The Sleeping Beauty Problem: Any halfers here?

[Dale]

I am a halfer.

Then every time you wake I will sell you the following bet for $0.40: a bet that pays $1 if the coin landed heads, $0 if the coin did not land heads. How much money do you expect to make?

 

[Ken G]

I'm saying that calculating P(heads| SB) awakened from the information in the problem objectively requires knowing (or deducing) a particular probabiiity distribution F on the situations { (heads, Monday, awakened) , (tails, Monday, awakened), (tails, Tuesday, awakened)}. The distribution F defines the "method of picking" the situation when SB is awakened. The distribution F implies a particular value for P(heads | SB is awakened) by using objective calculations for conditional probabilities.

Well that is certainly true, but surely the question asserts a simple procedure that determines F. The experimenters flip a coin, and if they get heads, they wake up SB only on Monday. If tails, they wake her Monday and Tuesday. Any time they wake her, they allow her to bet at some given odds that the day is Monday. They also apply amnesia elixir so she cannot remember if she has been wakened before. That's it, that's all you need-- with that situation, is it not perfectly clear she will make money, after many identically repeated experiments, if she accepts any odds more favorable to her than a 2/3 chance it is Monday, and will lose for any odds less favorable than that?

Now of course, the experimenters can cheat in any way they like. They can simply lie to her. But if the setup is what is given here, and the experimenters follow it, then there is no question how she makes money. We can do the same thing with repeated bets on "heads" or "tails," but I'm trying to make sure the two problems are correctly coupled. In other words, no one can self-consistently claim that the odds of "heads" are 50-50, without also claiming it is definitely Monday, because it should be clear that only on Mondays are those the correct odds of a "heads," given nothing but the setup I just described.

 

[Buzz Bloom]

How much money do you expect to make?

Where are you getting that from?

Hi Dale and steven:

I had a change of heart and am now a thirder. Please see my post # 498.

Regards,
Buzz

 

[Stephen Tashi]

Well that is certainly true, but surely the question asserts a simple procedure that determines F. The experimenters flip a coin, and if they get heads, they wake up SB only on Monday. If tails, they wake her Monday and Tuesday.

That correctly describes the situations can possibly be selected. However, it doesn't describe a probability distribution F telling how to determining which situation applies "when Sleeping Beauty is awakened" - unless you assume that each possible situation has an equal probability of being the one that applies. I agree that a typical Bayesian approach is to make such an assumption.

A "halfer" method of picking which situation applies when Sleeping Beauty is awakened is: Flip the coin and run the experiment. From those situations that arise the experiment after the coin flip, pick one of the situations at random, giving each situation an equal probability of being selected.

"Thirders" have various objections to the "halfer" method, such as "You aren't forcing each situation in an experiment to be selected when coin lands Tails. You might pick Monday and omit Tuesday". However, any random method of picking a single situation might omit one of the situations in a series of 3 independent selections.

 

[Stephen Tashi]

Hi Dale and steven:

I had a change of heart and am now a thirder. Please see my post # 498.

The "thirders" have a valid position if certain plausible assumptions are made.

 

[Ken G]

That correctly describes the situations can possibly be selected. However, it doesn't describe a probability distribution F telling how to determining which situation applies "when Sleeping Beauty is awakened" - unless you assume that each possible situation has an equal probability of being the one that applies.

The only probability distribution involved in waking her up is described by the coin flip, there is no ambiguity, no assumptions-- it's all spelled out in the instructions.

A "halfer" method of picking which situation applies when Sleeping Beauty is awakened is: Flip the coin and run the experiment. From those situations that arise the experiment after the coin flip, pick one of the situations at random, giving each situation an equal probability of being selected.

Each situation does have equal probability, that's what gives the thirder result. There are three situations, correct?

 

[Dale]

I observed before that IMO the Sleeping Beauty problem is not a good illustration of the concept of "credence",

I agree. It is not a good illustration. You have to apply the definition of credence in this problem, but I don't think many people get the "now I get it" feeling from the problem.

 

[Dale]

The definition of credence I quoted assumes a bet where the payoff from the event E is 1 unit of utility.and an buying the bet at price X doesn't affect the payoff.

That is correct. That is the definition of credence and that bet is the one that the halfer solution gets wrong. So do you now understand that 1/2 is not a valid solution to the problem?

 

[andrewkirk]

On my analysis, the Thirder argument has the flaw that it assumes each of the two wakings in the Tails situation have the same probability as the waking in the Heads situation. That is not the case.

The probability of being woken on Monday in the Heads situation is just the probability of Heads, which is 0.5.

The probability of being woken on Monday in the Tails situation is 1/2 the probability of Tails, which is 0.5 times 0.5 = 0.25. The same goes for Tuesday-Tails.

So we have three potential wakings, with probabilities 0.5, 0.25, 0.25. For the latter two, the coin is Tails, for the first one, it is Heads. When SB is woken, she just uses that information to work out the probability that the coin was Heads, which is 0.5.

The error in the Thirder position is to assume that the three probabilities are the same. There is no mathematical rule to justify that.

I can see why Nick Bostrom argues for the Thirder position, because he uses the same logic to argue that, if simulation sophisticated enough to produce consciousness in the simulants is possible, then we are almost certainly in a simulation now, because the simulators will create very many simulations, all equally likely to each other and to the possibility that we are not in a simulation. He omits from that calculation the probability of such technology being developed, just as the Thirder solution omits the probability of Tails.

We can use the same Bostrom argument against the Thirder position by noting that there are at least a thousand different sub-scenarios for the Heads-Monday option, being differentiated by which temperature band of the thousand bands of width one-millionth of a degree, centred around standard body temperature, SB's body temperature on waking in the Heads case falls into. We now have one thousand equally likely scenarios for the Heads case and only two for the Tails case (we don't do temperature binning for that case), so the probability of Heads is 1000/1002, a virtual certainty.

PS: I've only read the first page of this thread.

PPS: Strictly speaking, a probability is simply a measure on a set of possible outcomes where the measure of the entire set is 1. Subject to that restriction, we are free to define the measure however we wish, which means assigning any probabilities we wish, as long as they don't contradict the criteria for a valid measure.

So there's no such thing as a 'correct' probability.

 

[Ken G]

On my analysis, the Thirder argument has the flaw that it assumes each of the two wakings in the Tails situation have the same probability as the waking in the Heads situation. That is not the case.

It suffices to count the events after the experiment is repeated N>>1 times. There will be N wakings on Monday and N/2 wakings on Tuesday. SB knows only, in each waking, that she samples equally from both those sets, so she samples in each instance from 3N/2 wakenings, and on N of them, it is Monday. This suffices to tell her that there is a 2/3 chance it is Monday, or if you prefer, she will break even in the long run by taking 3 to 2 odds. Do you say that is not the case?

So we have three potential wakings, with probabilities 0.5, 0.25, 0.25.

So you are now claiming that there are the same number of wakings on Monday as on Tuesday? You appear to be arguing that SB should think there is a 50% chance it is Monday. Of course that's not true, as can be seen if we do it for 99 days, not just 2. On heads, we awake her only on day 1, on tails, we awake her every day for 99 days. Do you say she should assess a 50% chance that it is Monday, based on the argument you just gave?

 

[Stephen Tashi]

An answer can always be changed by doing something different.

You assume that there is something there that we can be different from.

You could change the odds of poker by assuming every hand has equal probability, but that's no way to win poker.

I agree that many books pose poker problems and omit to explicitly state that all deals are to be regarded as equally probable. A traditional interpretation is for the reader to assume the author of the problem means that to be the case. However, it's not traditional (except perhaps among ultra-Bayesians) that the existence of N possibilities in an arbitrary probability word problem can be taken to imply that each possibility has probability 1/N.

As I said before a person can take the position that the author of the Sleeping Beauty problem intended to say (or for us to assume) that the situation "When Sleeping Beauty is awakened" has an equal probability of being any of the 3 possible situations. Or a person can take the position that we are permitted to use a Bayesian approach and explicitly assume each situation has as equal probability.

To design an empirical test to decide whether the "halfer" or "thirder" or some other answer is correct, the problem gives enough information to simulate many runs of the experiment. In those runs, the events C =(tails, Monday, awake) and D = (tails, Tuesday, awake) are not mutually exclusive events. In fact P(C|D) = 1. Both events occur in the experiment when the coin lands tails.

However, when we contemplate how to implement a distribution F to simulate the event "Sleeping Beauty is awakened", we must have a procedure that stochastically selects a single situation when she is awakened. In implementing that procedure A and B are mutually exclusive events. This makes it clear that simulating the situation when Sleeping Beauty is awakened does not treat events in the same way as the probability space for the experiment, where A an B are not mutually exclusive events.

I think you take it for granted that the way to to implement F would be to pick a situation at random from those that occurred in the runs of the trials, giving each the same probability. Or we could step through each situation that did occur and compute some frequencies, which , I think, amounts to the same assumption. I agree these are reasonable approaches. However, they aren't specified in the statement of the problem.

 

[Ken G]

You assume that there is something there that we can be different from.

All halfers should answer this question. If the experiment is done for 99 days, and on heads, there is only an awakening on Monday, and on tails, there are 99 days of awakening, what should SB assess as her expectation that it is Monday? Can you possibly think there is any reasonable interpretation of that scenario where she does not expect it is vastly more likely that the day is not Monday? Yet if the day is not Monday, then the toss was a tails.

 

[andrewkirk]

SB knows only, in each waking, that she samples equally from both those sets, This suffices to tell her that there is a 2/3 chance it is Monday. Do you say that is not the case?

I would say there is a 0.75=0.5+0.25 chance it is Monday, being the sum of the probs for Heads-Monday and Tails-Monday.

If Tails leads to 99 extra awakenings rather than one, then the prob of Heads-Monday is 0.5 and that of Tails & <day =n>, where 1<=n<=100, is 0.5/100=0.005. So in that case the probability that it is Monday is 0.5 + 0.005= 0.505.

 

[Stephen Tashi]

So you are now claiming that there are the same number of wakings on Monday as on Tuesday?

I think you are claiming that "numbers of things" has a specific relation to probabilities, which it does when each thing is equally probable. If the 3 possible situations A = (heads, Monday, awake), B = (tails, Monday, awake), C = (tails, Tuesday, awake) have the same probability then you have a point. But must they have the same probability?

 

[Stephen Tashi]

All halfers should answer this question. If the experiment is done for 99 days, and on heads, there is only an awakening on Monday, and on tails, there are 99 days of awakening, what should SB assess as her expectation that it is Monday? .

I think the answer is ( (1/2)(1) + (1/2)(1/99) , which may offend people's intuition, but it doesn't contradict any information given in the problem.

Let me make it clear that I'm not a "halfer". My vote in the poll was for:
It depends on the precise formulation of the problem

 

[eloheim]

First let me apologize. I basically never leave a comment on a thread I haven't read all the way through. This time I'm going to make an exception to ask about my reasoning on the problem. I'm sorry, I read the first several pages thoroughly but 27 is just not in the cards for me right now, so please anyone feel free to ignore me if you want and I will not be offended.

Looking at the original formulation of the problem, as on the Wikipedia page. Say we have two princesses, one's a thirder and one's a halfer. Say both are put through the experiment many times (independently). Also say we award each princess a dollar for each time they correctly guess the coins toss result for that week.

The halfer thinks both heads and tails are equally likely, so she might as well guess heads every day, and it should make no difference from guessing tails, so that's what she does (always guess heads). The thirder, on the other hand, believes tails is twice as likely as heads, and so guesses tails every time. At the end of many iterations of the experiment, isn't it pretty clear that the thirder is going to have a lot more dollars than the halfer?

Is this a good argument that the thirders are right?

 

[andrewkirk]

At the end of many iterations of the experiment, isn't it pretty clear that the thirder is going to have a lot more dollars than the halfer?

Yes, but that's an expected value, not a probability, and the question is about a probability.

The situation you describe is one in which the probability of Heads is 0.5 and the Princess gets paid $1 over the course of the whole experiment if she correctly guesses Heads and $2 over the course of the whole experiment if she correctly guesses Tails. Hence the expected payoff for guessing heads is 50c and for guessing Tails it is $1. But those are not probabilities.

 

[Dale]

I think the answer is ( (1/2)(1) + (1/2)(1/99) , which may offend people's intuition, but it doesn't contradict any information given in the problem.

That would lead to some really bad betting

 

[Ken G]

Yes, it would offend my intuition to think I have a 50-50 chance of winning a poker hand with a pair of twos, but more to the point, I'd lose my shirt. So it's not about intuition, it's about payoff. There is no need to even invoke a concept of probability, one only needs to know the betting odds, which is about expected payoffs.

 

[eloheim]

Yes, but that's an expected value, not a probability, and the question is about a probability.

Hmm..I'll have to do a little reading about definitions in probability theory I think to fully understand this. Thanks for the reply.

 

[Stephen Tashi]

Yes, it would offend my intuition to think I have a 50-50 chance of winning a poker hand with a pair of twos, but more to the point, I'd lose my shirt. So it's not about intuition, it's about payoff. There is no need to even invoke a concept of probability, one only needs to know the betting odds, which is about expected payoffs.

You make an analogy to a different situation which does have as set of equilprobable outcomes. A person who has an intuitive objection the the "halfer"'s answers can reject the "halfer' model for the distribution of F and assume a different model. My only point is that assumptions are being made.

Beauty' can determine her betting strategy without making any assumption that the probability of heads changes to 1/3 after she awakens. She can compute her betting strategy using the assumption that the probabiiity of heads is 1/2 and using her knowledge of the conduct of the experiment. That computation does not imply that she should give even odds to (always) betting about what day of the week it is.

 

[Dale]

Beauty' can determine her betting strategy without making any assumption that the probability of heads changes to 1/3 after she awakens.

The betting strategy and the credence are linked by definition (specifically the definition of credence).

 

[Stephen Tashi]

The halfer thinks both heads and tails are equally likely, so she might as well guess heads every day, and it should make no difference from guessing tails, so that's what she does (always guess heads).

That isn't true because a "betting strategy" is more than a simple estimate of a probability. Using the fact that the probability of heads is 1/2, Sleeping Beauty would arrive at a betting strategy that did not give even odds to the coin landing heads. This because if it lands tails, she is required to make her bet on Monday and Tuesday and is thus penalized twice, but if it lands heads, she only wins once.

 

[Stephen Tashi]

The betting strategy and the credence are linked by definition (specifically the definition of credence).

The definition of credence describes buying a single bet - see the definition of credence.

 

[Dale]

The definition of credence describes buying a single bet - see the definition of credence.

Yes. Exactly. It is a single bet purchased each time she determines her credence.

 

[Stephen Tashi]

Yes. Exactly. It is a single bet purchased each time she determines her credence.

And since she knows she might have to buy the bet twice, she is purchasing an agreement that includes the possibility of being forced to make two bets.

 

[Dale]

And since she knows she might have to buy the bet twice, she is purchasing an agreement that includes the possibility of being forced to make two bets.

More like she is rational so given the same information she will make the same bet multiple times. That is also stipulated in the problem setup.

Do you now agree that the halfer solution is incompatible with the problem as stated, in particular with the definition of credence and the rationality of Beauty.

 

[andrewkirk]

@Dale Is there a definition of credence that has been agreed somewhere in the 537 posts? With a quick skim, all I could see was a link to this Wikipedia article on credence, which does not give a formal definition, instead just offering a few vague sentences and a single example, none of which seem to help much with its interpretation in this context.

Thanks

Andrew

 

[Dale]

Is there a definition of credence that has been agreed somewhere in the 537 posts?

In post 384 I have several links I found useful. The second one has a brief definition that we have seemed to settle on.

The Sleeping Beauty Problem: Any halfers here?

 

[Stephen Tashi]

Do you now agree that the halfer solution is incompatible with the problem as stated, in particular with the definition of credence and the rationality of Beauty.

No, why would I? (Goody, a chance to recapitulate.) So far nobody has shown the "halfer" solution contradicts information given in the problem. People have said they don't like the implications of the halfer solution - well, that's not a mathematical argument. Further, the halfer solution has nothing to do with a betting strategy. A betting strategy is computed using the value 1/2 for the probability of heads and the specification of the experiment - which says that a bet is offered each time Sleeping Beauty is awakened - not at randomly selected times that Sleeping Beauty awakens. A person's opinion about the probability distribution that describes the state of the experiment at a random awakening is irrelevant to computing a betting strategy. Both the "thirder" and "halfer" solutions for P(H|awake) are solutions. Which one you use, doesn't affect the betting strategy.

Since Sleeping Beauty can (rationally) determine a betting strategy by using the value P(Heads) = 1/2 and her knowledge of the conduct of the experiment, it is clear that her answer for "credence" is not a bet just on the value of P(Heads | Sleeping Beauty is awakened). She picks a strategy based on "what I ought say" given that she may have to say it twice - unless she takes the question "What is your credence that the coin landed heads?" literally. If she takes the question literally, she must be given a bet on that event alone, not a bet that involves that event plus some other consequences that affect the net payout.