The Sleeping Beauty Problem: Any halfers here?

[PeroK]

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

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

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

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

If not, please justify another answer.

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

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

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

 

[stevendaryl]

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

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

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

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

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

 

[Demystifier]

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

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

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

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

Isn't this equivalent to my post #67?

 

[PeroK]

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

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

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

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

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

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

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

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

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

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

 

[PeroK]

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

Is that the case without the amnesia drug as well?

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

 

[stevendaryl]

Isn't this equivalent to my post #67?

Yes. I posted before reading your post.

 

[Khashishi]

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

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

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

 

[PeroK]

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

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

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

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

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

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

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

 

[PeroK]

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

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

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

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

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

 

[stevendaryl]

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

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

 

[forcefield]

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

 

[PeroK]

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

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

 

[forcefield]

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

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

 

[lewando]

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

 

[Demystifier]

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

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

 

[PeroK]

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

You were the founder member!

 

[Demystifier]

You were the founder member!

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

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

 

[PeroK]

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

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

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

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

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

 

[Demystifier]

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

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

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

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

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

The best disproof of 1/2 so far.

 

[Marana]

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

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

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

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

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

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

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

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

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

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

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

 

[PeroK]

@Marana If you were the sleeper, please tell me what you would answer to these three questions:

You wake up:

What is the probability that the coin is heads?

If it's Monday, what would be your answer?

If it's Tuesday, what would be your answer?

 

[PeroK]

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

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

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

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

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

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

So, in effect, in your solution Tuesday and everything except the coin toss is irrelevant? Nothing that happens in the experiment can change the probability of 1/2?

You might as well just wake the sleeper on the Monday, tell her it's Monday and ask her. She says 1/2. Nothing else in the experiment makes any difference to this?

The coin is tossed. It's 50-50 heads or.tails. End of.

PS this is now the psychological issue I mentioned in an earlier post. Your a priori conviction that the answer of 1/2 must be correct is so strong that simple questions such as "is it Monday?" become invalid. Anything that disproves the a priori answer must be invalid.

 

[Demystifier]

when she wakes up it is not a random experiment.

Probability is not only about randomness, but also about absence of knowledge. Suppose that I pick one of the letters A or B, by will. Then I ask you, what is the probability that I picked A? What is your answer?

 

[PeroK]

Probability is not only about randomness, but also about absence of knowledge. Suppose that I pick one of the letters A or B, by will. Then I ask you, what is the probability that I picked A? What is your answer?

Or, someone has two children. The probability of two boys is 1/4.

If they have two girls then they come to see you on a Monday; otherwise, they come to see you on a Tuesday. Nothing random. Yet, if they come to see you on a Tuesday, the probability of two boys has increased to 1/3.

 

[stevendaryl]

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

Here's the way that I became a thirder, which I think is convincing (even if it is much more work than the original, one-line argument for 2/3 or 1/2).

Imagine that experimenters are doing this experimenter over and over, with lots of different test subjects (sleeping beauties). At any given moment, there will be some group of people who are either experiencing Day 1 or Day2 (rather than Monday and Tuesday, because the subjects may start the experiment on different days). Each such person has an associated coin flip result. So each subject has an associated label (known to the researcher, but not the subject): $(D,C)$ where $D$ is the day number, either 1 or 2, and $C$ is the coin flip result, either $H$ or $T$.

The researchers provide you with a list of all current subjects, and you pick one at random.

Here, there is a technical question about what it means to be in Day 2, and what it means to pick a subject at random.

First interpretation: You can only be experiencing Day 2 if your coin flip result was "heads". So your sample should include only those whose labels are:

• $(D=1, C=H)$
• $(D=1, C=T)$
• $(D=2, C=H)$

So there are no subjects with label $(D=2, C=T)$.

According to this interpretation, you are twice as likely to pick $C=H$ subject. Supposing that there are N experiments started each day, then let $N(D,C)$ be the number of subjects with label $(D,C)$. Then typically, you would expect that:

$N(1, C=H) = N(1,C=T) = N/2$ (of the new Day-1s, half typically are associated with a result of heads, and half are associated with a result of tails).
$N(2, C=H) = N/2$ (if a subject has label $(D=2, C=H)$ today, then she had label $(D=1, C=H)$ yesterday)
$N(2, C=T) = 0$ (no Day-2 for tails)

So on a typical day, there are $3N/2$ subjects, with the following statistics:

• $N$ have $C=H$
• $N/2$ have $C=T$
• $N$ have $D=1$
• $N/2$ have $D=2$

So for a randomly selected subject, the odds are

• $2/3$ that $C=H$
• $2/3$ that $D=1$.

Second interpretation: We can allow the possibility of $(D=2, C=T)$, but these subjects will know that that label applies to them, because their memories are not wiped after the first day. We'll call these subjects "informed".

With this interpretation, on a typical day, there are $2N$ subjects, with the following statistics:

• $N$ have $D=1$
• $N$ have $D=2$
  
• $N$ have $C=H$ (as before)
• $N$ have $C=T$ (including the $N/2$ "informed" subjects)

So for a randomly selected subject, the odds are:

• $1/2$ that $C=H$
• $1/2$ that $D=1$
• $1/4$ that the subject is "informed" (that is, $(D=2, C=T)$)
  
• $3/4$ that the subject is "uninformed"

So this interpretation restores the intuitive idea that there should be equal likelihood of heads and tails. However, we can compute conditional probabilities:

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

So under either interpretation, if the random subject is uninformed (doesn't know his coin result), then there is a 2/3 likelihood that her result was "heads".

 

[DrClaude]

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

It is never too late! I have added the poll.

 

[Demystifier]

It is never too late! I have added the poll.

That's great, thanks!

 

[PeroK]

One more thought. If the sleeper doesn't know about the drug, then every time she wakes she will think it is Monday and her answer will be 1/2.

The halfer's position is that knowing about the drug doesn't change this. Effectively, for reasons that I admit I don't follow, knowing about the drug and knowing that it might be Tuesday doesn't change the answer.

 

[Charles Link]

I side with the halfers. And here is an argument for why I side with the halfers: (I don't think it has been presented yet in this form=I didn't read every single post...) Suppose we weight the coin so that it has a $p= .99$ chance for heads. But suppose we also change the rules(as previously mentioned) so that she will be woken up, let's say 1000 times, to be interviewed if it comes up tails. I do think if she says .99 as the probability that it was heads that she has calculated it correctly. In all likelihood, (with 99% probability), it is the first Monday.

 

[PeroK]

I side with the halfers. And here is an argument for why I side with the halfers: (I don't think it has been presented yet in this form=I didn't read every single post...) Suppose we weight the coin so that it has a $p= .99$ chance for heads. But suppose we also change the rules(as previously mentioned) so that she will be woken up, let's say 1000 times, to be interviewed if it comes up tails. I do think if she says .99 as the probability that it was heads that she has calculated it correctly. In all likelihood, (with 99% probability), it is the first Monday.

Before you vote for 1/2, you may wish to consider this post:

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

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

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

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

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

 

[Charles Link]

Before you vote for 1/2, you may wish to consider this post:

The probability for Monday is 3/4 and for Tuesday 1/4.

 

[PeroK]

The probability for Monday is 3/4 and for Tuesday 1/4.

If that is true, then the probability of heads is:

$(3/4)(1/2) + (1/4)(0) = 3/8$

 

[Charles Link]

I think the alternative problem I posed (in post #99) answers the question: .99+.01(.001) it is Monday, .01(.001) it is the second day, .01(.001) that it is the 3rd day, etc.

 

[PeroK]

I think the alternative problem I posed answers the question: .99+.01(.001) it is Monday, .01(.001) it is the second day, .01(.001) that it is the 3rd day, etc.

That makes no sense to me. The problem is about a coin with 50-50 heads and tails. Please explain your solution for that, especially in light of post #100 and the fact that if it's Monday it's 50-50 heads/tails and if it's Tuesday it's 100% tails.

 

[Charles Link]

That makes no sense to me. The problem is about a coin with 50-50 heads and tails. Please explain your solution for that, especially in light of post #100 and the fact that if it's Monday it's 50-50 heads/tails and if it's Tuesday it's 100% tails.

It is somewhat of a puzzle. My logic of post #99 was somewhat of a response to the logic previously posted of adding extra days if it came up tails. I'm trying to demonstrate a possibility using a weighted coin, and extrapolating it to a non-weighted coin. $\\$ In some ways, the calculation that Sleeping Beauty does here is similar to what happens in science when we try to compute a probability under the assumption that the state we are considering is completely random. $\\$ I think my example in post #99 is worth consideration, because it illustrates the type of assessment that ultimately results: Is Sleeping Beauty likely to be awake on the Monday (with a weighted coin) of a very likely event, or did the unlikely occur, so that she is now part of a long chain of what would occur if the unlikely took place? I can also follow the logic of assigning a 1/3 to the heads condition, but, in some ways, this problem defies logic.