A variation of the sleeping beauty problem

Some time ago we had a discussion of the sleeping beauty problem
https://www.physicsforums.com/threads/the-sleeping-beauty-problem-any-halfers-here.916459/
which is a well known problem in probability theory. In that thread, there was no consensus whether the probability of heads is 1/2 or 1/3. Since the thread is closed now, I open a separate thread where I propose a variation of the problem, hoping that it will help to solve the problem in a way that everyone can agree on.

The rules of the game are the following. There are two examinees that cannot communicate with each other. A fair coin is tossed, and the result is not shown to examinees. If the coin landed heads, one examine is asked what is the probability that the coin landed heads. If the coin landed tails, both examinees are asked what is the probability that the coin landed heads. All these rules of the game are known to both examinees. The question is: If you are one of the examinees, and if you are asked what is the probability that the coin landed heads, what is your answer?

I claim that this version of the problem is equivalent to the original sleeping beauty problem, and that in this version of the problem it is clear that the correct answer is 1/3 (and not 1/2). Do you agree?

My argument is the following. For the sake of intuition, I first go to the extreme. Instead of considering ##n=2## examinees, I consider ##n=1.000.000## examinees. Either only one of them is asked the question (if the coin landed heads), or all of them are asked the question (if the coin landed tails). So if I'm asked the question at all, it is very unlikely that I am the lucky one that has been asked, while all others have not been asked. Given that I was asked the question, it is much more likely that I was not particularly lucky, i.e., that I am just one of ##n=1.000.000## examinees who were all asked the question. In other words, it is much more likely that the coin landed tails, so the probability that the coin landed heads is much smaller than 1/2. With this intuition in mind it is not difficult to compute that the probability of heads is ##1/(n+1)## (rather than 1/2), which in the case of ##n=2## gives the probability ##1/3##.

Demystifier said:

TL;DR Summary: A version of the "sleeping beauty" problem, in which a single sleeping examinee is replaced by two awake examinees.

Some time ago we had a discussion of the sleeping beauty problem
https://www.physicsforums.com/threads/the-sleeping-beauty-problem-any-halfers-here.916459/
which is a well known problem in probability theory. In that thread, there was no consensus whether the probability of heads is 1/2 or 1/3. Since the thread is closed now, I open a separate thread where I propose a variation of the problem, hoping that it will help to solve the problem in a way that everyone can agree on.

The rules of the game are the following. There are two examinees that cannot communicate with each other. A fair coin is tossed, and the result is not shown to examinees. If the coin landed heads, one examine is asked what is the probability that the coin landed heads. If the coin landed tails, both examinees are asked what is the probability that the coin landed heads. All these rules of the game are known to both examinees. The question is: If you are one of the examinees, and if you are asked what is the probability that the coin landed heads, what is your answer?

This is an elementary example of conditional probability. Some variation of which will be taught as part of an introduction to probability theory. There is no lack of consensus in this case. Except, where these problems get into the public domain and are analysed incorrectly. The same applies to Monty Hall etc.

Demystifier said:

I claim that this version of the problem is equivalent to the original sleeping beauty problem, and that in this version of the problem it is clear that the correct answer is 1/3 (and not 1/2). Do you agree?

In essence it is and helps to justify the 1/3 answer to sleeping beauty. Many laypeople who argue for 1/2 in sleeping beauty would argue for 1/2 in this case - which reveals a failure to understand elementary probability theory.

The real issue with the sleeping beauty is finding the flaw in the 1/2 argument. Sleeping beauty introduce a twist, which makes the flaw difficult to spot.

Demystifier said:

My argument is the following. For the sake of intuition, I first go to the extreme. Instead of considering ##n=2## examinees, I consider ##n=1.000.000## examinees. Either only one of them is asked the question (if the coin landed heads), or all of them are asked the question (if the coin landed tails). So if I'm asked the question at all, it is very unlikely that I am the lucky that has been asked, while all others have not been asked. Given that I was asked the question, it is much more likely that I was not particularly lucky, i.e., that I am just one of ##n=1.000.000## examinees who were all asked the question. In other words, it is much more likely that the coin landed tails, so the probability that the coin landed heads is much smaller than 1/2. With this intuition in mind it is not difficult to compute that the probability of heads is ##1/(n+1)## (rather that 1/2), which in the case of ##n=2## gives the probability ##1/3##.

Yes, that is a good way to highlight that the answer of 1/2 becomes somewhat absurd as something you would bet on.

It also highlights the problem with the argument to justify 1/2 in the sleeping beauty case that "it's an alternative view of probability". The problem is that different view becomes unreliable in terms of calculations, confidence and betting on given outcomes. For example, you would get people who say the result of any tennis match is 50-50. But, if you start betting on a club player against Djokovic on that basis, then you'll lose money. So, the 1/2ers in sleeping beauty eventually have to concede that either 1) betting on the outcome isn't necessarily a test of probability theory; or, 2) that losing money on bets doesn't prove you're wrong!

Demystifier said:

I claim that this version of the problem is equivalent to the original sleeping beauty problem, and that in this version of the problem it is clear that the correct answer is 1/3 (and not 1/2). Do you agree?

I am not sure. The difference as I see it is that in this problem one of the examinees may not be asked at all. So one could think of being asked the question as evidence. In the sleeping beauty problem this doesn't exist as far as I can see.

To put it another way. Imagine in the sleeping beauty problem you are awakened 1000 times on tails. How is being awakened evidence of tails?

jbergman said:

I am not sure. The difference as I see it is that in this problem one of the examinees may not be asked at all. So one could think of being asked the question as evidence. In the sleeping beauty problem this doesn't exist as far as I can see.

To put it another way. Imagine in the sleeping beauty problem you are awakened 1000 times on tails. How is being awakened evidence of tails?

Actually, thinking about it more, I am inclined to agree with your analysis.

I think https://www.quantamagazine.org/solution-sleeping-beautys-dilemma-20160129/ is a great article on the sleeping beauty problem.

They explain that depending on how you view the problem there really are two valid answers.

This answer on stackexchange also really goes into more detail to illustrate how different scenarios lead to different answers, https://stats.stackexchange.com/a/169582

jbergman said:

They explain that depending on how you view the problem there really are two valid answers.

If you take the halfer position and I take the thirder position, then I can win money off you by betting on the problem. That decides the issue, IMO. Ultimately, however you jazz it up, the halfer position cannot put its money where its philosophy is, if I can put it that way.

jbergman said:

They explain that depending on how you view the problem there really are two valid answers.

Or, to put it another way. You can have two valid answers to a problem that cannot be repeatedly tested. Like the probability that the universe is infinite. What probability means in that case is hard to define and you cannot, for the time being at least, resolve the issue by repeating some experiment.

The sleeping beauty problem is a concrete, well-defined experiment that could be carried out repeatedly. You could write a computer simulation. And the answer is clearly that in one-third of cases the coin is a one thing and in two-thirds of cases it is the other. You can't get a numerical answer of 1/2 from a computer simulation. And, criticallly, this simulation could be written by the woken princess. All she forgets is whether she has been woken previously. But, she has all the information about the experiment and could write a computer simulation. If she does that, she takes the thirder position. The halfer position, therefore, must forbid her to reason concretely about the problem (and/or write a computer simulation); but, instead, give the answer of 1/2 based on abstract philosophical reasoning.

The halfer position requires, therefore, that "probability" or "credence" be dissocciated from the results of repeated experiment. This may be a valid philosophical position, but it's not a valid position if probability is allowed to be empirical in nature and testable by experiment.

PS there's a parallel here with the Monty Hall problem. In that case, there are also two valid answers. One is that switching is better and the other is that switching makes no difference. The point is that these are valid answers to different versions of the problem. Switching is better in the actual game-show problem; and it makes no difference in the alternative "Monty fall" problem.

Once you establish which problem you are dealing with, there is only one answer. That's why doing repeated experiments (or a computer simulation) is critical. That forces you to establish the problem precisely.

You can try this with the sleeping beauty. There is simply no computer simulation that produces an answer of 1/2. Or, at least, if your simulation produces 1/2, then we can see how the problem has been altered. This, agaim, highlights that 1/2 is an answer backed only by philosophy, and not by hard calculation.

I think .5 is the answer.
The order of events:
Complete information about the experimental procedure is explained to SB
A fair coin is flipped, SB does not know the result
Sunday evening:
- some analyses explore SB's hypothetical credence of Heads assessed Sunday evening (and maybe Wednesday morning) when for both SB would answer p(H)=.5
- some analyses explore SB's hypothetical credence on Sunday evening as to how she would calculate and respond when awakened, knowing the procedure
Monday:
SB is awakened and interviewed, independent of the results of the coin flip
Q - What is the Monday awakening conditional upon?
A - Nothing, it is a given, it happens whether the coin is H or T
Tuesday:
SB is awakened a second time if the coin was Tails
Q - What is the Tuesday awakening conditional upon?
A - Only the state of the coin being Tails, which could have been flipped immediately before the Tuesday decision to awake or not, or flipped at any earlier time but the result not observed until immediately before the Tuesday awake decision
>> the results of the coin flip do not need to be known by the experimenters until Tuesday morning
- the only reason they flip the coin before the Monday awakening is to make it a past event prior to the Monday awakening (so the credence question may use the past tense - otherwise the answer would always be .5 for the probability of the future flip determining a possible Tuesday awakening), so the coin may have been flipped at any time before without determination of the result by the experimenters until Tuesday

===
It looks to me that if SB is awakened on Monday no matter what, it's a given and credence-wise means nothing, both flip results effect Monday the same, unconditionally, so they "cancel out"; the only open question is if SB gets awakened on Tuesday... there is a .5 chance of that (tails), so .5 chance of Heads

bahamagreen said:

It looks to me that if SB is awakened on Monday no matter what, it's a given and credence-wise means nothing, both flip results effect Monday the same, unconditionally, so they "cancel out"; the only open question is if SB gets awakened on Tuesday... there is a .5 chance of that (tails), so .5 chance of Heads

If we put you in the scenario and - forget the drug - you just pretend that you have no memory. Let's say the rule is tails it's both days. Would you be happy with the strategy to bet heads every time? You know all the rules of the game, but you don't know which day it is.

What's to stop you, when you get wakened, to write a computer simulation of the game and run it and see that 2/3 wakenings are the result of a tail? Why can't you do that? And, if you do write a computer simulation, would you ignore the result? Would you exclude the results of your computer simulation from the data you use to determine your credence?

bahamagreen said:

It looks to me that if SB is awakened on Monday no matter what, it's a given and credence-wise means nothing, both flip results effect Monday the same, unconditionally, so they "cancel out"; the only open question is if SB gets awakened on Tuesday... there is a .5 chance of that (tails), so .5 chance of Heads

PS let's say we are both in the scenario. You stick with your 50-50 credence and are happy to bet heads every time, and I run my computer simulation of the game and bet tails every time. Let's say the one who is right wins $100 and this gets put in their box. We both then forget what's happened.

The experiment is run hundreds of times. At the end, we get to look in our boxes to see how much money we have won. At that point, do you expect to have as much money in your box as I have in mine? I suggest that you know that I'll have won (probably) twice as much money as you! And, I suggest, therefore, that your credence of 50-50 was wrong and you didn't use all the information at your disposal.

Even if you are still surprised at this point, how do you explain how I managed to win more than you, given that the probability of heads was always equal to tails? That should be impossible. If it was 50-50, then I can't possibly have won more than you by betting on tails every time.

This is what I mean by putting your money where your philosphy is.

The Monday awakening is unconditional, independent of the coin flip result, independent of when the coin is flipped, and independent of when the result is observed. The experiment does not really begin until Tuesday, the Tuesday awakening conditional on the coin flip, independent of Monday. What is the Monday awakening conditional upon? Nothing except that it is Monday and SB will be awakened on Monday under all possible conditions. The experimenters do not even need to know the results of the coin flip until Tuesday. Monday has no bearing on anything.
If she is told Sunday that the coin flip will not occur until Tuesday morning, they may ask her credence Sunday of Heads coming up in that future coin flip and she would answer .5
When awakened she would know it was either Monday or Tuesday. Credence of the future flip Heads .5, credence of a flip preceding an awakening decision Heads would also be .5
They may tell her that the coin flip will occur Sunday evening but they won't look at the result until Tuesday morning to determine a second awakening. Same result as above - delayed observation is the same as late flipping, right?

What is the form of your proposed wager? Are you including Monday, and if so, why?

bahamagreen said:

What is the form of your proposed wager? Are you including Monday, and if so, why?

It's got to be every wakening. Why would it be anything else? If we are playing the game, we have no idea what day it is, so it's illogical to only bet on a Monday or only on a Tuesday. It has to be both days.

PS this is why I claim that halfers dissocciate probability from outcomes. The probability becomes just an abstract, untestable number.

PeroK said:

It's got to be every wakening.

Indeed. If we only bet on Tuesday then I know which day it is by whether you ask me to bet or not.

Ibix said:

Indeed. If we only bet on Tuesday then I know which day it is by whether you ask me to bet or not.

As an aside, here's the problem with the halfer position. It's based on the assumption that there is no new information. But, a computer simulation using only the rules of the game, shows that the answer is 1/3. This simulation uses only information that the SB has. So, there must be new information. That new information is that the game has started!!

So, if we go back to Sunday evening, there are two different questions:

1) Credence now that the coin is heads or tails: 50-50.
2) Credence during the game that the coin is heads or tails: 1/3-2/3. Because, at that stage, there is a chance that actions have been taken that depended on the coin. You don't know for sure whether actions have been taken. But, it's enough that you know that actions might have been taken and that it might be Tuesday. And, that is indisputably new information.

And, this is the answer you would give on the Sunbday evening - thinking ahead to the game, the answer will be 1/3 by then. Once you know the game has started.

Can you reconcile the two different questions having different answers?
Actions that depend on the coin do not include Monday. What happens Monday is strictly determined up to asking SB's credence; why would Monday be included?
How do actions that depend on the known outcome of the coin (Tuesday) change anything?
How is starting the game new information if informed of the procedure in advance?

Focusing on the coin, what possible conditional operations or manipulations to the coin change the p of Heads from .5 to something else? Is there agreement that flipping the coin Sunday without observing it until Tuesday is the same as just waiting until Tuesday to flip and observer it (same p of Heads)?

bahamagreen said:

Can you reconcile the two different questions having different answers?
Actions that depend on the coin do not include Monday. What happens Monday is strictly determined up to asking SB's credence; why would Monday be included?
How do actions that depend on the known outcome of the coin (Tuesday) change anything?

That's the nature of conditional probability. The simple fact is that 2/3 of awakenings take place on tails. How do you explain that? It can only be explained if actions are taken depending on the coin flip.

bahamagreen said:

How is starting the game new information if informed of the procedure in advance?

Because you still have to start the game. At that point actions may have taken place that were dependent on the toss of the coin. You know in advance that the game will start at some point - but until it has started, nothing has been done conditionally on the result of the toss.

To be honest, your objections here would apply to any attempt to do probability theory. These objections are tantamount to saying that the toss of a coin is always either 50-50 or definitely heads or tails. That 0, 1/2 and 1 are the only allowable probabilities for a coin.

If that's true, then you've thrown out not just the SB problem, but the entire theory of conditional probabilities.

bahamagreen said:

What happens Monday is strictly determined up to asking SB's credence; why would Monday be included?

How can Monday not be relevant? On Monday, SB thinks it might be Tuesday.

bahamagreen said:

Focusing on the coin, what possible conditional operations or manipulations to the coin change the p of Heads from .5 to something else? Is there agreement that flipping the coin Sunday without observing it until Tuesday is the same as just waiting until Tuesday to flip and observer it (same p of Heads)?

I think the simplest way to understand is to have Sleeping Beauty decide to systematically say "heads" every time she's wakened and repeat the experiment ##2N## times, where ##N## is large. In ##N## trials she says it once and is correct. In ##N## trials she says it twice and is wrong both times. (Edit: that's the long run average, anyway.) Thus "heads" is correct in ##N## of her ##3N## repetitions of the word - 1/3 of them.

One of the links in the OP of the thread linked in the OP of this one notes that Sleeping Beauty should say both before and after the experiment that the probability of heads is 1/2 (until shown the coin, anyway). It's only during the experiment thst she should say 1/3.

Ibix said:

One of the links in the OP of the thread linked in the OP of this one notes that Sleeping Beauty should say both before and after the experiment that the probability of heads is 1/2 (until shown the coin, anyway). It's only during the experiment thst she should say 1/3.

That's a very sharp observation. It shows the contemporaneous nature of the information involved. It's only during the game that things are done based on the coin toss. The information that SB has that she is in the game is relevant. It doesn't apply before or after.

That information (that she is in the game and knows it) is the information that the half position does not accept - and that's where the error is.

I going to go look at theory of conditional probabilities ... (Wikipedia)

P(A|B)=(P(A^B))/P(B)

Where "^" is intersection
and
B is Monday awakening
A is Tuesday awakening

P(B)=1
P(A)=.5
P(A|B)= P(.5^1)/1=.5

PeroK said:

That's a very sharp observation.

It is rather neat. The link to which I referred, if anyone wants to read the argument.

bahamagreen said:

I going to go look at theory of conditional probabilities ... (Wikipedia)

P(A|B)=(P(A^B))/P(B)

Where "^" is intersection
and
B is Monday awakening
A is Tuesday awakening

P(B)=1
P(A)=.5
P(A|B)= P(.5^1)/1=.5

So P(A|B) is the probability that she's awake on Monday given that she's awake on Tuesday? Shouldn't that be zero? Or am I misunderstanding what you mean by P(Monday awakening|Tuesday awakening)?

bahamagreen said:

I going to go look at theory of conditional probabilities ... (Wikipedia)

The probability that you are supposed to be calculating is:

Probability that the coin was heads, given that someone has just woken her up.

Is that the probability that you are calculating?

its the probability of awakenings before Wednesday
... from Ibix's link...
"I’ve just observed an event (D = waking up before Wednesday) that is twice as likely under the tails scenario..."
I'm not seeing the twice as likely; the Monday waking is certain (p=1), the Tuesday waking based on coin toss. p=.5
From Wikipedia,
"In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) is already known to have occurred."
The credence question is about the coin toss result event "already known to have occurred".

bahamagreen said:

The credence question is about the coin toss result event "already known to have occurred".

But what's the event known to have occurred? Sleeping Beauty has no way to know if Monday's waking has occurred.

bahamagreen said:

its the probability of awakenings before Wednesday
... from Ibix's link...
"I’ve just observed an event (D = waking up before Wednesday) that is twice as likely under the tails scenario..."
I'm not seeing the twice as likely; the Monday waking is certain (p=1), the Tuesday waking based on coin toss. p=.5

Let me try an example, but I fear we are too far down the rabbit hole now.

Suppose you start a job for an eccentric boss. You have two days work - Monday and Tuesday. Your boss tosses a coin on Sunday night. If it's heads, he pays you $100 on Monday and nothing on Tuesday. If it's tails, he pays you $100 on Monday and another $100 on Tuesday.

Ibix's claim is that if the coin is tails, then you get paid twice as often: there are two payment events. In that sense, you are twice as likely to get paid for your day's work if it's tails.

What you're saying is that you get paid at least once in any case, so it's not more likely. If it's heads, you get paid and if it's tails you get paid.

The question is whether getting paid twice is relevant. In this example it is relevant. In one case, you earn $200 and in the other case only $100.

Now let's suppose you have a really poor memory and can't remember what day it is. You don't know whether it's Monday or Tuesday. Your boss turns up at the end of the day and gives you $100. What's the probability that the coin was Heads? That's the SB problem.

Ibex
Once started, the only event known to have occurred is the coin toss and the Monday waking.
She may awake either two or three times, that last time on Wednesday in which she will not know if it is Monday, Tuesday, or Wednesday until they announce the end of the project.
When she awakes at any time during the project, she knows:
- if it is Wednesday, the Monday waking has occurred
- it it is Tuesday, the Monday waking has occurred
- if it is Monday, the Monday waking has occurred (just now)

Perok
"Now let's suppose you have a really poor memory and can't remember what day it is. You don't know whether it's Monday or Tuesday. Your boss turns up at the end of the day and gives you $100. What's the probability that the coin was Heads?"

If I don't know the result of the coin toss and I don't know what day it is and only know that I got paid, the payment does not inform me which day it is nor inform me whether the coin toss came up heads or tails.
I know I get paid on Monday independent of the toss, and may get paid Tuesday dependent on the toss. Like the SB scenario, I collapse Monday and omit it because it distinguishes nothing. What happens Tuesday is fully dependent on the coin toss, so .5
How could I evaluate the p of heads other than .5? How could I believe events post coin toss change probability of an extant completed coin toss?
When the scenario is posed as payments, bets, etc. are you doing a probability calculation that counts Monday twice as if the two scenarios Monday only and Monday plus Tuesday make up three things? I see Monday as independent - same thing happens for both toss results.

bahamagreen said:

How could I evaluate the p of heads other than .5? How could I believe events post coin toss change probability of an extant completed coin toss?

That's the "extreme" halfer position, which rejects all (conditional) probability theory. In other words, the 1/3 answer must be wrong in all cases, because it's impossible.

There's no point in arguing about the SB problem if you don't believe in probability theory in the first place!!

Before you give up on me, could you show the 1/3 calculation?

bahamagreen said:

When she awakes at any time during the project, she knows:
- if it is Wednesday, the Monday waking has occurred
- it it is Tuesday, the Monday waking has occurred
- if it is Monday, the Monday waking has occurred (just now)

Ok. So with this phrasing, the question is how likely is it that the last time you woke up was Monday? The answer to this is not 1 - what is it?

bahamagreen said:

Before you give up on me, could you show the 1/3 calculation?

Let's run the experiment 6 times:

Week 1: Heads
Monday: Princess is woken

Week 2: Heads
Monday: Princess is woken

Week 3: Tails:
Monday: Princess is woken
Tuesay: Princess is woken

Week 4: Heads:
Monday: Princess is woken

Week 5: Tails
Monday: Princess is woken
Tuesday: Princess is woken

Week 6: Tails
Monday: Princess is woken
Tuesday: Princess is woken

Summary:

Coin tosses 6 times, 3 heads and 3 tails. Probability of heads = 1/2

Princess woken 9 times: 6 times the coin was tails; 3 times the coin was heads. Probability of heads given awakening event: 1/3

Inescapable conclusion: if the Princess bets Heads every time she is woken up, she would win $300. If she bets tails every time she is woken, she would win $600.

Thanks, that is very clear. You are distinguishing the probability of heads given awakening event as a separate thing from the probability of the tossed coin being heads. Looks to me like the word credence (mental acceptance as true or real) has two different domains depending on whether inside the experiment vs outside abstractly before or after. I can see both. I can see the "6 times the coin was tails; 3 times the coin was heads" for the inside domain vs "Coin tosses 6 times" for the outside domain.
I'm still thinking about it.

The issue isn't whether the coin is heads or tails. What matters is how often Sleeping Beauty guesses correctly.

bahamagreen said:

I'm still thinking about it.

Note that the probability that the coin is heads or tails is not 1/2 when the experiment is taking place. It's either definitely heads or definitely tails. This does not change, but as the princess gains information, she can get closer to the correct answer - which is either heads or tails. What's really happening with conditional probability is not that an outcome is changing, but the princess is getting information about what that outcome was in this particular case.

As an example, let's say we roll a die and it comes up 5.

Your initial credence is 1/6. This is the best you can do. Note that it does not describe the state of the die (which is definitely a 5), 1/6 describes your knowldedge about the die.

Now, you are told that the die shows a prime number. Or, some event takes place that only takes place if the die is a prime number.

Your credence should now be 1/3. Again, this does not describe a change in the state of the die, but a change in the state of your knowledge about the die.

Then you are told it is an odd number. This rules out 2. And, again, your credence changes to 1/2.

If the princess says 1/3, she knows that the coin is either definitely heads or definitely tails. And, in fact, that is a valid answer. The state of the coin is either definitely heads or definitely tails. The answer 1/2 also never describes the state of the coin, except before it's tossed. Neither answer 1/2 or 1/3 ever describes the state of the coin; only the limited knowledge that the princess has. 1/2 represents zero knowldege about the coin; and 1/3 represents slightly more knowldege about the coin.