Undergrad The Sleeping Beauty Problem: Any halfers here?

[Dale]

So far nobody has shown the "halfer" solution contradicts information given in the problem.

It contradicts the stipulation that Beauty is rational by requiring her to knowingly lose money on bets.

If Beauty's credence for the coin being heads is 1/2, then she will necessarily buy a $1.00 bet that it is heads at a price of $0.40. She would expect to lose money on this bet, but would buy it anyway. That is irrational.

Which one you use, doesn't affect the betting strategy.

It had better affect it since that is the definition of credence.

 

[Stephen Tashi]

If Beauty's credence for the coin being heads is 1/2, then she will necessarily buy a $1.00 bet that it is heads at a price of $0.40.

Are you talking about a situation in the experiment? Rationally, she doesn't base her purchases of bets during the experiment on an estimate of P(Heads| awakened) (which has no unique answer that she can compute). She bases her answers in the experiment on a betting strategy, which is indeed computed using the fact that P(Heads) = 1/2.

She would expect to lose money on this bet, but would buy it anyway.

You need to define the bet you are talking about.

 

[Dale]

P(Heads| awakened) (which has no unique answer that she can compute).

This is wrong. I showed how to compute it uniquely in post 255.

computed using the fact that P(Heads) = 1/2

Yes, P(H) =1/2 is given.

You need to define the bet you are talking about.

This is completely standard usage (unlike you I am not trying to change the meaning of the terms here): A $1.00 bet that the coin landed heads, means that if the coin landed heads she gets a prize of $1.00 and if not she gets $0.00.

Such a bet, purchased at $0.40, would be expected to lose money. Yet she would necessarily (and irrationally) buy it if her credence that the coin landed heads is 1/2.

 

[Stephen Tashi]

This is completely standard usage (unlike you I am not trying to change the meaning of the terms here): A $1.00 bet that the coin landed heads, means that if the coin landed heads she gets a prize of $1.00 and if not she gets $0.00.

Then , unlike the situation in the experiment, the price paid for the the bet cannot have the consequence that if the coin doesn't land heads, she will purchase the bet twice at that price and win nothing each time.

Such a bet, purchased at $0.40, would be expected to lose money. Yet she would necessarily (and irrationally) buy it if her credence that the coin landed heads is 1/2.

If you want to talk about Sleeping Beauty's credence that the coin lands heads (without any other consequences that affect the net payout) then please define a bet to offer Sleeping Beauty that has those properties.

A rational Sleeping Beauty is not a "thirder" or a "halfer" and knows she cannot compute P(Heads | | Aawkened) because finding that probability is an ill-posed problem unless she makes some assumptions. So when the experimenter proposes any bet about whether the coin landed heads" she bases her decisions on her betting strategy - assuming the experimenter will accept her stated credence as implying a price she will pay for the bet.

If Sleeping Beauty was rational and honest (or naive), she would point out to the experimenter that no bet has been offered to her that is simply "You get $1 if the coin landed heads" without any other strings attached. However if Sleeping Beauty is rational and devious, she will not point this out to the experiment and (if the experimenter is you) the experimenter will accept her answers as indicating her credence that the coin landed heads.

 

[Dale]

Then , unlike the situation in the experiment, the price paid for the the bet cannot have the consequence that if the coin doesn't land heads, she will purchase the bet twice at that price and win nothing each time.

Why not? If she expects to make money on the bet then she should want to purchase it as often as possible. Do you have a reference that says that a person cannot rationally purchase a good bet more than once?

If you want to talk about Sleeping Beauty's credence that the coin lands heads (without any other consequences that affect the net payout) then please define a bet to offer Sleeping Beauty that has those properties.

The bet I described is exactly the one in the definition of credence. The price is below the "buy or sell" price. There is nothing in the definition restricting the number of times the bet is to be purchased or any of the other objections you raise or properties you require. You are adding those to bend the situation to your liking. Just apply the definition and you get the thirder result

she cannot compute P(Heads | | Aawkened)

False

no bet has been offered to her that is simply "You get $1 if the coin landed heads" without any other strings attached.

False

 

[Ken G]

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.

The issue is only one thing-- what is her betting strategy. You said in the case of 99 days, she should bet at 50-50 odds that it is Monday, that this is a good bet for you. I say, she will lose her shirt.

 

[Stephen Tashi]

Why not? If she expects to make money on the bet then she should want to purchase it as often as possible. Do you have a reference that says that a person cannot rationally purchase a winning bet more than once?

Look at situation in the experiment. If she offers to pay X for the bet "You get $1 if the coin lands heads" every time it is offered then, assuming P(HEADS) = 1/2, her expected expenditure on bets is (1/2)X + (1/2)2X = (3/2)X. He expected gain is (1/2)(1). So the actual expected expenditure is not X, but (3/2)X and her expected gain is 1/2. She knows the price she offers for the bet will be paid once for an even bet (on Monday) and, with probability 1/2, paid again for a sure losing bet on Tuesday. She is not paying the same price twice for the same bet.

If Sleeping Beauty actually thought the probability that the coin landed heads was 1/3, she would do the above calculation with P(HEADS) = 1/3 and conclude her expected expenditure is (1/3)X + (2/3)2X = (5/3)X and her expected gain is 1/3. So when asked for her credence she would answer 1/5.

The bet I described is exactly the one in the definition of credence. There is nothing in the definition restricting the number of times the bet is to be purchased

As pointed out above, it isn't the same bet that is purchased.

 

[Dale]

She is not paying the same price twice for the same bet.

Yes, she is. It is exactly the same bet word for word. And it could be resolved identically and immediately each time. It is exactly the same bet as defined every time she is asked her credence.

 

[Stephen Tashi]

Then why don't you answer my question-- what is her betting strategy, if she bets on it being Monday?

Asume she pays X for the bet "You get $1 if today is Monday" and must do this every time the bet is offered. The bet is offered each time she is awakened. Taking P(Heads) = 1/2, her expected net expenditure is (1/2)X + (1/2)2X = (3/2)X. Her expected gain is (1/2)1 + (1/2)(1) = 1. She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".

Of course if she uses P(Heads) = 1/3, she computes a different fair price for the bet - a wrong one, I think.

 

[Stephen Tashi]

Yes, she is. It is exactly the same bet word for word. .

"Word for word" isn't sufficient to show two bets are the same. The expected payoff needs to be the same.

 

[Dale]

She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".

As you have described it, that is in fact her credence since it gives the price she would buy or sell the defined bet.

 

[Dale]

"Word for word" isn't sufficient to show two bets are the same. The expected payoff needs to be the same.

Every bet has the same expected payoff. Procedurally they can all be done the same too. They are the same linguistically, financially, and procedurally.

 

[Stephen Tashi]

As you have described it, that is in fact her credence since it gives the price she would buy or sell the defined bet.

We're going round and round. The experimenter offers SB a bet stated as "If the coin landed heads you win $1". On Monday, this is fair bet. On Tuesday, it is a sure loser. SB knows that if she says she is wiling to pay X for the bet, she will do so on both bets whenever the second bet is offered. If you want to define a bet to measure SB's credence for the event "the coin landed heads" then please define a bet where she does not incur the possibility of having to buy a second bet with a different expected payoff.

 

[Dale]

The experimenter offers SB a bet stated as "If the coin landed heads you win $1". On Monday, this is fair bet. On Tuesday, it is a sure loser. SB knows that if she says she is wiling to pay X for the bet, she will do so on both bets whenever the second bet is offered.

Yes. So she had better factor all that into the price she is willing to pay for the bet.

If you want to define a bet to measure SB's credence for the event "the coin landed heads" then please define a bet where she does not incur the possibility of having to buy a second bet with a different payoff

The bet to measure her credence is already defined. No exceptions are listed in the definition.

Stop trying to change the definition, just use it as stated.

 

[Stephen Tashi]

Every bet has the same expected payoff.

No.

If SB were awakened at random days in random experiments, the bet "You get $1 if the coin lands heads" could be assigned a single expected payoff because the 3 situations where the bet happens are random variables chosen by "random sampling with replacement"

However, in computing the betting strategy, the systematic plan of the experiment is considered. After the coin toss, the situations that arise are determined, not random.

The bet to measure her credence is already defined. No exceptions are listed in the definition.

Here's a bet: "You get 1$ if the coin lands heads - and we toss another coin and you lose $10 if it lands heads".

Do you really think the fair price for that bet is the same as the fair price for "You get $1 if the coin lands heads"?

 

[andrewkirk]

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?

thanks for that. From the Stanford article we get the following definition, which I'm reproducing here so everybody can see it without having to follow links:

Your degree of belief [credence] in E is p iff p units of utility is the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E.

With that in place, we now need to know exactly what SB is betting on. In particular, we need to know whether she knows, before the experiment begins, that she will be offered a bet every day or, if not, on what basis it will be determined whether she will be offered a bet. If she does not know that, and remember it each time she wakes up, none of the analysis I've read so far about betting applies.

Let's assume that she does know, and always remembers, that each day she will be offered the opportunity to pay $x$ dollars for a chance to name Heads or Tails and then immediately win $1 if what she named was correct, and also that $x$ will be the same every day! Then it's a simple exercise in expected values of net payoffs, as per my previous post.

The expected payoff, over the course of the entire experiment, of a strategy that always guesses Heads (recalling that Heads is associated with only being woken on Monday, on the wiki presentation) is (giving the probability for the case where the coin was actually Heads in the first term):
$$0.5\times (1-x)+ 0.5\times 2 \times(0-x)=0.5-1.5x$$

So, for the bet to be worth taking, it must have an expected payoff greater than 0.

For Heads, this means that $0.5-1.5x>0$ so that $x<1/3$

If SB is offered the Heads bet with $x=0.33$, and remembers the full details of the betting program, including that she'll be offered the same bet, at the same price, every day, she will take the bet.

So on that definition, and with those very specific rules about betting, (which were not stated in the original problem statement), we might say that the credence in the proposition that the coin came up Heads is one third.

A different conclusion is reached if SB has not been told that she will be offered a bet every day. Then she cannot do the above calculation. Her expected payoff will be the expected value
$$0.5\times (1-x)\times P_1+ 0.5\times (0-x)\times P_2+ 0.5\times (0-x)\times P_2P_3+ 0.5\times (0-x)\times (1-P_2)P_4
$$

where $P_1$ is the probability of being offered a bet on Monday, given that the coin landed Heads,
$P_2$ is the probability of being offered a bet on Monday, given that the coin landed Tails, and
$P_3$ is the probability of being offered a bet on Tuesday given that a bet was offered on Monday, and that the coin landed Tails.
$P_4$ is the probability of being offered a bet on Tuesday given that a bet was NOT offered on Monday, and that the coin landed Tails.

This is positive if $x<\frac{P_1}{P_1+P_2+P_2P_3+(1-P_2)P_4}$. SB would make an assumption for these three probabilities in order to decide whether to bet. One possible assumption is that the probability of being offered a bet on any day is the same as any other day and independent of whether bets have been offered on other days. If so, the credence is 1/3 again.

However, an equally plausible is the assumption that $P_1=P_2=P_4$ but that only one bet will be offered, so that a bet will not be offered on both Monday and Tuesday. In that case $P_3=0$ and the credence is
$$\frac{P_1}{3P_1-P_1{}^2}=\frac1{3-P_1}$$
which will be somewhere between a third and a half.

So it depends on SB's knowledge of what betting opportunities will be made available to her.

 

[Stephen Tashi]

we might say that the credence in the proposition that the coin came up Heads is one third.

That's a subject of debate in recent posts. SB can calculate her betting strategy using P= 1/2 and never worry about calculating P(heads | awakened). In fact P(heads|awakened) has no unique value and is irrelevant to planning a betting strategy. So SB can determine what she ought pay for the bet "The coin lands heads" without computing a posterior probability that the coin lands heads. Her fair price for the bet is dishonest as a report of her credence because, according to the definition of "credence", she should imagine a bet where "You get $1 if the coin lands heads" is bet with no strings attached. However, in computing her betting strategy, as you illustrated, she considers that it might obligate her pay price X for another bet on Tuesday that is worded the same, but is a sure loser.

 

[Ken G]

Asume she pays X for the bet "You get $1 if today is Monday" and must do this every time the bet is offered. The bet is offered each time she is awakened. Taking P(Heads) = 1/2, her expected net expenditure is (1/2)X + (1/2)2X = (3/2)X. Her expected gain is (1/2)1 + (1/2)(1) = 1. She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".

Of course if she uses P(Heads) = 1/3, she computes a different fair price for the bet - a wrong one, I think.

But you are mixing up the probability of heads in one way of framing the calculation, with her credence, which can be shown to be quite different. Most obvious is the 99 day version. When you say P(Heads)=1/2, you are talking about the probability that heads was flipped originally, which is of course indeed 1/2 (but that's not her credence, as we shall see). So in a single 99-day experiment, her expected expenditure is (1/2)X+ (1/2)99X = 50X. Her expected gain is 1, as you say. Ergo, she should state her credence that it is Monday as 1/50, as X=1/50 is the fair bet that it's Monday-- that is what you would get and that's correct. The problem is, the use of P=1/2 in that calculation is simply not her credence that heads was flipped, that's the probability of a heads flip taking the experiment from a kind of disinterested external perspective, analyzing the situation from the start (perhaps, from the perspective of the experimenters trying to figure out if they will bilk money out of SB). What you are not seeing yet is that if she has a credence of 1/50 that it is Monday (which we have agreed is true), and the only possible way it could be any other day is if the coin flip was tails, this clearly implies she strongly suspects the coin flip was tails, every time she is awakened. That simple truth is quite independent from the use of P(heads) = 1/2 in the above calculation.

To see the calculation from SB's perspective, it looks like this. Her credence that it is Monday is 1/50, and if it is Monday, her credence that the coin was heads is 1/2. That contributes 1/100 to her credence that the coin is heads. Her credence that it is any other day is 49/50, and if it is any other day, her credence that the coin was heads is 0. Hence her total credence that the coin was heads is 1/100, not 1/2. Furthermore, she can correctly determine her expected payoff on her bet that it is Monday by taking X = 1/50. So in the long run, for the 99 day version, she can break even if she always pays 1/50 for the bet "today is Monday" at $1 payoff, and she can also break even if she always pays 1/100 for the bet "the coin was heads" at $1 payoff. If she pays 1/2 for that latter bet, she is a chump, because her expected payoff is (1/2)*1, but her expected cost is 1/2*1 + 1/2*99 = 50. She's losing $49 every time she accepts that strategy, this is an error. So notice that even though the 1/2 appears in the calculation I just did, that's the probability of heads taken from the start of the experiment, it is not her credence on the bet "it was a heads." That's clear because if she thinks it's her credence, she gets bilked, and that proves it is not the correct credence. You should see the contradiction in thinking her credence that it is Monday is 1/50, but her credence that the coin was heads is 1/2.

 

[Ken G]

So it depends on SB's knowledge of what betting opportunities will be made available to her.

But that's obvious, it would be like playing poker and not being sure the dealer isn't dealing from the bottom of the deck. Of course that will mess up your chances, but that's just clearly not the way poker is played, it would be a cheat. SB only needs to know she is not being cheated, just as any poker player must assume, it goes without saying. This is also why the correct answer to the question is not "it depends on the precise formulation," because if you are playing bridge and you want to know the chances of dropping the queen if the defense holds 4 cards, you can look up that credence in a bridge book, and I guarantee you that you will never see "the answer depends on whether or not the dealer was cheating." SB must be told what the betting strategy is, and it must be the truth, and if that betting strategy is that she will be offered the bet every time she wakes, her credence is 1/3-- it's not an assumption, anything else would be a cheat.

 

[andrewkirk]

SB only needs to know she is not being cheated, just as any poker player must assume.

only needs to know that, in order to reach what conclusion?

If the conclusion is to be that she is willing to pay 33c for a bet that pays $1 if the coin landed Heads and zero otherwise, she needs more information than that - specifically, what are the probabilities of me being offered the same bet on every other day.

I'm afraid I know nothing about poker, so I can't really assess the extent to which the rest of your post relates to the SB situation.

 

[Stephen Tashi]

But you are mixing up the probability of heads in one way of framing the calculation, with her credence, which can be shown to be quite different.
.

Before we get into the details, let me clarify my viewpoint. My viewpoint is that the posterior probability of heads P(heads | SB is awakened) cannot be calculated from the information in the problem. Asking for it's value is an ill-posed problem and is irrelevant for planning a betting strategy. As you pointed out, planning a betting strategy should use P(Heads = 1/2)., not P(Heads|wake). The value P(heads)= 1/2 is used regardless of whether one is a "halfer" or "thirder" or agnostic about P(heads|SB awakened).

In my view, when SB answers "What is your credence that the coin landed heads?", she gives a dishonest answer if she reports the number based on her betting strategy. However, many posters in this thread are willing to accept her report as her credence that the coin landed heads, so they don't object to it. Her report is dishonest because she is not considering a simple bet "You get $1 if the coin lands heads". Instead, when planning her strategy, she considering that paying a price X for the bet obligates her to pay that price every time the bet is offered. Two different versions of the bet might offered. On Monday, it is an even bet. On Tuesday (if offered) the bet is a sure loser.

Since calculating P(Heads| awake) is an ill-posed problem, SB cannot offer a "credence" for the event (Heads | awake) unless she makes some assumptions. She does not need to calculate P(Heads| awake) unless the experimenter is stickler and can propose a bet on "You get $1 if the coin lands heads" that is a "pure" bet - i.e. a bet with no conditions that she might have to buy another bet with different expected payoff. (I myself haven't been able to formulate a "pure" bet. that could be offered to SB during the experiment.)

From my reading, the general opinion of those who have studied the SB problem is that one cannot distinguish between the "halfer" and "thirder" positions by bets that can be offered during the experiment, provided we assume SB is rational and plans a betting strategy that is independent of her opinion of P(Heads|awake).

Both "thirders" and "halfers" are incorrect to assert that their answer for P(Heads|awake) is the unique correct answer. Both the "thirder" and the "halfers" are correct that that one can create a probability model that is consistent with their answer for P(Heads|awake) and does not contradict the information given in th SB problem. The computation of a betting strategy is independent of such a probability model.

 

[Stephen Tashi]

So it depends on SB's knowledge of what betting opportunities will be made available to her.

Most posters in this thread ( including myself) assume that when she is awakened SB knows the how the experiment is conducted and that she will be asked about her credence for "the coin landed heads" every time she is awakened - but you're right, the Wikipedia statement of the problem doesn't make this crystal clear. It does say that the amnesia drug makes her forget "her previous awakening". It doesn't say she forgets other facts.

 

[Ken G]

Before we get into the details, let me clarify my viewpoint. My viewpoint is that the posterior probability of heads P(heads | SB is awakened) cannot be calculated from the information in the problem. Asking for it's value is an ill-posed problem and is irrelevant for planning a betting strategy. As you pointed out, planning a betting strategy should use P(Heads = 1/2)., not P(Heads|wake). The vale P(heads)= 1/2 is used regardless of whether one is a "halfer" or "thirder" or agnostic about P(heads|SB awakened).

Yes, it seems that the whole concept of "probability" is creating problems, but the correct betting odds should be clear enough-- 1/3 heads is the break-even betting strategy, and that's all "credence" means.

In my view, when SB answers "What is your credence that the coin landed heads?", she gives a dishonest answer if she reports the number based on her betting strategy.

Let us simply define credence by her fair betting strategy, that's really all the concept is supposed to entail.

On Monday, it is an even bet. On Tuesday (if offered) the bet is a sure loser.

Yet that is always true. If you are playing bridge, and you analyze a finesse as having a 50-50 chance, of course in one deal, it's a sure bet, and another, it's a sure loser. But that's just not what credence means, credence is not some kind of "actual probability", it is simply the correct betting strategy given the information you have. It makes no difference if she takes the bet every time or not, the correct betting strategy is always 1/3 heads because she never gets any additional information in one trial versus another. You don't have to finesse every time you get a certain hand in bridge-- yet the odds are still 50-50 if that's all you know. I suspect the problem in this discussion is the halfers are seeking some kind of "actual probability" of heads, and indeed we both just did a calculation using P(heads)=1/2, but that isn't what credence is-- credence is simply the fair bet, any time the bet is made given no information beyond what is supplied in the experiment.

From my reading, the general opinion of those who have studied the SB problem is that one cannot distinguish between the "halfer" and "thirder" positions by bets that can be offered during the experiment, provided we assume SB is rational and plans a betting strategy that is independent of her opinion of P(Heads|awake).

This is what is false. There is clearly a single break-even betting strategy in the experiment as described, and it is clearly 1/3 heads. The 99 day version makes this very clear-- you already agreed the fair bet in that experiment is X=1/50 for "today is Monday", so how can the fair bet be 1/2 heads when heads can only pay off on Monday?

 

[Ken G]

only needs to know that, in order to reach what conclusion?

In order to reach the conclusions that the fair bet is 1/3 "the coin was heads," and the fair bet is 2/3 "today is Monday."

If the conclusion is to be that she is willing to pay 33c for a bet that pays $1 if the coin landed Heads and zero otherwise, she needs more information than that - specifically, what are the probabilities of me being offered the same bet on every other day.

But the problem does clearly specify this. Using the Wiki version:
"Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends."

So there it is, SB is interviewed every time, so her credence is about a bet she is given every time. It's clear as a bell, no additional assumptions needed. Of course the experimenter cannot couple the offering of the bet to the outcome of the coin, that would clearly be cheating and would do violence with the entire concept of credence. If credence were dependent on the uncertain and dishonest wiles of the experimenter, then you are saying the odds of winning any game of chance (the details of poker are of course irrelevant) cannot be known unless you know you are not being cheated. That goes without saying, it is simply what "odds" mean. If you assess the chances of a sports team winning, you never say "assuming none of the players are throwing the game intentionally," because that also goes without saying in the concept of odds.

 

[Stephen Tashi]

you already agreed the fair bet in that experiment is X=1/50 for "today is Monday", so how can the fair bet be 1/2 heads when heads can only pay off on Monday?

I'm not sure which experiment you're talking about. Which post was it?

 

[andrewkirk]

The experiment already stipulates that SB is given the same bet every time, that's what credence means

Not according to the SEP definition. There is no mention of 'every time' in that definition, because it only envisages one bet. This situation has the potential for two bets, where the number of times the bet is taken is correlated with the result of the bet. That is a completely different situation from the one envisaged in the SEP definition, which implicitly assumes that there is only one bet. Hence there is room for multiple different interpretations as to how the single-bet SEP definition might be generalised to apply to this case.

For convenience, here is the SEP definition again:

Your degree of belief [credence] in E is p iff p units of utility is the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E.

 

[stevendaryl]

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 would say that's a conclusion, not an assumption. If you repeat the experiment many, many times, for many, many weeks, then, letting N be the number of weeks, then on the average:

• There will be N/2 times when Sleeping Beauty is awake and it's Monday and it's heads.
• There will be N/2 times when SB is awake and it's Monday and it's tails.
• There will be N/2 times when SB is awake and it's Tuesday and it's tails.
• There will be N/2 times when SB is asleep and it's Tuesday and it's heads.

So the three possibilities: (Monday, Heads), (Monday, Tails), (Tuesday, Tails) occur equally frequently. So if Sleeping Beauty's credence for each possibility is the same as the relative frequency of that possibility in repeated experiments, then she has to give equal likelihood to those three possibilities.

That's the relative frequency argument: the relative frequency of heads among the events in which she is awake is 1/3.

 

[stevendaryl]

To me, the halfer position has a very implausible consequence. There are three possibilities given that Sleeping Beauty is awake:

1. (Heads, Monday)
2. (Tails, Monday)
3. (Tails, Tuesday)

The subjective probabilities of the three, given that she is awake, and given that today is either Monday or Tuesday, must equal 1. So

P(H & Monday | Awake) + P(T & Monday | Awake) + P(T & Tuesday | Awake) = 1

So if P(H & Monday | Awake) = 1/2, and P(T & Tuesday | Awake) > 0, then it follows that

P(H & Monday | Awake) > P(T & Monday | Awake)

This in turn implies:

P(H & Monday) > P(T & Monday)

which implies:

P(H | Monday) > P(T | Monday)

This is what's bizarre to me about the halfer position. They want to say that P(H | Awake) = P(H), because being awake doesn't tell you anything new about whether it's heads or not. But the consequence of that position is that P(H | Monday) > P(T | Monday). If being awake tells you nothing about whether it's heads or tails, why would you say that it being Monday tells you something? How can the fact that it's Monday make heads more likely than tails?

What's especially bizarre about this is that, as far as the thought experiment goes, it doesn't make any difference whether you flip the coin on Monday morning or on Tuesday morning, since it's not necessary to consult the result until Tuesday morning. So in the case where the coin flip is on Tuesday morning, the halfer position implies that:

• If Sleeping Beauty is told that today is Monday, and that tomorrow morning, a coin will be flipped, she will say that it is more likely that the result will be heads than tails.

That's completely bizarre.

 

[Dale]

we need to know whether she knows, before the experiment begins, that she will be offered a bet every day

The bet is implied by the definition of credence. So it is every time she is asked about her credence, which is every interview.

So on that definition, and with those very specific rules about betting, (which were not stated in the original problem statement)

They were stated in the original problem. Beauty is asked about her credence in each interview so the implied bet is necessarily offered each interview also.

 

[stevendaryl]

Not according to the SEP definition. There is no mention of 'every time' in that definition, because it only envisages one bet. This situation has the potential for two bets, where the number of times the bet is taken is correlated with the result of the bet. That is a completely different situation from the one envisaged in the SEP definition, which implicitly assumes that there is only one bet. Hence there is room for multiple different interpretations as to how the single-bet SEP definition might be generalised to apply to this case.

I'm not so sure that it matters. It is true that the number of opportunities to bet depends on the coin result. However, each individual opportunity to bet seems to fit the definition.