real life this is achieved only within epsilon of certainty. (Btw, here's where the continuity argument enters in: how crucial is the assumption of absolute certainty versus near certainty? What hinges on that?) In this protocol, if the initial test result is positive, then the test is repeated k times (k=2 or 10, or whatever you deem necessary), either with a new sample or from an aliquot of the original sample, I don't think it matters which. Here the repetition is because of the obstinacy of the head of the test lab and their predilection for amnesia drugs; in real life the reasons would be something like the very high cost in anguish and/or money of a false positive, however unlikely. You, as a recorder of test results, see a certain number of test samples come through the lab. The identities of the samples are encrypted, so your epistemic state with regard to any particular test result is identical to that for any other test sample and its result.
So now the question comes down to this: upon any particular awakening, how is the test subject's epistemic state at any particular awakening significantly different from the lab tech's epistemic state regarding any particular test sample? There is a one-to-one correspondence between test samples being evaluated and questions to the patient about their prognosis. Should they give the same answer, or is there a reason why they should give different answers? Just as with the patient, the lab tech knows that any randomly chosen individual has a 50% chance of of giving a positive test result, but does she give the same answer to that question as to a different question: given that she has a particular sample in her hands, what is the probability that the person it belongs to will test positive? She knows that she has k times as many samples in her lab that will test positive than otherwise, but she has no way of knowing whether the sample in her hands is an initial sample or a replicate. It seems to me that halfers might be claiming these two questions are the same question, while thirders claim that they are different questions with different answers. Is this a fair description? If not, please clarify.
What you say is true for any outside observers, and for Sleeping Beauty after the experiment is over and the logbooks analyzed. But while Sleeping Beauty is in the experiment, this option is simply not available to her. The scenario has been carefully constructed to make this so, that's what makes it an interesting problem. The whole point of the amnesia drug in the SB setup (or downloadable avatars, or forking universes, random passersby, whatever) is that she has NO justification nor even a method for NOT treating any of her awakenings as separate variables, because the information that could allow her to do this is unavailable to her. By construction--and this is the defining feature of Sleeping Beauty--all Sleeping Beauty's awakenings are epistemically indistinguishable. She has no choice but to treat them all identically.
This phenomenon is a common occurrence in queueing systems where there's a very definite and well-understood difference between omniscient "outside observers" and epistemically indistinguishable "arriving customers", who can have different values for the probability of observing the system in state X, where the system is executing a well-defined random process, or even a combination random-deterministic process.
Thanks for your detailed response. I'll make a few comments now, and address more of it later (short on time).
No, I was just saying that this, lim N-> infinity n1/(n1+n2+n3), is not actually a probability in the sleeping beauty case.
No, I wouldn't say that. My argument is that you should use probability laws to get the answer. If you take ratios... (read more)