The Sleeping Beauty Problem: Any halfers here?

[PeterDonis]

if on Monday you wake Sleeping Beauty with probability $\alpha$, and on Tuesday you wake her with probability $\beta$,

Well, another Bayesian approach would be to use this information to determine the priors for P(Monday|Awakened) and P(Tuesday|Awakened), instead of just using a maximum entropy prior. In other words, using information about the process that is going to generate the single trial to assign prior probabilities. E. T. Jaynes talks about this sort of thing in his book on probability theory. (For example, he discusses using information about the details of the process by which a coin is flipped, e.g., using a robotic coin flipper, to assign priors for heads and tails.) If this is done, then yes, your formula makes sense.

 

[stevendaryl]

Did I misread the problem? I thought that Sleeping Beauty gets awoken Monday, regardless of heads or tails, but for Tuesday only if it's tails.

Yes, I was asking a related question: If on Monday, you wake her up with probability \alpha, and Tuesday wake her up with probability \beta (wiping her memory), what should be her subjective likelihood that it's Monday when she wakes up? Obviously if \alpha = 0 and \beta > 0, then the likelihood that it is Monday would be zero. If \alpha > 0 and \beta = 0, then the likelihood that it is Monday is 1. If \alpha = \beta, then the likelihood that it is Monday is 1/2.

The actual Sleeping Beauty case is \alpha= 1, \beta = 1/2. Except that, instead of asking the probability of it being Monday, they ask the probability that the coin toss result was (will be?) heads.

 

[StoneTemplePython]

This kind of thing happens from time to time in probability and seems to be a linguistic problem.

The issue people get hung up on is whether they are minimizing the error between their estimate of the coin's probability of heads, $\hat{p}$, and its actual $p$, or are they minimizing the error of how often they will be wrong (which depends on how often they are woken up). Put differently, the payoff function isn't clear.

Probability has its roots in betting, and it's wise to reformulate this problem into betting as a few have done. In general betting reformulations are helpful, but especially so when people aren't totally sure what the payoff function they are optimizing, is. The converse is that if you have a problem that you cannot even approximately formulate as a betting problem, then you know you're in trouble.

- - - - -
n.b. There's actually a good book out there (that I've only read a little of thus far) by Shafer and Vovk which insists that all of probability can be reformulated in terms of betting, including the foundational stuff from Kolmogorov (who was Vovk's Phd advisor)

 

[Charles Link]

This kind of thing happens from time to time in probability and seems to be a linguistic problem.

The issue people get hung up on is whether they are minimizing the error between their estimate of the coin's probability of heads, $\hat{p}$, and its actual $p$, or are they minimizing the error of how often they will be wrong (which depends on how often they are woken up). Put differently, the payoff function isn't clear.

Probability has its roots in betting, and it's wise to reformulate this problem into betting as a few have done. In general betting reformulations are helpful, but especially so when people aren't totally sure what the payoff function they are optimizing, is. The converse is that if you have a problem that you cannot even approximately formulate as a betting problem, then you know you're in trouble.

- - - - -
n.b. There's actually a good book out there (that I've only read a little of thus far) by Shafer and Vovk which insists that all of probability can be reformulated in terms of betting, including the foundational stuff from Kolmogorov (who was Vovk's Phd advisor)

I was also of the opinion that this particular problem may not have a perfect answer, but your @StoneTemplePython explanation that "the payoff function isn't clear" does a good job of summing up the dilemma.

 

[Stephen Tashi]

I was also of the opinion that this particular problem may not have a perfect answer,

On the one hand, there are problems in the sense of real life (or imagined real life) problems. On the other hand there are problems in the sense of well posed mathematical problems.

If the Sleeping Beauty problem is a well posed mathematical problem and it has two different answers then mathematics is in trouble.

If the Sleeping Beauty problem is considered as a real life problem, then there can be different ways of modelling it as well posed mathematical problems and the fact that people get different answers isn't disturbing to mathematics. Different models for the same imagined real life problem are a problem for physics, economics, psychology etc.

The wikipedia article says:

The Sleeping Beauty puzzle reduces to an easy and uncontroversial probability theory problem as soon as we agree on an objective procedure how to assess whether Beauty's subjective credence is correct. Such an operationalization can be done in different ways:

Are any of the participants taking a contrary view? Does anyone present think the Sleeping Beauty problem indicates an inconsistency in probability theory?

 

[Marana]

I went through the numbers, and it's 2/3. If you do the experiment over and over, with different starting times, then of the "active" Sleeping Beauties (the ones in day 1 or day 2 of the experiment), 1/3 are on day 1 after a coin flip of heads, 1/3 are on day 2 after a coin flip of heads, and 1/3 are on day 1 after a coin flip of tails. If the actual Sleeping Beauty thinks of herself as a random choice among those, then she would come up with 2/3 heads.

It's sort of similar to the situation where there is some country where people have a 50% chance of producing one offspring, and 50% chance of producing two offspring. If you take a random adult and ask the probability that it will have two offspring, the answer is 50%. If you take a random child and ask what is the probability that their parent had two children, it's 2/3.

This seems like a restatement of the frequency of awakenings argument, side by side instead of back to back. I still think that relative frequency of awakenings is an odd choice for probability of coin tosses if awakenings are not experiments. Just like we wouldn't use relative frequency of awakenings without the drug.

Sleeping beauty is similar to the random adult who answers 50%, while an outside observer is similar to the child who answers 2/3. From sleeping beauty's perspective, she flips a coin with a 50% chance of producing one awakening and 50% chance of producing two awakenings. But an outside observer observes a random awakening, of which 2/3 are tails.

An outside observer is capable of choosing a random awakening. Sleeping beauty is totally incapable of choosing a random awakening for herself, as she is locked into the experiment structure.

To call it a misuse, you need to say what reason is there not to. What harm comes from it?

To me, the best example of a counter-intuitive result coming from the thirder position is to change it to a lottery. A person has a one in a million chance of winning. But you can make it subjectively 50/50 by waking the winner a million days in a row. That's strange.

Yeah, I mentioned earlier that everybody always agrees on how to bet. The strategy is set in stone on sunday before the coin flip even happens. So it is questionable if there is any importance to whether or not subjective probability should be said to change or not change when awakening, or any need to define it.

But even if 1/3 is a harmless answer, some of the arguments for it can still have flaws. The frequency of awakenings argument lacks justification, and I think there is a major flaw in the most popular argument for 1/3, that would be bad even if 1/3 is an ok answer.

How do you get that? That's truly nonsensical, for the following reason (pointed out by @PeroK): If the memory wipe happens on the morning of the awakening, right before sleeping beauty wakes up, then there is no need to even toss the coin until Tuesday morning. So on Monday, the coin hasn't even been tossed (under this variant). How could the knowledge that today is Monday tell you about a coin that has not yet been tossed?

Like I said, I don't agree with defining the probability that way. I think probability of heads is 1/2 and probability of heads when you learn "it is monday" is also 1/2. But the typical halfer solution is instructive nevertheless.

I've been saying that waking up is not a random experiment, and that nobody here would even think of using relative frequency of awakenings to stand for probability of coin tosses if not for the drug tempting us. The typical halfer solution occurs when you force it to be a random experiment: flip a coin and then randomly select a day. So P(H)=1/2 and P(H|M)=2/3. The result makes sense because monday is more likely to be selected on heads. Again, I disagree with the model, and therefore I don't define the probability that way, but the technique is fine.

The most popular thirder technique is not fine. It attempts to have it both ways: on the one hand it doesn't use the structure of a random experiment, and on the other hand it uses conditioning purely on the time. Unlike the halfer argument above, the thirder argument does not make waking up an experiment: the coin is the experiment and you may wake up twice for one coin. The argument is equivalent to the "it is 2:00", "it is 2:01" situation. Conditioning doesn't work like that! When you learn "it is 2:01" it is a different kind of information, and using conditioning would be a contradiction that rendered probability meaningless. The most popular thirder argument is disastrous.

That's not the same thing. You can eliminate that problem by making statements about connections between events. "The first time I looked at the clock, it was 2:00." "The second time I looked at the clock, it was 2:01". No contradiction.

This doesn't work because we already know what will happen on monday. No amount of re-wording will change the fundamentals. Monday is the first awakening no matter what. The only thing "it is monday" tells us is that it is, in fact, monday right now.

In contrast "it is tuesday" also tells you that you are awake on tuesday, which was not known beforehand and changes the probability.

 

[stevendaryl]

Like I said, I don't agree with defining the probability that way. I think probability of heads is 1/2 and probability of heads when you learn "it is monday" is also 1/2.

But that belief has strange consequences, if you're going to use the usual rules of probability theory.

You say: P(Heads | Monday) = P(Heads)

But surely, P(Heads | Tuesday) = 0 (in earlier posts, I got heads and tails confused, and said P(Tails | Tuesday) = 0).
It's impossible for sleeping beauty to be awake if today is Tuesday and the coin result was heads.

The rules of conditional probabilities are:

P(Heads) = P(Heads | Monday) P(Monday) + P(Heads |Tuesday) P(Tuesday) = P(Heads) P(Monday) + 0

So P(Monday) = 1.

So it follows that Sleeping Beauty should have a subjective likelihood of 1 that today is Monday. That can't be right.

Let me ask you what the "halfer" answer is to this question:

If Sleeping Beauty is awakened on Monday with probability \alpha, and on Tuesday with probability \beta, with her memory wiped after Monday, then upon awakening, what would you say is her subjective likelihood of it being Monday? (The actually Sleeping Beauty problem has \alpha = 1 and \beta = 1/2)

You have been saying that it's not a true random experiment, and it's not for Sleeping Beauty. But surely you can reason as follows:

• If \alpha = 0 and \beta > 0, then Sleeping Beauty will know for certain that today is Tuesday upon wakening.
• If \beta = 0 and \alpha > 0, then Sleeping Beauty will know for certain that today is Monday upon wakening.
• If \alpha = \beta, then Sleeping Beauty has no reason to prefer Monday over Tuesday.

So even though you might say that probability isn't defined in this case, since you can't randomly select what day it is, it makes sense for Sleeping Beauty to have different levels of belief that is Monday or Tuesday, depending on the values of \alpha and \beta. Subjective (Bayesian) probability is a way to quantify this. What I would say is the thirder position is that

P(Monday | Awake) = \frac{\alpha}{\alpha + \beta}

This is the same as the relative frequency answer. I assume that the halfer position would be that the probability is undefined? Even when \alpha = 0, so it's impossible for her to be awake on Monday?

 

[Dale]

Oh my, there is so much misinformation and misunderstanding in his thread. I am amazed.

@Marana a rational person must bet according to the probabilities. It is irrational to bet differently than the probability. If the rational betting strategy is agreed then the rational probability clearly follows. Any discrepancy is a mistake.

@Boing3000 waking up with amnesia is information, and information does change conditional probability. I don't know whether you don't understand that waking up with amnesia (waking) is information or if you are refusing to condition on that information, but either way is wrong.

@PeterDonis (and others earlier) Bayesian reasoning is fine here, and we can use the Bayesian idea of probability without resorting to a fictitious ensemble of sleeping beauty experiments. The probability that Beauty is asked to compute is P(heads|awake), not P(monday|awake), so the prior is P(heads) which is fixed by the fact that the coin is fair.

 

[Boing3000]

waking up with amnesia is information, and information does change conditional probability. I don't know whether you don't understand that waking up with amnesia (waking) is information or if you are refusing to condition on that information, but either way is wrong.

So you can say you have amnesia because you remember what you don't ? Come on Dale be serious.
The whole point of the drug in that experiment is to place Beauty in a forever Monday, and that she has no new information to collect by awakening.
That she could eventually remember Sunday and the explanation (thus that she would be drugged or not) is of no help to her. She might as well be questioned Sunday evening.

So maybe you should be more specific on what "i am refusing". Because you seem to refuse what amnesia is.

 

[stevendaryl]

So you can say you have amnesia because you remember what you don't?

He's saying that "waking up with amnesia" is information. The fact that you are awake tells you that either the coin was not Heads, or today is Monday.

 

[stevendaryl]

@PeterDonis (and others earlier) Bayesian reasoning is fine here, and we can use the Bayesian idea of probability without resorting to a fictitious ensemble of sleeping beauty experiments. The probability that Beauty is asked to compute is P(heads|awake), not P(monday|awake), so the prior is P(heads) which is fixed by the fact that the coin is fair.

Just the prior of heads doesn't answer the question, though. We can write:

P(heads|awake) = P(heads | awake & monday) P(monday | awake) + P(heads | awake & tuesday) P(tuesday | awake)

The second term is zero (since it's impossible for it to be heads if Sleeping Beauty is awake on a Tuesday). So we have:

P(heads | awake) = P(heads | awake & monday) P(monday | awake)

At this point, I would say that P(heads | awake & monday) = P(heads). Knowing that you are awake and that it is Monday doesn't tell you anything about whether the coin is heads or tails. So we have finally:

P(heads | awake) = P(heads) P(monday | awake) = 1/2 P(monday | awake)

So the two conditionals I think are closely related.

 

[Boing3000]

He's saying that "waking up with amnesia" is information.

Please tell me how YOU would KNOW that you have amnesia, after waking up. Don't hand wave, tell me the experiment you can do to TEST your amnesia.

The fact that you are awake tells you that either the coin was not Heads, or today is Monday.

That "you are awake" tell's you your are awake... some day (Monday if you trust your memory). It tell you nothing NEW about the coin THAT DAY.

There will be a lot of water running under that bridge, before hell bent thirders acknowledge that this is what render that problem useless...

 

[Dale]

So you can say you have amnesia because you remember what you don't ? Come on Dale be serious.
The whole point of the drug in that experiment is to place Beauty in a forever Monday, and that she has no new information to collect by awakening.
That she could eventually remember Sunday and the explanation (thus that she would be drugged or not) is of no help to her. She might as well be questioned Sunday evening.

So maybe you should be more specific on what "i am refusing". Because you seem to refuse what amnesia is.

Waking up with amnesia is an observable fact. Furthermore it is an observable fact with a strong correlation to the quantity of interest. It is legitimate information in the sense of probabilistic inference.

 

[stevendaryl]

Please tell me how YOU would KNOW that you have amnesia, after waking up.

In the Sleeping Beauty thought experiment, you don't know that you have amnesia; you know that either (1) It is Monday, or (2) It is Tuesday and you have amnesia and the coin result was tails. In case 1, the probability of heads is 1/2. In case 2, the probability of heads is 0. If you don't know which case you're in, then your subjective probability of heads is a weighted average between 1/2 and 0.

 

[stevendaryl]

Waking up with amnesia is an observable fact. Furthermore it is an observable fact with a strong correlation to the quantity of interest. It is legitimate information in the sense of probabilistic inference.

Actually, what you know in the Sleeping Beauty problem is that you have no memory of Monday. That could be because of amnesia, or it could be because today is Monday.

 

[stevendaryl]

That "you are awake" tell's you your are awake... some day (Monday if you trust your memory). It tell you nothing NEW about the coin THAT DAY.

Using all-caps doesn't make your case stronger. Being awake tells you that either you are in one of two situations:

1. It is Monday.
2. It is Tuesday and the coin toss result was tails.

Sleeping Beauty can reason:

If I'm in situation 1, then it's equally likely that the coin toss was heads or tails. So the probability of heads would be 1/2.
If I'm in situation 2, then it's impossible that the coin toss was heads. So the probability of heads would be 0.
Since I don't know which situation I'm in, I need to do a weighted average:

P(heads| awake) = P(heads | situation 1) P(situation 1 | awake) + P(heads | situation 2) P(situation 2 | awake)

Since P(heads | situation 2) = 0, and P(heads | situation 1) = 1/2, this implies:

P(heads | awake) = 1/2 P(situation 1 | awake)

So the halfer position is equivalent to saying P(situation 1 | awake) = 1. The thirder position is equivalent to P(situation 1 | awake) = 2/3.

 

[Boing3000]

Waking up with amnesia is an observable fact.

Just tell me HOW. Don't put a sentence with "is". Put a sentence with HOW.
Just so you know there is no calendar no watch in the room.

 

[stevendaryl]

Just tell me HOW. Don't put a sentence with "is". Put a sentence with HOW.
Just so you know there is no calendar no watch in the room.

I would say that knowing that you have amnesia is irrelevant. What's important is knowing that you are awake and have no memory of Monday. That tells you that either it is Monday, or it is Tuesday and the coin toss result was tails.

 

[Boing3000]

Actually, what you know in the Sleeping Beauty problem is that you have no memory of Monday. That could be because of amnesia, or it could be because today is Monday.

Finally ! Whatever case is IDENTICAL to Beauty, you cannot project in her frame: a forever monday. Only experimenter have more information.

 

[Boing3000]

I would say that knowing that you have amnesia is irrelevant.

It is irrelevant, because you don't get to forget that. This does NOT change.

What's important is knowing that you are awake and have no memory of Monday.

Which is is case for EVERY day. This does NOT change

That tells you that either it is Monday, or it is Tuesday and the coin toss result was tails.

That is trivially known on Sunday. This does NOT change

So why put beauty to sleep if nothing will change ? That problem have a problem. Until you can clearly specify how Beauty can "update her information" by awakening, you are making o logical mistake.

 

[PeterDonis]

a rational person must bet according to the probabilities

But just knowing the probabilities isn't enough. You also have to know the payoffs attached to each possible outcome. In other words, you are betting on the weighted expectation of the payoff.

@Demystifier in post #67 described two different payoff schemes, one of which implies betting as a thirder, the other of which implies betting as a halfer.

 

[stevendaryl]

It is irrelevant, because you don't get to forget that. This does NOT change.

Which is is case for EVERY day. This does NOT change

That is trivially known on Sunday. This does NOT change

So why put beauty to sleep if nothing will change ? That problem have a problem. Until you can clearly specify how Beauty can "update her information" by awakening, you are making o logical mistake.

Underlining doesn't make your point any better than all-caps. Tell me what you disagree with in the following:

1. If Sleeping Beauty is awake, she knows that either it is Monday, or that it is Tuesday and the coin toss result was tails. [edit: I did say "heads"]
   
2. If it is Monday, then the probability of the coin toss being heads is 1/2.
3. If it is Tuesday, then the probability of the coin toss being heads is 0.
4. If she doesn't know whether it is Monday or Tuesday, then she should use a weighted average, to get something between 0 and 1/2.

The only way to get the probability of heads to be 1/2 is if the probability of it being Tuesday is 0.

 

[stevendaryl]

But just knowing the probabilities isn't enough. You also have to know the payoffs attached to each possible outcome. In other words, you are betting on the weighted expectation of the payoff.

@Demystifier in post #67 described two different payoff schemes, one of which implies betting as a thirder, the other of which implies betting as a halfer.

The betting that supports being a thirder is: Bet at each awakening, and after the experiment ends pay off all bets.

The betting that supports being a halfer is: Bet at each awakening, and after the experiment ends, only honor the last bet (or the first; it doesn't matter).

 

[stevendaryl]

Finally ! Whatever case is IDENTICAL to Beauty, you cannot project in her frame: a forever monday. Only experimenter have more information.

It's not forever Monday for her---it's that she doesn't know whether it's Monday or Tuesday. There is a difference between (falsely) believing that it is Monday and not knowing whether it is Monday.

Anyway, you seem to use all-caps or underlining, rather than mathematics to get your answers, it would be more helpful if you could derive them, since what's being asked for is a numerical answer.

I can't imagine how anyone would dispute the following:

1. P(heads | awake) = P(heads | monday & awake) P(monday | awake) + P(heads | tuesday & awake) P(tuesday | awake) : That's just a theorem of conditional probability.
2. P(heads | tuesday & awake) = 0: That's part of the problem statement.

So the only possible dispute, I would think is over the two numbers:

1. P(heads | monday & awake)
2. P(monday | awake)

The thirder position is that the first number is 1/2 and the second number of 2/3. The halfer position is that the first number is 1/2 and that the second number is what? Undefined?

 

[Boing3000]

Underlining doesn't make your point any better than all-caps.

No, what does make those point better is that you cannot invalid them, and start talking about font style.

Tell me what you disagree with in the following:

All point are false:
1) she is always awake.
2) there is no if ...
3) there is no if ...
4) ... because she doesn't know.

The only way to get the probability of heads to be 1/2 is if the probability of it being Tuesday is 0.

Stop clutching to that straw. Everyone knows that.

The sad thing is that that Beauty doesn't know what day it is. So there is no point to asking her the question.

 

[stevendaryl]

All points are false.

That's just strange. Let's take the first one:

• If Sleeping Beauty is awake, she knows that either it is Monday, or that it is Tuesday and the coin toss result was tails. [edit: I did say "heads" by mistake]

That is basically the statement of the Sleeping Beauty problem. How can it be false?

You're not making any sense at all.

 

[stevendaryl]

The sad thing is that that Beauty doesn't know what day it is. So there is no point to asking her the question.

This gets back to your not actually understanding (or you understand and don't accept) the concept of conditional logic. She doesn't know what day it is, but she can do conditional reasoning: For any statement X, she can reason:

P(X) = P(X | it is Tuesday) P(it is Tuesday) + P(X | it is Monday) P(it is Monday)

If you're denying that, then you're tossing out the mathematics of conditional probability, and replacing it by the all-caps method of reasoning, which has not been shown to be sound.

 

[PeterDonis]

she is always awake

Yes, so the only conditional probabilities we need to compute are the ones that are relevant if she is awake. That's why every term in the breakdown that @stevendaryl gave (and which I gave in an earlier post) has a factor that conditions on her being awake.

 

[Boing3000]

That's just strange. Let's take the first one:

• If Sleeping Beauty is awake, she knows that either it is Monday (and the coin had been tossed (or not, with unknowable probability), or the coin and been tossed, to no avail until tomorrow) ), or that it is Tuesday and the coin toss result was heads tails.

That is basically the statement of the Sleeping Beauty problem. How can it be false?

What if the alternative to the if ? Is she is asked question in her sleep ?
I have yet again correct the information you cannot understand about the real statement of Sleeping Beauty problem (using fancy font)

You're not making any sense at all.

I may well be, but then you cannot even explain your statement that she can test her amnesia or "update her information". A simple question, that you cannot even provide any answer for.

 

[stevendaryl]

Yes, so the only conditional probabilities we need to compute are the ones that are relevant if she is awake. That's why every term in the breakdown that @stevendaryl gave (and which I gave in an earlier post) has a factor that conditions on her being awake.

Thank you. To me, it seems that the conclusion that P(heads | awake) < 1/2 follows just from the laws of conditional logic:

P(heads | awake) = P(heads | Monday & awake) P(Monday | awake) + P(heads | Tuesday & awake) P(Tuesday | awake)

If P(heads | Monday & awake) = 1/2 and P(Monday | awake) < 1, then it follows that P(heads | awake) < 1/2.

The exact answer P(heads | awake) = 1/3 requires additional justification, but concluding that it's less than 1/2 doesn't seem to.