Undergrad The Sleeping Beauty Problem: Any halfers here?

[Marana]

This makes no sense. First, you can consider the relative frequencies hypothetically, even in the case of a single experiment. That's standard probability theory.

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

Going directly from the definition, it does not apply to sleeping beauty. Even in a hypothetical sense, it can't be repeated. Tuesday tails must follow monday tails.

This is not the same as a coin flip, which hypothetically could be repeated as much as you want.

 

[stevendaryl]

One approach is to base the answer on a solid random experiment. On wednesday at noon we know we will believe in 1/2. By the principle of reflection we should believe 1/2 now. Another approach is to argue that there is no relevant information change. Halfers would say that waking up only tells you "I wake up at least once", which you already knew. It is true that there are other changes, like not knowing what day it is, but it isn't clear why that is relevant, or why it would lead to 1/3. Another option is to say the probability is not currently well defined.

I already mentioned this, but it's possibly worth repeating: The halfer answer can be thrown into doubt by considering a similar thought-experiment that sounds like the probabilities should be the same, but which justifies the thirder answer.

Instead of considering just three possibilities--two wakenings for heads, one for tails--let's make it a little more symmetric by throwing in a second wakening for the tails case as well. But in the tails case, the subject is told the coin flip result on his second wakening.

Then letting X be the statement "the subject has not been told the result", we have:

P(X) = 3/4 (because there are 4 awakenings, and X is true for 3 of them)

P(H|X) = P(H)/P(X) = \frac{1/2}{3/4} = 2/3

I think both the Bayesian and frequentist accounts would agree with this answer. And it seems that the way in which it differs from the original Sleeping Beauty problem is irrelevant to the probabilities.

 

[stevendaryl]

What I believe is that if you analyse the problem using probability theory (in particular relative frequencies), then you get a completely self-consistnet answer of 1/3. If you try to answer 1/2, then the probability of heads is inconsistent with the probability of 2/3 that it is Monday. And, if the probability is not 2/3 that it is Monday, then there is something very badly wrong.

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

 

[PeroK]

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

As I said, you can certainly state that probability theory does not apply in this case. And, the fact that 1/3 is a self-consistent answer is then irrelevant.

But, you can't then take another tack and use some wooly intuitive alternative to probability theory to give an answer of 1/2, which is not even self consistent.

Finally, as is covered in the Wikipedia analysis. If the sleeper procedes on the basis of 1/3, she will make the right decision in terms of prizes and fates. But, if she procedes on the the basis of 1/2 she will make less favourable decisions. And, since no one has actualy lied to her: i.e. she has not been given any false information, why does she get things wrong?

The random observer could win more prizes by using 1/3 heads than she could be using 1/2. So, why is she disadvantaged if no one has given her false information and she has as much information as the random observer?

Finally, I see no reason why you couldn't repeat the experiment many times, awarding her a prize every time she guesses heads/tails correctly.

The sleeper who uses 1/3 and always guesses tails will win more than the sleeper who uses 1/2 and guesses equally heads/tails. And that, IMHO, is as good a measure of whether you have the right calculation or not (again, with the proviso that no one has actually lied to you).

 

[stevendaryl]

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

Going directly from the definition, it does not apply to sleeping beauty. Even in a hypothetical sense, it can't be repeated. Tuesday tails must follow monday tails.

This is not the same as a coin flip, which hypothetically could be repeated as much as you want.

Right. There is something a little weird about a question like "What is the probability that today is Monday?" You can't randomly pick what day it is.

Your point about this being neither purely Bayesian nor purely frequentist is right. We're really being asked to compute a subjective probability, which makes it sound Bayesian. But the most straightforward way of computing it is to use a frequentist definition of probability.

 

[PeroK]

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

If I were the sleeper I would bet on it being Monday with 2/3 probability. Or, more simply, bet on its being tails.

I would be happy to take bets with anyone on that. There's no way I could (on average) lose. Every time I get woken I bet tails on a 50-50 bet and I win 2/3 of the time. After being a sleeper for a year or two, I retire on my winnings. I could even give odds 55-45 and still come out ahead.

The halfer who bets equally on heads/tails can only break even.

And, the confident halfer who bets heads all the time will lose!

PS I don't even begin to understand why relative frequency wouldn't apply to that situation. Gee, I'd be happy to find a philosopher who was willing to lose all his money that way!

PPS Although maybe the amnesia drug would have side-effects!

 

[Demystifier]

A Decision Theory Approach

A part of the problem is that, in the usual formulation of the Sleeping Beauty problem, the probability and degree of belief are abstract ambiguous concepts the meaning of which is not completely clear. To overcome this ambiguity let me reformulate the problem as a decision problem. With a given rules of game, what is wise for the Beauty to do? I will present two versions of the game rules, one corresponding to 1/3 and the other to 1/2.

Let me explain the rules through a dialog:
Experimenter: Hi Beauty, do you want to play a game with me?
Beauty: First I need to know the rules.
Experimenter: I shall flip a coin. If you guess correctly what the coin has shown, I will give you 100$. If you guess incorrectly what the coin has shown, you will give me 100$.
Beauty: Sounds boring. Is there a catch?
Experimenter: Yes. I will flip the coin only ones, but you will make two guesses in the case of tails and one guess in the case of heads. For each guess you will be awaken from a sleep and you will not remember anything about previous (if any) awaking.
Beauty: Sounds interesting, but something is still not crystal clear to me. In the case of tails, will the 100$ be payed for each guess?
Experimenter: Hmm, you are a smart beauty, I didn't think of it.
Beauty: Well, I need to know, my strategy will depend on it.
Experimenter: OK, suppose that I propose the rule A: The 100$ is payed for each guess.
Beauty: Then I would be very happy to play this game, because I have a strategy which puts me in an advantage over you.
Experimenter: And what if, instead, I propose the rule B: The 100$ is payed for the first guess only.
Beauty: Then it is a fair game, where nobody has any advantage over the other one.

So what is the best Beauty's strategy in the case of rule A? What about the case of rule B? The answers should be obvious.

In the case of rule A the Beauty's strategy is always to say "tails". (It corresponds to the 1/3 answer in the usual formulation of the problem.)

In the case of rule B, there is no special strategy for Beauty because any guess is as good as any other. (It corresponds to the 1/2 answer in the usual formulation of the problem.)

 

[PeroK]

A Decision Theory Approach

A part of the problem is that, in the usual formulation of the Sleeping Beauty problem, the probability and degree of belief are abstract ambiguous concepts the meaning of which is not completely clear. To overcome this ambiguity let me reformulate the problem as a decision problem. With a given rules of game, what is wise for the Beauty to do? I will present two versions of the game rules, one corresponding to 1/3 and the other to 1/2...

As I understand it, everyone is supposed to agree on this, because this formalises things.

In my opinion, with rule B there is really no point in the second awakening. Unless we have rule A, then it's not the problem as stated.

What I can't grasp is the halfer objection to this as a logical formulation of the problem. It's exactly the way I think of probabilities: calculations based on the available information.

For me, the halfers must fall into two categories: those that disagree philosophically with your formulation; and, those who simply miscalculate on the basis of rule A.

 

[Demystifier]

As I understand it, everyone is supposed to agree on this, because this formalises things.

Yes.

In my opinion, with rule B there is really no point in the second awakening.

Well, in that case the rule B can be modified, e.g. such that another coin is flipped to decide whether the payment will be according to the first or second guess (but not both). It doesn't affect the conclusions and strategies.

 

[Demystifier]

For me, the halfers must fall into two categories: those that disagree philosophically with your formulation; and, those who simply miscalculate on the basis of rule A.

I want to believe that only the first category is represented in this thread.

 

[PeroK]

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

Suppose the sleeper is told that she will always be woken on the Monday and there is a 50-50 probability of being woken on the Tuesday. Let's assume the decision mechanism is not specified.

Now we have no coin that must already be heads or tails to distract us.

In being woken, she must calculate it's 2/3 Monday and 1/3 Tuesday.

If not, please justify another answer.

If the decision is made via the outcome of a pre tossed coin, then this is just one of many possible mechanisms.

How could the specific mechanism affect the probability in this case?

And, if the mechanism does affect the outcome, how do you calculate in more complex cases where there is a probability $p_n$ of being woken on day $n$, where there is no intuitive a priori answer such as 1/2?

 

[stevendaryl]

Finally, as is covered in the Wikipedia analysis. If the sleeper procedes on the basis of 1/3, she will make the right decision in terms of prizes and fates.

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

 

[Demystifier]

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

Isn't this equivalent to my post #67?

 

[PeroK]

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

PS I'm not sure I understand this. My reply below was based on misreading this post, I think. In any case, my reply simply restates how I believe the sleeper can make money given the scenario. As I understand it, everyone is supposed to agree with this is any case.

I'll keep my money until it's time to bet. So that I don't guess what day it is by looking at how much money I've got, I'll have an automatic system that simply gives me $1 any time I want (although, with this game, I'm on for more than a $1 a time!).

I get woken, I get my $1, I put down my bet and I say "tails". That is it. The rest is obfuscation!

And, I'd have a side bet on the day of the week as well, if I could.

The whole point of my argument is that if the coin is initially heads, you only get one bet. The counterargument says you must lose $2 on heads, because you only get to bet once if it's heads. Umm ... isn't that just about the definition of the conditional probability that if I get woken it's only half the chance it's heads than tails? That is absolutely the definition of "probability that it's heads".

This is really about putting your money where your philosophy is!

 

[PeroK]

For me, it is quite obvious that the correct answer is 1/2.
There are 3 cases: Head-Monday, Tail-Monday, Tail-Tuesday
Head vs Tail are 50% 50%
so we have: Head-Monday 50%, Tail-Monday 25%, Tail-Tuesday 25%
So probability it was head is 50% or 1/2.

Is that the case without the amnesia drug as well?

Isn't it strange that if it's Monday, it's twice as likely to be a Head as a Tail?

 

[stevendaryl]

Isn't this equivalent to my post #67?

Yes. I posted before reading your post.

 

[Khashishi]

Rather than calculating conditional probabilities, it's simpler if we just calculate all the total probabilities from the experimenter's point of view.
From the experimenter's point of view, there's a 50/50 chance of heads/tails.
So, there's a 50% chance of waking Sleeping Beauty once, and 50% chance of waking them twice.

So, Sleeping Beauty will be asked to answer once or twice. If they answer heads (or tails), there's a 50% chance that they are right.
The issue is that the trials are not independent. If Sleeping Beauty answers heads, they're either right once (50% chance) or wrong twice (50% chance). If they answer tails, they are right twice (50% chance) or wrong once (50%).

If they want to make a fair wager, they'd better bet more on tails, because even though the probability is the same, they stand to lose twice on the same flip on heads. Therefore, level of belief is NOT equal to level of fair wager.

 

[PeroK]

Rather than calculating conditional probabilities, it's simpler if we just calculate all the total probabilities from the experimenter's point of view.
From the experimenter's point of view, there's a 50/50 chance of heads/tails.
So, there's a 50% chance of waking Sleeping Beauty once, and 50% chance of waking them twice.

So, Sleeping Beauty will be asked to answer once or twice. If they answer heads (or tails), there's a 50% chance that they are right.
The issue is that the trials are not independent. If Sleeping Beauty answers heads, they're either right once (50% chance) or wrong twice (50% chance). If they answer tails, they are right twice (50% chance) or wrong once (50%).

If they want to make a fair wager, they'd better bet more on tails, because even though the probability is the same, they stand to lose twice on the same flip on heads. Therefore, level of belief is NOT equal to level of fair wager.

If you take the idea of a random observer who happens on the experiment. Why is their level of belief different from the sleeper?

Also, unless someone has lied to you, why would your level of belief be different from the probabilities you can calculate.

If I understand your position, it is that that the sleeper would say her belief is that heads is 50-50, even though she can calculate she would lose money by betting in this?

Personally, I can't see how belief can be mathematically different from the odds I can calculate. For example, I can't imagine a situation where I believe that a coin is definitely heads but am unwilling to bet on it - unless I know or suspect I have been lied to.

 

[PeroK]

The sleeping beauty problem is a well known problem in probability theory, see e.g.
https://en.wikipedia.org/wiki/Sleeping_Beauty_problem
http://allendowney.blogspot.hr/2015/06/the-sleeping-beauty-problem.html
https://www.quantamagazine.org/solution-sleeping-beautys-dilemma-20160129
or just google.

Allegedly, there are many "thirders" who think that the correct answer is 1/3, but also many "halfers" who think that the correct answer is 1/2. For me, it is quite obvious that the correct answer is 1/3. Is there anybody here who is convinced that the correct answer is 1/2? If you are one of them, what is your argument for 1/2?

Sadly, dispiritingly, shamefully(?) it appears that on the whole of PF, in terms of thirders, it's only thee and me.

I thought this was a clear-thinking scientific forum, but, in this case, we appear to be outnumbered by woolly philosophical thinking.

In the spirit of some of the posts above I believe that everyone on PF is a thirder, even though I can calculate that halfers are in the majority.

 

[stevendaryl]

Sadly, dispiritingly, shamefully(?) it appears that on the whole of PF, in terms of thirders, it's only thee and me.

You don't think 98% agreement is good enough to join your club?

 

[forcefield]

The coin flip creates two equally likely scenarios. Hence the probability is 1/2.

 

[PeroK]

You don't think 98% agreement is good enough to join your club?

Given the meagre membership, I think we can lower the entrance criteria in your case.

 

[forcefield]

The coin flip creates two equally likely scenarios. Hence the probability is 1/2.

I was only a part-time halfer: I should add that she has information about the relative probabilities of her being awake in the two scenarios. Hence 1/3.

 

[lewando]

I like to think the SB is the inquisitive type and would be happy to submit to a large series of trials. SB will precede each trial with a simple wager to probe her experience. The wager (claim, really) is this: “the coin flip will be heads”. She will always make the same wager before each trial. At the end of each trial (Wednesday), she will be informed of, and record, the result of the actual coin flip and will have access to this information at all times throughout future trials. As the number of completed trials becomes large enough to be statistically significant, her answer during the interview phase will become more correct.

 

[Demystifier]

You don't think 98% agreement is good enough to join your club?

It is always healthy to question one owns convictions. For that purpose, in post #67 I have given a case in which 1/2 is correct. I hope it will not me banish from the club.

 

[PeroK]

It is always healthy to question one owns convictions. For that purpose, in post #67 I have given a case in which 1/2 is correct. I hope it will not me banish from the club.

You were the founder member!

 

[Demystifier]

You were the founder member!

The best way to become the club president is to banish the founder.

Now seriously. I am sorry that I didn't open a poll, I guess it's too late now. The possible poll answers would be:
- 1/2
- 1/3
- It depends on the precise formulation of the problem.

 

[PeroK]

Okay, here's my last idea to disprove the answer of 1/2:

You wake sleeper and she gives the answer of 1/2. Then:

a) You ask her: if it's Monday, what would be your answer? She says 1/2.

b) You ask her: if it's Tuesday, what would be your answer? She says 0.

The only combination of those probabilities that gives 1/2 is that it must be Monday. Therefore, if the sleeper is consistent she must conclude that it is definitely Monday.

 

[Demystifier]

Okay, here's my last idea to disprove the answer of 1/2:

You wake sleeper and she gives the answer of 1/2. Then:

a) You ask her: if it's Monday, what would be your answer? She says 1/2.

b) You ask her: if it's Tuesday, what would be your answer? She says 0.

The only combination of those probabilities that gives 1/2 is that it must be Monday. Therefore, if the sleeper is consistent she must conclude that it is definitely Monday.

The best disproof of 1/2 so far.

 

[Marana]

Okay, here's my last idea to disprove the answer of 1/2:

You wake sleeper and she gives the answer of 1/2. Then:

a) You ask her: if it's Monday, what would be your answer? She says 1/2.

b) You ask her: if it's Tuesday, what would be your answer? She says 0.

The only combination of those probabilities that gives 1/2 is that it must be Monday. Therefore, if the sleeper is consistent she must conclude that it is definitely Monday.

I may be different than a typical halfer, because I don't think we can use anything like this.

You are treating it as though MH, MT, and TT are the possible outcomes of a random experiment, each with probability 1/3. In that case, it is certainly true that the probability of heads is 1/3, and the probability of heads after conditioning on Monday is 1/2. But this is not an accurate description of sleeping beauty: when she wakes up it is not a random experiment. Tuesday always follows monday, tuesday tails always follows monday tails. Sleeping beauty doesn't lose her memory of this fact, nor does she lose her memory of what week is coming. The usual probability methods you are using need to be justified somehow for this situation which is not a random experiment, and I don't see how.

A typical halfer may also treat it like a random experiment, except they would break it into two stages. First the coin toss, then the waking up. In other words, the typical halfer is saying that waking up can only be viewed as a selection of the possible days to wake up. So they would say that MH has probability 1/2, MT has probability 1/4, and TT has probability 1/4. Probability of heads is 1/2, and probability of heads after conditioning on monday is 2/3. I disagree with this for the same reason I disagree with thirders: it isn't a random experiment.

So to me, the question is how do we define probability for such a weird situation which isn't a random experiment at all? It may be that there is currently no definition for that situation, and it may be that we don't need one. Notice that our strategies are already set in stone on sunday, before the coin flip. Whether or not there is new, relevant information on monday (which I can't imagine what it could be) it definitely won't change our strategy, calling into question if there is any importance to it. Halfers and thirders will always agree on how to act for any betting setup, even if they disagree on how probability should be defined. A halfer may argue "I prefer to bet on tails because if it is tails the bet is offered twice", a thirder may say "I prefer to bet on tails because I am defining the probability as 2/3 for tails", but they will both bet the same thing.

But to me, the most natural way to define it is to start with something solid, a genuine random experiment. The coin flip, which has sample space of H and T, each with probability 1/2. Then if you believe there is no new, relevant information on monday, you could hold onto that 1/2 probability when waking up. Or you could use the principle of reflection, knowing that you would believe 1/2 at noon on wednesday. I admit, neither of these are fully convincing.

According to the way I calculate it, it is not possible to condition on it being monday or tuesday. Which I think makes sense, because when using conditioning you can't have impossible events become possible. "It is monday" can't be followed by "it is tuesday." The probability for "it is tuesday" became 0 when you learned "it is monday". Losing memory of monday does not fix this problem, so conditioning is not justified.