It will surprise many people, but nobody is using the correct wording of the problem. They are including elements that Adam Elga introduced in order to explain his thirder solution simply. The two wordings have the same answer, of course, but the objections Halfers raise to his solution all relate to the parts he introduced.

Here is the actual wording from Elga’s paper, with two edits that remove irrelevant information: “Some researchers are going to put you to sleep. During the [time] that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are [awake], to what degree ought you believe that the outcome of the coin toss is Heads?”

In the original wording, my “[time]” was two days. But it didn’t actiually say the wakings would occur on different days, and no order was specified. My “[awake]” was “first wakened,” which is a bit ambiguous. Since no order is specified, it can’t mean “Monday, but not Tuesday.” And since Elga’s solution involved telling SB information after her first answer, it seems obvious that it means “before podssibly learning something.”

There is another way of implementing this, where the “new information” becomes obvious, and the Halfer’s methods work against them.

Once SB is asleep, flip two coins, a Quarter and a Nickel. If either is showing Tails, (1) wake her, (2) ask her the question, and after she answers, (3) put her back to sleep with amnesia. Then, TURN THE NICKEL OVER, and repeat the same three steps. The question is “to what degree ought you believe that the Quarter is showing Heads?”

Note that the Quarter never changes, so what it is currently showing is the same as the outcome of its flip. But even though the Nickel might not be showing the what its flip-outcome was, the prior distribution for the two coins is found by Halfer techniques. Each of {HH, HT, TH, TT}, where the Quarter is listed first, has a probability of 1/4.

But the “new information” that SB has is that HH is eliminated. She wouldn’t be awake, and facing the question, it both coins afre currently showing Heads. The remaining three combonations each update to a conditional probability of 1/3. Since only one of them has the Quarter showing Heads, the answer to the question is 1/3.

]]>Some argue that Beauty, upon awakening, learns no new information, and therefore the priori probabilities of the initial coin flip are valid. While it is true that B learns no new information, that is NOT what determines the posterior probabilities; they are changed by the simple act of waking up – thus entering phase two of the experiment, the first one being the coin flip. For example, consider the same rules but for flipping the coin a second time on tails to determine whether Beauty should be awoken on Monday or Tuesday. Now the answer is heads with 1/2. The equations remain the same, but Pr(C=h|D=mon) evaluates to 2/3 instead of 1/2.

(1) and (2) work for all possible scenarios with respect to awakening Beauty. For Pr(C=h|D=mon):=1 however, signifying Beauty is never awoken on Monday for tails, they evaluate to x=y, meaning Monday and heads are synonymous, thus C taking on its priori probabilities, 1/2.

]]>Let C be the coin flip {h,t}, and D the day {mon,tue}. Then

(1) y := Pr(C=h) = Pr(C=h|D=mon)*Pr(D=mon) + Pr(C=h|D=tue)*Pr(D=tue) = 1/2*x + 0*(1-x) = x/2

(2) x := Pr(D=mon) = Pr(D=mon|C=h)*y + Pr(D=mon|C=t)*Pr(C=t) = 1*y + 1/2*(1-y) = x/4 + 1/2

(1) and (2) defines an equation system in x and y with solution 2/3 and 1/3 respectively. QED.

“:=” means “defined as” – introducing a definition.

Thus, there is no need for convoluted explanations betting on outcomes, or sidetracking with modified problems. All the information needed for solving the problem at hand is readily available in the problem description.

]]>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.

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