I The Sleeping Beauty Problem: Any halfers here?

[PeterDonis]

It might mean that some of the ways of translating an informal problem into a mathematical one are just wrong.

It might. Or it might mean the informal problem was ambiguous. In the end that's a judgment call.

 

[stevendaryl]

Yes, that's true. In other words, the original problem statement is vague, and making it precise requires picking one set of assumptions about the nature of probability and credence, out of multiple mutually incompatible possibilities of which different people will pick different ones.

That's in the nature of thought experiments. That's sort of the point of them---you have an informally described situation, and the challenge is to formalize them sufficiently well that are amenable to being solved through mathematics, logic and possibly experiment. The fact that different people choose different formalizations does not by itself mean that all ways of formalizing it are equal. Some ways of formalizing it just don't stand up to scrutiny.

That's sort of the point of exploring slight tweaks of the problem statement. If a principle of reasoning is used in the original problem, and applying the same principle in the tweaked case clearly gives the wrong answer, that suggests that the principle of reasoning is flawed.

It's certainly possible that such a problem is inherently ambiguous; there are multiple, completely legitimate interpretations that lead to different answers. But just the fact that someone proposes a solution doesn't automatically make it correct.

 

[stevendaryl]

It might. Or it might mean the informal problem was ambiguous. In the end that's a judgment call.

Right. To me, the test that a solution is a legitimate interpretation is precisely if the solution approach is robust with respect to slight tweaks of the problem.

 

[Boing3000]

It's interesting that some of the wilder defences of the 1/2 position seem to have taken this to the extreme that even if you look at the coin and see a head, say, then the question over the probability that it is heads lands heads when flipped remains 1/2.

I hope that your are not referring to my posts, first because I don't defend 1/2 over 1/3 because they are both valid.
Secondly because, as stevendaryl, you insert error of omission, that I corrected above in bold
In this case the probability is associated with the action not with a state, or lack of information about that state.

And that the certainty now that it is a head is not actually a probability.

I'll let you slip hairs further to decide if certainty equals 1.0 probability.
What I am sure is that a coins once landed on head, never ever turn into a tail, nor became a bowl of petunia. In my book it is not a probability.

This also leads to the extreme position that the probability of a fair coin flipped, will land on being heads is absolutely 1/2 and can never change, even to 0 or 1, under any circumstances.

Coin never are in any other states than 0 or 1. There is a 1/2 possibility that you observe it being tail or head.
There are even circumstances when flipping is unfair, and nobody said otherwise. You can very much train yourself or a robot with precise mechanics to approach 0 or 1 probability when flipped expertly (not flipped "fairly"), even if the coin itself is fair. That's what knife throwers do (with sharp coins).
You can also build any type of dependencies between a series of events leading in all types of probabilities. This is basic knowledge.

These reveal fundamental disagreements over the nature of probability and credence that wouldn't be removed even if the problem were stated more precisely, as you suggest.

I have seen no such fundamental disagreements.

 

[stevendaryl]

I think that there is a related problem to the Sleeping Beauty problem whose solution is implicit in any solution of the Sleeping Beauty.

Suppose that Sleeping Beauty is awakened with probability \alpha on Monday, and with probability \beta on Tuesday (after a memory wipe). What is her subjective probability that today is Monday, given that she's awake?

I claim that it is \frac{\alpha}{\alpha + \beta}. You can justify that using relative frequency: If you do it over and over again, N times (once a week for N weeks) then she'll be awake approximately (\alpha + \beta) N times, and of those, \alpha N will be Mondays. So the relative frequency of her being awake on Mondays as a fraction of the total times that she is awake is \frac{\alpha}{\alpha + \beta}.

If halfers have some other answer, then please say what it is. I'm assuming that the only other possible answer is: Undefined. However, that doesn't seem right, because in the case \alpha = 0, then Sleeping Beauty knows for sure that it's Tuesday, and if \beta = 0, then she knows for sure that it's Monday.

 

[PeterDonis]

What is her subjective probability that today is Monday, given that she's awake?

Yes, this is what I was calling P(Monday|Awakened) in earlier posts. Your formula gives the answer 2/3. I argued in a previous post that this answer can only be justified (based on relative frequencies) if Beauty is told that the experiment will be run a large number of times, and she will be given the amnesia drug at the end of each run so that during each run she doesn't know how many runs have been made. For a single run, I argued that a maximum entropy prior could be used, which gives the answer 1/2 (and therefore the answer 1/4 to the question of what P(Heads) is, which would I guess be a "quarterer" answer).

This line of argument does seem to make the halfer position hard to maintain, since by the equations in my earlier post, P(Heads) = 1/2 requires P(Monday|Awakened) = 1, i.e., Beauty would need to be certain that it was Monday when she was awakened.

My relevant earlier posts are here:

https://www.physicsforums.com/threa...m-any-halfers-here.916459/page-7#post-5780588

https://www.physicsforums.com/threa...m-any-halfers-here.916459/page-8#post-5781206

 

[PeroK]

In other words, you stuck to the problem as stated, except for embellishments. Which just converse my point.

Only if those calculations are the ones that correspond to the vague ordinary language in the original problem statement. Which you argue for by adding embellishments. Which again concedes my point.
.

In this thread, I believe I am representing mainstream mathematical thought. By undermining those of us who are doing this, and saying nothing to counter the tide of personal theorising, you are, I believe, acting against the spirit of PF.

These last comments - effectively denying us any attempt to illuminate a problem in any way and attacking us for doing so are particularly harmful.

I wish I could understand your motivation for attacking only those of who are presenting material in a measured analytical fashion in this thread.

You have lost a great deal of my respect for you.

Sadly, those are my last words on this thread.

 

[PeterDonis]

In this thread, I believe I am representing mainstream mathematical thought.

If "mainstream mathematical thought" includes the frequentist interpretation of probability, then I suppose you are. But, as I said before, not everyone is a frequentist--not even all "mainstream" mathematicians, AFAIK.

the tide of personal theorising

If you think particular posts in this thread are personal theorizing, by all means report them.

attacking

I don't think pointing out assumptions or limitations in a particular line of thought is "attacking". I also don't see what's so difficult about admitting that the original specification of some problem in a Wikipedia article is not very precise (note that I only came into this thread after that option was added to the poll). I'm not saying you personally wrote an imprecise or vague specification of the problem.

You have lost a great deal of my respect for you.

I'm sorry you feel that way. My intention has not been to "attack" anyone, nor to prevent anyone from illuminating the problem.

 

[stevendaryl]

Yes, this is what I was calling P(Monday|Awakened) in earlier posts. Your formula gives the answer 2/3. I argued in a previous post that this answer can only be justified (based on relative frequencies) if Beauty is told that the experiment will be run a large number of times, and she will be given the amnesia drug at the end of each run so that during each run she doesn't know how many runs have been made. For a single run, I argued that a maximum entropy prior could be used, which gives the answer 1/2 (and therefore the answer 1/4 to the question of what P(Heads) is, which would I guess be a "quarterer" answer).

Well, if on Monday you wake Sleeping Beauty with probability \alpha, and on Tuesday you wake her with probability \beta, it seems that her credence for today being Monday would have to depend on \alpha and \beta, at least in the extremes:

\alpha = 0, \beta > 0 \Longrightarrow P(Monday | Awakened) = 0
\alpha > 0, \beta =0 \Longrightarrow P(Monday | Awakened) = 1
\alpha = \beta > 0 \Longrightarrow P(Monday | Awakened) = 1/2

My formula interpolates smoothly between those extremes, even if you don't buy the relative frequency argument in the case of one trial.

 

[Charles Link]

Well, if on Monday you wake Sleeping Beauty with probability \alpha, and on Tuesday you wake her with probability \beta, it seems that her credence for today being Monday would have to depend on \alpha and \beta, at least in the extremes:

\alpha = 0, \beta > 0 \Longrightarrow P(Monday | Awakened) = 0
\alpha > 0, \beta =0 \Longrightarrow P(Monday | Awakened) = 1
\alpha = \beta > 0 \Longrightarrow P(Monday | Awakened) = 1/2

My formula interpolates smoothly between those extremes, even if you don't buy the relative frequency argument in the case of one trial.

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.

 

[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.

 

[stevendaryl]

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.

I gave you the formula:

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

That formula is a theorem of conditional probability (together with the assumption that it's either Monday or Tuesday). What else do I need to explain about it?

 

[Stephen Tashi]

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

The main source of confusion is that "the" Sleeping Beauty Problem has not been stated as a specific mathematical problem. People are offering solutions without stating which mathematical interpretation of the problem they are solving.

Questions about "What should sleeping beauty think?" are not well posed mathematical problems because what a person "should" think is subjective.

A question of the form "What is the probability that...?" has some chance of being a well posed mathematical problem if the event in the question is precisely defined.

 

[PeterDonis]

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

Yes, I agree with that: unless Beauty is certain that it's Monday when she's awakened, the conditional probability P(Heads|Awake) must be less than 1/2.

 

[stevendaryl]

What if the alternative to the if ? Is she is asked question in her sleep ?

You don't have to have an alternative in order for a statement of the form "If A then B" to make sense. A statement of the form "If A then B" is true in case B is true or A is false.

In our case, A is "Sleeping Beauty is awake". B is "It is either Monday, or it is Tuesday and the coin toss result was tails"

But you're saying "If A then B" is false in this case. By the laws of logic, that is only possible if A is true and B is false. So you are saying that it is false that "It is either Monday or it is Tuesday and the coin toss result was tails"?

 

[PeterDonis]

Questions about "What should sleeping beauty think?" are not well posed mathematical problems because what a person "should" think is subjective.

I wouldn't quite say it's "subjective" (although the term "subjective credence" is indeed used in the Wikipedia article's statement of the problem). The issue is that it depends on which (objective) thing is the one the person being asked the question is supposed to use to determine their answer.

For example, the interpretation of "subjective credence" as which bets the person would or would not take means that the person has to know the payoffs, as I pointed out in response to @Dale . And the original problem statement does not include payoffs, so it is incomplete on this interpretation. But the conditional probabilities, which are also involved in a rational determination of which bets to take, are still perfectly objective--unless one takes there to still be some uncertainty possible about P(Monday|Awake). And once the payoffs are specified, the rational calculation of which bets to take is also perfectly objective.

 

[Stephen Tashi]

But the conditional probabilities, which are also involved in a rational determination of which bets to take, are still perfectly objective-.

Can we clarify things by having a statement of which conditional probabilities are to be computed?

In particular, what precisely are the "given" events involved in the conditional probabilities?

 

[JeffJo]

I'm going slow to prevent any possibility disagreement. Consider these variations:

1) The day doesn't matter. So roll a die at the same time you flip the coin (don't show it to her). If they are (Heads, Even), let her sleep through Tuesday, as in the original experiment. If (Heads, Odd), she sleeps through Monday, and awakened Tuesday. Does anybody think her answer changes? I hope not. Call it Z.

2) The coin result only affects the question. So in addition to adding the die, flip two coins instead of one. Say, a dime and a quarter. Wake her once if they are the same, and twice if they are different; the day she sleeps is still determined by the die. Then ask her for her confidence that the coins showed the same face.

2A) If you show her the dime, and it is heads, we have variation (1). And its answer, Z.

2B) If you show her the dime, and it is is tails, we have a problem with an equivalent solution, so the same answer, Z.

2C) If you don't show her the dime, she can treat it as a random variable. The law of total probability says her answer is Z*Pr(DIME=H)+Z*Pr(DIME=T)=Z.

3) Make it simpler. Write (H,Mon), (H,Tue), (T,Mon), and (T,Tue) on four cards. Pick one at random when you flip the coin. Don't show it to Beauty, but she sleeps through the day on the card, if the coin matches the card. Ask her for her confidence in that match. This is really the same problem as (3), so the answer is, again, Z.

But this problem can be solved, because an awake Beauty does have new information. One card is ruled out, she just doesn't know which. But she can identify it with a change of variables. Instead of "Mon" and "Tue", use "Today" and "Other Day". Instead of "H" and "T", use "Up" and "Down".

Now the four cards say (Up,Today), (Up,Other), (Down,Today), and (Down,Other). She was dealt one at random, and now knows it wasn't (Up, Today). She is asked for her confidence that the card she was dealt says "Up", and that is unambiguously 1/3.

 

[stevendaryl]

Can we clarify things by having a statement of which conditional probabilities are to be computed?

In particular, what precisely are the "given" events involved in the conditional probabilities?

There are three (interdependent) boolean variables that completely describe the situation on any day:

1. heads (if heads is false, that means the result was tails)
   
2. monday (if monday is false, that means today is tuesday)
   
3. awake (if awake is false, that means she is asleep)

I would say that the number to be computed is:

P(heads | awake)

If you're willing to formulate the problem in terms of those three variables, then you can immediately write down the equation:

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

The real dispute seems to be over the conditioning on Sleeping Beauty being awake. As far as Sleeping Beauty is concerned, she never consciously experiences not being awake, so there is no reason to mention that variable at all. I think that's pretty silly. In conditional probability, there is no harm in conditioning on an always-true condition: P(X | true) = P(X). So it can't hurt anything to include the "awake" condition.

But the halfers would say not to mention that variable at all. In that case, then the formula becomes:

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

At this point, the halfers are facing a mathematical contradiction: They would say:

P(heads) = 1/2
P(heads |monday) = 1/2
P(heads | tuesday) = 0 (it's impossible)

So the formula boils down to

1/2 = 1/2 P(monday)

which implies that P(monday) = 1; it's impossible for it to be Tuesday. That's a nonsensical result. Of course, it's possible for it to be Tuesday.

 

[PeterDonis]

I would say that the number to be computed is:

P(heads | awake)

This is the one we have been focusing on, but note that unless we interpret "subjective credence that the coin came up heads" to mean this conditional probability, it is not necessarily one that we need to compute

For example, if we interpret "subjective credence" to mean which bets would or would not be taken, then we have to know the payoffs, as I said before, and we have to compute every conditional probability associated with a payoff, in order to compute a weighted expectation. Keeping the payoff definitions as general as possible, we would have

Payoff(Bet Heads) = P(Heads|Monday) P(Monday|Awake) Payoff(Bet Heads|Monday & Heads) + P(Tails|Monday) P(Monday|Awake) Payoff(Bet Heads|Monday & Tails) + P(Tails|Tuesday) P(Tuesday|Awake) Payoff(Bet Heads|Tuesday & Tails)

where no "Tuesday & Heads" term appears because that possibility is ruled out by the problem statement. Note that, because of the weighting by payoffs, this computation is not the same as the computation of P(Heads|Awake). A similar formula would apply for Payoff(Bet Tails).

 

[Stephen Tashi]

There are three (interdependent) boolean variables that completely describe the situation on any day:

Ok, but we should really start by defining the "probability spaces" that are involved.

Presumably, combinations of values of the variable define events in a probability space. But talking about value of a variable on "any day" seems to imply the events in the probability space involve days or sequences of days.