First, consider a meaningless change: SB will be assigned, but not told, a random coin result C from the set {Heads,Tails}, and a random day D from the set {Monday,Tuesday}. She will be put to sleep on Sunday night, and the coin will be flipped. She will be wakened at least once, and maybe twice, during the next two days. She will he left asleep only on day D, and only if the actual coin flip was C. Anytime she is wakened, she is asked for her assessment of the probability that the coin result was C.

The change is meaningless because it is the original question when D=Tuesday and C=Heads, and the other combinations are symmetric variations that must have the same answer.

Then, use four volunteers and only one coin. Accomplish the random assignments mentioned earlier by randomly distributing the four possible combinations of {C,D} among them. On each day, three of the volunteers will be wakened, but only one of those three will have been assigned the C that matches the actual flip. So each can confidently answer 1/3 to the question asked.

]]>In my previous post I have presented the probabilities for both definitions of the event Red. It is clear that we both agree that in case we define Red as the event “SB will be led into a red room in this line of history” (call it Red1) P(Red1)= 1/2 and that if we define the event Red as “SB observes (at her current awakening I may add) that she is led into a red room” (call it Red2), that P(Red2)=1/4. My argument is that you can’t define the events unless you have already defined a random experiment whose outcomes constitute the sample space these events belong to. This is why I have described two random experiments, one corresponding to the first interpretation of event Red (Red1) and the other to the second interpretation (Red2). The subtle point is (and to my opinion the main reason for all the controversy around SB problem) that although the first random experiment actually takes place (there is a coin flip whose outcome is random), the second random experiment never actually takes place. It is used by SB to model her uncertainty on which stage of the actual experiment she is at her current awakening. So it is perfectly consistent to model her ignorance upon her current state as if the stage she is at has been randomly selected and assign P(Red)=1/4 and P(Blue)=3/4 for her current awakening. However, she knows that such random selection never actually takes place. Thus, if she is actually led on a Blue room she cannot update to P(Heads|Blue2)=1/3 but she still can use P(Heads|Blue1)=1/2.

Best regards,

Yannis ]]>

SB has to base her assessment on what she sees, not what fortune tellers may know. She only sees that she is led into a red room, not what may happen on another day.

]]>>> that event Red, what is P(Red)?

> P(Red)=P(Heads)=1/2

Incorrect. In three out of four possible situations, she will be led into a blue room. In one, she will be led into a red room. So P(Red)=1/4.

What you ignore, is that there are two different days that occur after a “heads” result. You consider “Room=Red” to be true on some Mondays when she is led into a blue room. Yes, she will be led into a red room the following day, but when she observes “Room=Blue,” it is not true FOR THAT OBSERVATION regardless of whether it is true on another day.

Similarly, in the original experiment, she will be awakened in one of three possible situations. When she finds herself awake, each is equally likely to be her current situation. Your error is considering a different situation to be her current situation, just because it is guaranteed to happen. The error is that it ISN’T happening.

]]>“The experiment is essentially the same; except SB is always wakened. In the cases where she was wakened in the original, she will be led into a room that is painted blue for her interview. On Tuesday after Heads, she is led into a room that is painted red. Once she is awake:”

Then, you pose 5 questions on certain probabilities.

To answer your questions:

1) What is the probability she will be led into a red room? That is, if we call that event Red, what is P(Red)?

P(Red)=P(Heads)=1/2

2) What is the probability she will be led into a blue room? That is, what is P(Blue)?

P(Blue)=1

3) If she is led into a red room, what is the probability of Heads? That is, what is P(Heads|Red)?

P(Heads|Red)=1

4) If she is led into a blue room, what is the probability of Heads? That is, what is P(Heads|Blue)?

P(Heads|Blue)= P(Heads)=1/2

5) Do Red and Blue represent a partition of the sample space?

No they do not.

In order to define a sample space you have to define a random experiment whose outcomes constitute this space. In the experiment you have defined (which is an extension of RE1), the only random element is the original coin toss. Thus, the probabilities in question are nothing like the ones you have calculated (the correct values are given in my answers). The confusion arises by the fact that you are actually referring to a different random experiment you have not defined (and is actually an extension of RE2). Thus, like in RE2, once awakened, SB models her situation as the outcome of a random experiment in which her current state is the result of a random procedure on top of the original coin toss (notice that such random procedure does not take place and SB is aware of this fact). Therefore, it could be either Heads and Blue with probability ¼, or Heads and Red also with probability ¼, or Tails and Blue with probability ½. Thus, P(Red)=1/4 and P(Blue)=3/4 (notice that in the context of this random experiment Red and Blue represent a partition of the sample space). Of course, in the context of this random experiment the probabilities you have calculated are correct, including P(Heads|Blue)=1/3. The subtle part is that once SB is awakened in a blue room she cannot update P(Heads) to 1/3, because she knows that a Blue event (as defined by the later random experiment) has not actually occurred. She knows that (apart from the role of the coin toss) her awakening on that blue room was the result of a deterministic procedure. Thus, no evidence on a Blue event (as defined by the later random experiment) has been produced. On the contrary only evidence of a Blue event as defined by the former random experiment has been produced hence P(Heads|Blue)= P(Heads)=1/2. The same applies for the Red room, she can only use the conditional probabilities of the first random experiment to update. Hence, once awakened in a red room P(Heads|Red)=1.

Best regards,

Yannis ]]>

“–Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be wakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be wakened and interviewed on Monday only. If the coin comes up tails, she will be wakened and interviewed on Monday and Tuesday. In either case, she will be wakened on Wednesday without interview and the experiment ends. Any time Sleeping Beauty is wakened and interviewed, she is asked, “What is your belief now for the proposition that the coin landed heads?”–”

If the experiment is repeated one thousand times then, on average, she will be awakened and asked about her belief fifteen hundred times during the experiments. On average, five hundred of those fifteen hundred times she is asked (i.e. one third of the times) the coin will have landed heads and so the proposition “the coin landed heads” will be true. On average, one thousand of those fifteen hundred times she is asked (i.e. two thirds of the times) the coin will have landed tails and so the proposition “the coin landed heads” will be false.

So sleeping beauty knows several facts. She knows that in repeated trials of the experiment, on average, one out of every three times she is awakened the proposition “the coin landed heads” will be true. She knows she was just awakened. So it is reasonable for her to give one third as an answer. She also knows that in repeated trials of the experiment, on average, one out of every two times that the experiment is run the proposition “the coin landed heads” will be true. She knows the experiment is being run. So it is reasonable for her to give one half as an answer. Her answer to the question should be: My belief is that, on average, the proposition is true one out of every two times the experiment is run and, on average, one out of every three times I am awakened and asked my belief about it.

]]>Rather than go into your unexplained, and incorrect, definition of “new information,” and your arbitrary shifting between different probability models, let me describe a simple analogy: The experiment is essentially the same; except SB is always wakened. In the cases where she was wakened in the original, she will be led into a room that is painted blue for her interview. On Tuesday after Heads, she is led into a room that is painted red. Once she is awake:

1) What is the probability she will be led into a red room? That is, if we call that event Red, what is P(Red)?

2) What is the probability she will be led into a blue room? That is, what is P(Blue)?

3) If she is led into a red room, what is the probability of Heads? That is, what is P(Heads|Red)?

4) If she is led into a blue room, what is the probability of Heads? That is, what is P(Heads|Blue)?

5) Do Red and Blue represent a partition of the sample space?

If the answer to #5 is “yes” – and it obviously is – then the Law of Total Probability applies. That means the P(Heads) must equal P(Heads|Red)*P(Red)+P(Heads|Blue)*P(Blue). Four of these five terms are pretty easy to assign values to: P(Red)=1/4, P(Blue)=3/4, P(Heads|Red)=1, and P(Heads)=1/2. That means (1)*(1/4)+P(Heads|Blue)*(3/4 )=1/2, so P(Heads|Blue)=1/3. Since this changed, being led into a a blue room, or a red room, clearly represents “new information.” My final question is, what part of the procedure makes SB’s information “new” in one case, but not the other? Specifically, why does it matter what could happen in the situation that SB knows IS NOT HER CURRENT SITUATION that affects her answer?

My point is that it does not matter what happens in the “other” situations. There are initially four equally-likely situations, and all that matters to SB is that she has both necessary and sufficient conditions to define a subset of those four that represent her current situation. The two problems are identical.

And please note that this does not involve a “frequency argument.” Getting the same answer does not mean I use the same argument. I agree about building the model forward in time, but you have to include every possibility that exists, not just those that represent what SB sees in the cases of interest.

]]>In regards to the original SB problem, learning “internally” that it is Monday is not actually new information! If the coin is heads, she will be awoken Monday with probability 1. If the coin is tails, she will be awoken Monday and Tuesday with probability 1. This is a completely sensible probability model for the event that she will be awoken Monday. Thus, P( H | awake on Monday) = 1/2 by bayes theorem. If she does not know which day it is, which is the original intent of the problem, then her subjective beliefs regarding Heads or Tails should still be 1/2 since she has no new information on which to update her beliefs. Being awoken is not an event that is able to provide information via bayes theorem. As you see, I am basically using the axiomatic approach to define a valid probability model for the events combined with bayesian updating. The other way to think about this is that the actual information she is using to update with is that she has been awoken at least once. This is a completely valid approach using conditional probability.

Where the frequency argument gets into trouble is that it is inadvertently assessing the long-run frequency of (# times heads and awoken) / # times awoken = 1/3, which is not the same probability that the question is asking for. Of all the times she is awoken, the proportion of times it will be when the coin is heads is subtly different than your beliefs as to whether or not the original coin flip was heads. These are actually two very different probability spaces. It’s irrelevant to the original question of the problem whether she thinks it is more likely to being questioned when the toss was heads or tails.

In regard to really old post from JeffJo on 10/26/11 7:15, the argument appears to be flawed. The events Monday and Tuesday are not the only information you have to condition on, you also have to condition on being awake. When you argue that

P(H) = P(H|M) * P(M) + P(H|T)*P(T), I believe it is improper to argue that P(M) is her belief that it is Monday. Instead, P(M) should be the probability that she is awoken on Monday. Thus,

P(H) = P(H| awoken on Monday) * P(awoken on Monday) = 1/2.

You have to build your probability model forward in time to make sure you are modeling the appropriate sequence of random events. Start with the coin toss and then what will happen as a result of the coin toss. Otherwise you will get confused as to what are random events, so like ending up thinking that P(M)=2/3. You should have a prior probability for heads or tails, then build a probability model for the events that will occur under each of those two scenarios, this is the likelihood. The implied likelihood you have created doesn’t make sense. I believe the formulation of what constitutes the probability model for the data, the likelihood, is key to solving the paradox.

Let me try coming at this from a different direction. The twist that makes the Sleeping Beauty Problem controversial is that it is really two problems in one: an external problem, as seen by Sunday Sleeping Beauty (SSB) or the director of the experiment, and an internal problem, as seen by Awake Sleeping Beauty (ASB). We can directly apply probabilities, based on intuitive statements like P(H)=P(T)=1/2, only to the external problem. Internally, the occurrences are derived from combinations of the external occurrences, and so may or may not have intuitive values.

But how to derive the internal probabilities, whether or not they are the same, is controversial. To distinguish these POVs, I’ll call their probabilities Pe(*) and Pi(*). Errors are made in the literature when the author assigns internal probabilities based on the external POV. It started with Lewis, who literally assumed Pi(C=H)=Pi(C=T)=1/2 in order to “prove” that Pi(C=H)=Pi(C=T)=1/2. You can’t prove anything by assuming the answer, whether or not it is correct.

The biggest difference between the two problems is that externally, the two-day experiment constitutes one trial. So T1 and T2 represent the same outcome. But internally, the T1 and T2 that were part of the same trial externally must be considered different trials. SB can only observe (this is what I mean by that word) that one, and only one, day applies to her at the moment. As a result, the number of trials is potentially different, and this famously leads to another controversy about how to test the answer with frequencies.

When you addressed this difference with a new RE, you claimed you did it to better define the events. But in reality what you did was reduce both the internal and external experiments to utilizing the same trial. That in turn allowed you to associate both the external, and internal, definitions to the same, single trial. That RE is:

RE2: Choose a coin result C from {H,T}. If C=H, set day D=1; otherwise, choose a day D from {1,2}. Wake SB on day D and ask her for Pi(C=1).

This RE still has to represent what we know to be the same true from *either* POV in the original problem, except what is affected by the number of trials. (I.e., by altering the number of trials you changed the absolute values of the external probabilities.)

So, the relative values of the external probabilities must still be the same for RE3 to represent the original problem. In the original problem, Pe(H1)=Pe(T1). You treated both POVs as the same, with P(H1) = 2*P(T1) both internally and externally. Literally, if the original problem were to be repeated many times, an external observer expects to see just as many H1 interviews as T1 interviews, irrespective of the frequency controversy I mentioned above. But RE2 has twice as many external H1s as T1s, so it cannot be a correct representation of the original problem. In its place, I propose:

RE3: Choose a coin result C from {H,T}. Choose a day D from {1,2}. Assign X=S (sleep) if the result is H2; otherwise assign X=W (wake). Wake SB according to X on day D and ask her for Pi(C=1).

I’ll keep Pe(*) and Pi(*)separate. It can’t be wrong to do so, and it may prove important.

You assumed Pe(C=H)=Pe(C=T)=1/2, and Pe(D=1)=Pe(D=2)=1/2. So will I. Now is it straightforward to derive Pe(H1W)=Pe(C=H)*Pe(D=1)=1/4; and the same for Pe(T1W), Pe(T2W), and Pe(H2S). Other combinations have probability 0. And the important thing is that these values maintain the relative values all of the external probabilities. They don’t maintain absolute values, because RE3 alters the definition (and the number) of “trials.” But by having all the same relative values, it is externally equivalent to the original. By having the internal events determined from the external ones the same way as the original, it is also internally equivalent.

But now we can see that Pi(H1)=Pe(H1|X=W)=Pe(H1W)/[Pe(H1W)+Pe(T1W)+Pe(T2W)]=1/3.

The only conceivable objection to RE3, is that an interview is no longer certain. It isn’t a valid objection, but that never stops people in this kind of controversy. This objection was the purpose of my quadruplet experiment, which extends RE3 in such a way to guarantee an interview (the same way the original did – there might be two) while maintaining the same information as in RE3 for any one of the girls. That RE was:

RE4: Secretly, and randomly, label four quadruplets H1,H2,T1, and T2. Choose a coin result C from {H,T}. On day 1 wake and interview quadruplet H1 or T1 according to C, and on day 2 wake and interview quadruplet T2 if C=T. Have them determine P(H|I get interviewed).

For any quadruplet who is wakened, her situation is identical to that in RE3 since she was randomly assigned day 1 or day 2. And I showed that RE3 is identical in principle to the original problem, while RE2 is not. RE4 is also shown to be identical by having the quadruplets calculate an answer jointly. That answer must be the same regardless of which quadruplet is being interviewed, so it is the same as the original SB’s answer must be.

+++++

> If one adopts your line of thought he could also argue that since H1, T1 and T2 are subjectively equivalent, ..

No, one can’t; and no, they aren’t. You are taking those statements out of context (yes, I repeated them out of context, but I assumed that context was understood when I did so).

T1 is subjectively equivalent to H1 under the condition that {D=1} has occurred, because with that condition the only difference is the subjectively equivalent {C=T} or {C=H}. But we can use this fact to derive the equation P(T1)=P(H1) without any such conditions.

T1 is subjectively equivalent to T2 under the condition that {C=T} has occurred, because with that condition the only difference is the subjectively equivalent {D=1} or {D=2}. But we can use this fact to derive the equation P(T1)=P(T2) without any such conditions.

The transitive property does not apply to these two subjective equivalences, because the conditions that are needed to make them subjectively equivalent are different. The transitive property does apply to the two equations because they are not conditional.

> It is wrong to assume this, because H1 will certainly occur in case of Heads, whereas in case of Tails either T1 or T2 will occur. Thus, although T1 and T2 are interchangeable, H1 and T1 (or H1 and T2) are not.

And this line of reasoning is fallacious. First, you are again arguing about internal events based on the external POV. You are relying on the temporal difference between Monday and Tuesday; “temporal” is not an internal property. This is the difference I am calling “observational.” Since SB can’t distinguish the days, she can only observe that “today” is either Monday, or Tuesday, but not both. This is subjectively equivalent to knowing the coin will be Heads, or Tails, but not both.

The equivalent internal statement is “H1 must surely apply to this interview, if Heads applies to it; whereas in if Tails applies either T1 or T2 could have. Thus, although T1 and T2 are interchangeable, H1 and T1 (or H1 and T2) are not.”

Second, you could just as easily say that “T2 must surely apply to this interview, if Tuesday applies to it, whereas in if Monday applies either H1 or T1 could have. Thus, although H1 and T1 are interchangeable, T1 and T2 (or H1 and T2) are not.” Since this says just the opposite of what the same line of reasoning said before, there must be a fallacy somewhere. That fallacy is that, where you concluded that “H1 and T1 are not interchangeable,” all you can actually deduce is that “This statement does not imply whether other pairs of events are interchangeable or not.”

> That’s why the only correct application of PoI is the one Elga does for T1 and T2 and it breaks down if you try to apply it to H1 and T1 because learning that it is Monday is not equivalent to learning that a ‘H1 or T1′ event occurred.

It is correct for H1 and T1. It does not “break down,” because learning *INTERNALLY* that it *IS* Monday, is equivalent to learning that a {H1 or T1} event occurred for this *INTERNAL* trial. What occurs for other trials is irrelevant.

JeffJo

]]>I will begin with your last post since it is very crucial to resolve this misunderstanding. You said:

>Let me just reiterate one point: a TRIAL of the RE we need for this problem is not the three-day experiment the director sees. It is a single day that ASB sees because she is awake.

I completely agree that “a TRIAL of the RE we need for this problem is not the three-day experiment the director sees”. In fact, when you are referring to ASB I refer to RE2 and by its definition it is clear that a TRIAL of RE2 ends with the selection of a single day where SB is awake. On the other hand, when you refer to SSB I refer to RE1 which spans all three days. In fact, my whole approach is based on the above point.

>Most of your arguments have ignored this fact, and so are invalid.

None of my arguments ignores this fact. I always clarify which RE I am referring to and treat TRIALS accordingly.

>In particular, the RE you suggested – which correlates what must be different trials to ASB – cannot be used to answer the question.

I have suggested two REs one that can be used to model SB’s situation upon awakening (RE2) and the trivial one (RE1) that can be used to model SB’s understanding of the whole experiment (which is always valid not only on Sunday. Even when she wakes up she can still use RE1 to calculate that her credence on ‘a Monday interview will occur during the experiment’ should be 100%.). RE2 defines T1 and T2 as disjoint events and does not correlate different TRIALS. In an RE2 TRIAL either a T1 or a T2 event can occur (but not both) and I have never suggested otherwise.

Taking the above under consideration lets get back to your 4:12pm post.

I said: “I also don’t see how relating the events to observations can be either meaningful or helpful?” and you replied:

>And I believe it is because you don’t want to try. To ASB, in a valid RE, {H1,T1,T2} are disjoint events. As I’ve described.

It must be clear by now that I always argued that to ASB, in a valid RE, {H1,T1,T2} are disjoint events. However, I still can’t see what ‘observations’ have to do with this.

I said: “If you take away the coin toss my point becomes even more clear. I never claimed that Elga’s definition was similar to the ‘an interview will take place on Wednesday’ definition.” and you replied:

>You implied it when you said T1 and T2 cannot be both disjoint, and products of the same RE.

I never said that. On the contrary I have proposed RE2 where T1 and T2 are both disjoint events, and products of the same RE. My entire line of arguments is actually based on this fact!

You continue:

>In my simplification, all five days are only “outcomes of the same trial of the RE,” as you put it in your previous reply, from the experiment director’s POV. But the correct RE is in ASB’s POV; observing an interview in her “now.” What the director sees is irrelevant.

I can’t see why you think I ever said or suggested any of the above. I never used the experiment director’s POV. I am explicitly defining the REs I am using and this is why I do not need to refer to POVs. RE2 is a random experiment that can be used to model ASB’s situation. Thus, ASB can use it to calculate her credence on what day of the week it is ‘now’. Moreover, ASB can still use RE1 to to calculate her credence on ‘a Monday awakening will occur during the experiment’. In my approach, the focus is not on the POV but on what you are modelling. You can say that when ASB is using RE1 she adopts SSB’s (or experiment director’s) POV but this is besides the point. If I try to use your terminology I could say that when ASB uses RE2 to calculate her credence on what day of the week it is ‘now’ she is using her own POV and not the director’s or SSB’s and clearly in that case the days of the week are NOT “outcomes of the same trial of the RE”.

>Yep – but this is from ASB *OBSERVATIONAL* point of view. In this RE, what happens on other days are different trials even though they are “outcomes of the same trial of the RE” in the director’s POV.

Exactly! We do not disagree on that. In fact it is what I tried to demonstrate by using the different REs and their corresponding sample spaces. In the {M1, T1,…,F1} sample space the events are disjoint (you call that “ASB *OBSERVATIONAL* point of view”), whereas in the {M,T,…,F} sample space they are not and they can all occur in the same TRIAL (you call that “the RE” in the director’s POV”).

I said: “If you accept the RE I propose” and you responded:

>I don’t. It is from the director’s POV, and artificially forces T2 to not happen if T1 does. In ASB’s RE, they occur in different trials.

I have explained in this post that RE2 is from ASB’s POV and its T1 and T2 events occur in different TRIALS.

I also said: “Thus, if you are trying to prove something about M1,T1,…,F1 events you cannot use M,T,…F events (as Elga does implicitly in his proof)” and you replied:

>And my entire point is that you can, because the RE is ASB’s observation.

But the RE that is ASB’s observation is the one using the {M1,T1,…,F1 } sample space. It is absurd to mix it with M,T,…F events which correspond to a different RE (the director’s POV according to your terminology).

What about the correct application of PoI? Have you accepted my arguments or you will address them on a different post?

Best regards,

Yannis