Undergrad The Sleeping Beauty Problem: Any halfers here?

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The Sleeping Beauty problem raises a debate between "thirders," who argue the probability of the coin landing heads is 1/3, and "halfers," who believe it is 1/2. Proponents of 1/3 argue that the princess's amnesia prevents her from gaining new information, thus her a posteriori probability remains unchanged. Conversely, halfers contend that since she learns nothing new upon waking, her initial probability of 1/2 should hold. The discussion also explores various scenarios and thought experiments to illustrate the implications of the problem, emphasizing the importance of conditional probabilities. Ultimately, the debate hinges on interpreting the information available to the princess at the moment she is awakened.

What is Sleeping Beauty's credence now for the proposition that the coin landed heads?

  • 1/3

    Votes: 12 33.3%
  • 1/2

    Votes: 11 30.6%
  • It depends on the precise formulation of the problem

    Votes: 13 36.1%

  • Total voters
    36
  • #151
For a bit of fun, I imagined @Marana in the role of experimenter and me in the role of sleeping beauty:

Me: You woke me up!
Marana: Yes, I need to ask you what is your credence that the coin is heads.

Me: What day is it?
Marana: I can't tell you that.

Me: Is it Monday?
Marana: I can't tell you that.

Me: Did you give me that drug already?
Marana: I can't tell you that either.

Me: I believe there is likelihood of 1/3 that it's heads.
Marana: That's wrong!

Me: Why is it wrong?
Marana: How did you get 1/3?

Me: I imagined that we did this experiment many times and ... therefore, by using relative frequencies I got 1/3.
Marana: But, your not allowed to think like that.

Me: How can you stop me thinking?
Marana: Okay, I can't stop you thinking, but it's wrong for you to think like that.

Me: Why is it wrong?
Marana: Because relative frequencies are invalid for you.

Me: Would you be allowed to use relative frequencies. Perhaps to explain this experiment to an observer?
Marana: Yes, of course.

Me: So, it's right for you to think like that but it's wrong for me to think like that?
Marana: Yes.

Me: But, I'm quite good at probabilities. Why can't I use my natural abilities here and now?
Marana: Because you can't.

Me: So, basically, I'm expected to give the dumb answer of 1/2 (because, hey, nothing can change a coin after it's been tossed) and not the clever answer of 1/3?
Morana: I'm afraid so.

Me: That's not fair!
Marana: That's the way it is.
 
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  • #152
forcefield said:
The coin flip creates two equally likely scenarios. Hence the probability is 1/2.
forcefield said:
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.
I was only part-time thirder: That relative probability is actually 1:1. The fact that SB is awake twice in case of tail and only once in case of head does not increase belief in tail. So I finally voted for 1/2.
 
  • #153
Marana said:
Sleeping beauty can consider herself to be randomly selected from all such sleeping beauties. That would give a probability of 1/2. The question is whether, from that starting point of 1/2, there is any way to update to 1/3. That's what I haven't seen so far.

I went through the numbers, and it's 2/3. If you do the experiment over and over, with different starting times, then of the "active" Sleeping Beauties (the ones in day 1 or day 2 of the experiment), 1/3 are on day 1 after a coin flip of heads, 1/3 are on day 2 after a coin flip of heads, and 1/3 are on day 1 after a coin flip of tails. If the actual Sleeping Beauty thinks of herself as a random choice among those, then she would come up with 2/3 heads.

It's sort of similar to the situation where there is some country where people have a 50% chance of producing one offspring, and 50% chance of producing two offspring. If you take a random adult and ask the probability that it will have two offspring, the answer is 50%. If you take a random child and ask what is the probability that their parent had two children, it's 2/3.

Equivalently, she can consider her experiment (the coin toss followed by either M or MT) to be randomly selected from all such experiments, and use the relative frequency of 1/2. This is justified because the experiment is an experiment. Again, the question is whether she can change from that 1/2.

Yeah, in the repeated sleeping beauty experiment, there are two different relative frequencies to compute:
  • Let's call a coin flip "active" if it's being used in a sleeping beauty experiment, and it was flipped less than two days ago (regardless of the result).
  • Let's call a sleeping beauty "active" if she is experiencing day 1 or day 2.
Then:
  • 2/3 of active sleeping beauties have an associated coin toss result of heads
  • 1/2 of all active coin flips are heads
So, if you point to a random coin flip result (recorded on a piece of paper) and ask a Sleeping Beauty what are the odds that it is heads, she should answer 1/2. But if you tell her that it's her result, she should say 2/3.

Relative frequency can be used, but not haphazardly. It needs to be justified... and we all agree on that! Because clearly using relative frequency of heads awakenings is the wrong way to compute the subjective probability of a heads coin toss without the drug. That is a misuse. And I'm arguing that, even with the drug, it is still a misuse to use relative frequency of heads awakenings as the subjective probability of a heads coin toss. But it is a proper use to use relative frequency with experiments, since they are experiments.

To call it a misuse, you need to say what reason is there not to. What harm comes from it?

To me, the best example of a counter-intuitive result coming from the thirder position is to change it to a lottery. A person has a one in a million chance of winning. But you can make it subjectively 50/50 by waking the winner a million days in a row. That's strange.

It is at issue in the most popular argument for 1/3. The argument goes: after conditioning on "it is monday", we must get 1/2.
That is, P(monday and heads)/P(monday) = P(monday and tails)/P(monday) = 1/2. Therefore P(monday and heads) = P(monday and tails).

I assume we all agree that you wouldn't use regular conditioning on "it is now 2:00" and "it is now 2:01" in my example.

That's not the same thing. You can eliminate that problem by making statements about connections between events. "The first time I looked at the clock, it was 2:00." "The second time I looked at the clock, it was 2:01". No contradiction.

I argue that the same holds here. You can't use regular conditioning on "it is monday", and the argument for 1/3 doesn't work.

I think that's barking up the wrong tree. Instead of Monday, think about Tuesday. You ask Sleeping Beauty what the probability of getting heads was. Now you tell her that today is Tuesday. Then she knows that she got heads. Of course, telling her "Today is Tuesday" gave her new information, and it allowed her to update her likelihood estimates. So if you have a theory of subjective probability that can't accommodate such an update, then I would say something is wrong with that theory.

This isn't really a new thing: none of us condition on the flow of time and immediately throw out every probability computation. It's unspoken, but we avoid it intuitively. Until the sleeping beauty problem, a bizarre setup that makes us want to use it.

I don't think that there is anything wrong with conditioning on the flow of time. If it's possible to forget what day it is, then knowing what day it is is additional information, and it can change your subjective likelihood estimates. It's a language problem for how to deal with words like "now" in a consistent way, not a problem with the applicability of probability.

Rather than using conditioning, what we usually do (correctly) is to know that, by itself, the passage of time doesn't change things that don't change with time.

But that just doesn't seem true. If today is Tuesday, she knows that the coin toss result was heads. So knowing it's Tuesday changes the odds from whatever to 1.

Now, if you did really want to condition on the time, you could treat the time as if it was randomly selected. That's fine. Then you would be conditioning on "it is time X and time X was randomly selected". Using this method you would get P(heads) = 1/2 and P(heads|monday) = 2/3.

How do you get that? That's truly nonsensical, for the following reason (pointed out by @PeroK): If the memory wipe happens on the morning of the awakening, right before sleeping beauty wakes up, then there is no need to even toss the coin until Tuesday morning. So on Monday, the coin hasn't even been tossed (under this variant). How could the knowledge that today is Monday tell you about a coin that has not yet been tossed?

It makes a lot more sense to say that knowing it's Monday doesn't tell you anything about the likelihood of a coin toss in the future. So P(H | Monday) = 1/2. Knowing that it's Tuesday tells you exactly what the coin toss result was: P(H | Tuesday) = 1. If you don't know whether it's a Monday or a Tuesday, then the probability of heads would be: P(H) = P(H|Monday) P(Monday) + P(H|Tuesday) P(Tuesday) > 1/2

But thirders are trying to have it both ways. They do not want to treat time as randomly selected or static, yet they want to condition on it.

If you want to randomly select a time, as well as randomly select heads or tails, then there are 4 possibilities, equally likely:
  1. Heads and Monday
  2. Heads and Tuesday
  3. Tails and Monday
  4. Tails and Tuesday
Then you rule out 4 on the basis that the rules say no memory wipe in that case. (So if Sleeping Beauty doesn't know whether it's Monday or Tuesday, then she can't be in situation 4). Eliminating situation 4 still leaves the other three equally likely.
 
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  • #154
Boing3000 said:
A reasonable person don't wager.
Look, for once, I'll try to be dead serious. This problem is in the "positive trolling" category, exactly like the chicken and eggs. Those problems have perfectly fine and simple answer, until you add some "hidden" confusion based on tricky language where everyone feels untitled to plug their own meaning.
In sleeping beauty, the trolling start when you speak about days name. This is irrelevant because of the drug and the closed room.

So if the coin is fair, and the experiment is fair, Beauty does not even have to be put to sleep to answer the question. Actually the coins does not even have to be flipped before Tuesday. So asking the question on Monday is definitely some sort of lying if the experimenter choose to do so.I agree, but then it is irrelevant to the question "What is your credence now for the proposition that the coin landed heads?".I have no idea how Beauty or anyone else could do that calculation. "Waking up" have never changed anything to how you process the world, except maybe on the February 2 (arghh I lied, I am not serious anymore)If the question was "how many time do you wake up", I would agree Beauty should answer 3/2.
But that is not the question, isn't it ?

Well, what is your calculation for this modified Sleeping Beauty problem?
  1. She gets a memory wipe in either case (heads or tails).
  2. But if the coin result was tails, then on Tuesday she is given a note saying that it is, in fact, Tuesday.
Then before she checks whether there is a note, then do you agree that there are 4 possibilities, equally likely?
  1. Result was heads, and today is Monday?
  2. Result was heads, and today is Tuesday?
  3. Result was tails, and today is Monday?
  4. Result was tails, and today is Tuesday?
Just from symmetry, wouldn't you give all 4 possibilities equal likelihood?

Now, she checks for a note. If there is no note, then she knows she is in situation 1-3. The lack of a note eliminates situation 4, but it shouldn't affect the likelihood of the other 3 possibilities.
 
  • #155
PeroK said:
for me, relative frequency is equivalent to probability

If you are dealing with a large enough sample space for the law of large numbers to be relevant, then this makes sense. So if Beauty were told that the experiment would be run, say, 1000 times, and after each run ended she would be given the amnesia drug so she would never know, when awakened, which run it was, then using the relative frequency to conclude that P(Monday|Awakened) = 2/3 and P(Tuesday|Awakened) = 1/3 (using my notation from previous posts) would make sense. And that would give P(Heads) = 1/3.

But the original statement of the problem doesn't say that; it says the experiment gets run once and that's it. And I don't think it is a slam dunk that relative frequencies are equal to probabilities if we just have a single sample. The frequentists say it is, but not everyone is a frequentist.

From a Bayesian point of view, if we're only making a single run of the experiment, then the only basis we have for assigning any value at all to P(Monday|Awakened) and P(Tuesday|Awakened) is prior probabilities. If we say that, given only a single run of the experiment, we have no useful information about the distribution of awakenings, then the obvious prior to use is the maximum entropy prior, which is P(Monday|Awakened) = P(Tuesday|Awakened) = 1/2. And if we use those values, then, by my previous posts, we get P(Heads) = 1/4! So this argument is an argument against both the thirder and the halfer. (And we would need to add another option to the poll. :wink:)
 
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  • #156
PeterDonis said:
If you are dealing with a large enough sample space for the law of large numbers to be relevant, then this makes sense. So if Beauty were told that the experiment would be run, say, 1000 times, and after each run ended she would be given the amnesia drug so she would never know, when awakened, which run it was, then using the relative frequency to conclude that P(Monday|Awakened) = 2/3 and P(Tuesday|Awakened) = 1/3 (using my notation from previous posts) would make sense. And that would give P(Heads) = 1/3.

But the original statement of the problem doesn't say that; it says the experiment gets run once and that's it. And I don't think it is a slam dunk that relative frequencies are equal to probabilities if we just have a single sample. The frequentists say it is, but not everyone is a frequentist.

From a Bayesian point of view, if we're only making a single run of the experiment, then the only basis we have for assigning any value at all to P(Monday|Awakened) and P(Tuesday|Awakened) is prior probabilities. If we say that, given only a single run of the experiment, we have no useful information about the distribution of awakenings, then the obvious prior to use is the maximum entropy prior, which is P(Monday|Awakened) = P(Tuesday|Awakened) = 1/2. And if we use those values, then, by my previous posts, we get P(Heads) = 1/4! So this argument is an argument against both the thirder and the halfer. (And we would need to add another option to the poll. :wink:)

That analysis doesn't essentially depend on the peculiarities of this experiment. So, that is a critique of probability theory more generally, and specifically in the case where there is only one experiment.
 
  • #157
PeroK said:
that is a critique of probability theory more generally, and specifically in the case where there is only one experiment.

Yes, that's true.
 
  • #158
PeroK said:
If you know how to write a computer programme, you could do a computer simulation of the problem and you would, to your surprise, see the answer 1/3 appear rather than 1/2.
I have encoded your logic (if I get it right, your toss 5 & 6 does not follow the same rule)
This is javascript push F12 and cut and paste it into your console
Code:
(function BeautyThirders(run) {
    var counts = { H: 0, T: 0 }
    var experiments = [
        function Head(counts) {
            counts.H++;
        },
        function Tail(counts) {
            counts.T++;
            counts.T++;
        }
    ];
    while (--run >= 0) {
        experiments[Math.floor(Math.random() * experiments.length)](counts);
    }
    alert("H/T = " + (counts.H / counts.T).toFixed(2));
})(10000);

PeroK said:
So, in that little simulation, we ended up with 3 Heads and 3 Tails for the coin. Beauty was woken 9 times (6 times on a Monday and 3 on a Tuesday). When she was woken, it was as a result of Heads 3 times and Tails 6 times.
Of course, but I am afraid you miss my point about this debate/poll. (I voted for the third option)
Beauty cannot count. She is drugged especially to avoid her to do so.
Given the procedure described into the Wikipedia article the situation is simple and straightforward. The credence that a coin lands Head (or tail for that matter) does not change if Beauty goes to sleep or not.

PeroK said:
Therefore, if she counted each time she was woken whether it had been a Head or a Tail, she would have got 6-3 tails. And, that is what's called a relative frequency and is the basis of what a probability is.
She cannot count, so unless some modification is done to the definition of the problem, I reject this arguments.

PeroK said:
So, halfers don't allow two bets on the same coin, but thirders do.
That's also how I interpret the difference and why I vote for the latest possibility.
When people whose bread and butter is probability cannot even agree on a simple problem, then my credence that this problem is badly defined is above 99.9%

PeroK said:
Finally, the thirders position is: if you don't allow the Tuesday bets to stand, why have them in the first place? Isn't that then a different problem?
Precisely, why indeed have question mark in the definition ? Why speak about bet when there is none ? The credence is 1/2. Now indeed, choosing Head instead of Tail may make you loose twice (an hypothetical bet), and that's why the thirder's view exist.

And that's how the problem should be fixed. The credence that the coin is Head(or Tail) has nothing to do with the credence of loosing one or two bets.
Here is a modified version of the code, which is bet oriented, and don't drug Beauty.
The frequency Beauty bets on Head is the second parameter (1 (always),2(half), 3 ..., > tries == never). So obviously, if you want to win, better never bet heads
Code:
(function BeautyBets(run, frequency) {
    var counts = { WakeUps : 0, Wins : 0 }
    var experiments = [
        function Head(counts) {
           //Monday
            if (++counts.WakeUps % frequency == 0)
                counts.Wins++;
            else
                counts.Wins--;
        },
        function Tail(counts) {
           //Monday
            if (++counts.WakeUps % frequency != 0)
                counts.Wins++;
            else
                counts.Wins--;
           //Tuesday
            if (++counts.WakeUps % frequency != 0)
                counts.Wins++;
            else
                counts.Wins--;
        }
    ];
    while (--run >= 0) {
        experiments[Math.floor(Math.random() * experiments.length)](counts);
    }
    alert(counts.Wins.toString() + "wins for " + counts.WakeUps + " bets (" + (counts.Wins / counts.WakeUps).toFixed(2) +")");
})(10000, 3);
 
  • #159
Boing3000 said:
Given the procedure described into the Wikipedia article the situation is simple and straightforward. The credence that a coin lands Head (or tail for that matter) does not change if Beauty goes to sleep or not.

Perhaps not, but it does change when the experimenter looks at it. It changes from 1/2 to either 0 or 1.

When Beauty is wakened the coin is not 50-50 heads or tails. It is either definitely Heads or definitely Tails.

Her credence is now definitely wrong, and she knows that. So she must, if she can, look for clues to help her guess what the coin really is.
 
  • #160
Boing3000 said:
Given the procedure described into the Wikipedia article the situation is simple and straightforward. The credence that a coin lands Head (or tail for that matter) does not change if Beauty goes to sleep or not.

Once again, if instead of wiping her memory in the case of heads, suppose they wipe her memory on Monday regardless, but if the coin flip was tails, her memory is restored 10 minutes after wakening.

Then before the 10 minutes is up, do you agree that Sleeping Beauty has to consider all four of the following possibilities equally likely?
  1. It's Monday, and the result was heads.
  2. It's Monday, and the result was tails.
  3. It's Tuesday, and the result was heads.
  4. It's Tuesday, and the result was tails.
After the 10 minutes is up, she knows whether she is in case 4 or not. If case 4 is ruled out, then what are the updated probabilities for 1-3?
 
  • #161
PeroK said:
Her credence is now definitely wrong, and she knows that.
Wrong how ? What happens that could make her change her mind about how coins works ?
PeroK said:
So she must, if she can, look for clues to help her guess what the coin really is.
No she does not. But if the problem definition is changed, not just to include vaguely some asymmetrical bet, but to attach some new credence to how she must bet to win. This is not equivalent to what the coin is.
Drugged or not, she knows the bet is asymmetrical, she knows that even though coins works 1/2, that she will not be asked about that, but asked about "being awake in the context of this asymmetrical setup" what do you bet ?

So one last time. The thirder are not reading a problem about "what the coin really is". They are reading a problem about what to say about a coin to win some bet
I have trivially modified the code so you can change the amount bet in each case. And non surprisingly if Head bet value is more than twice Tail bet, then it became more "interesting" to always bet on Head. Yet still nothing have happened to the coins, frequencies of days and what not.
It is also not surprising that the only invariant of this algorithm is when frequency is 2 (1/2). Meaning 0 gain or loss. Meaning the only "physical" thing is a fair coin have two side.
Code:
(function BeautyBets(run, frequency, headAmount, tailAmount) {
    var counts = { Bets: 0, Wins: 0, Flips: 0 }
    var experiments = [
        function Head(counts) {
            counts.Flips++;

            if (++counts.Bets % frequency == 0)
                counts.Wins += headAmount;
            else
                counts.Wins -= headAmount;
        },
        function Tail(counts) {
            counts.Flips++;

            if (++counts.Bets % frequency != 0)
                counts.Wins += tailAmount;
            else
                counts.Wins -= tailAmount;

            if (++counts.Bets % frequency != 0)
                counts.Wins += tailAmount;
            else
                counts.Wins -= tailAmount;
        }
    ];
    while (--run >= 0) {
        experiments[Math.floor(Math.random() * experiments.length)](counts);
    }
    alert("Money " + counts.Wins);
})(10000, 2, 2.1, 1);
 
  • #162
Boing3000 said:
Wrong how ? What happens that could make her change her mind about how coins works ?

No she does not. But if the problem definition is changed, not just to include vaguely some asymmetrical bet, but to attach some new credence to how she must bet to win. This is not equivalent to what the coin is.

I don't understand what you mean. Surely, if the walls of Sleeping Beauty's room are painted red if the coin flip is heads and blue otherwise (and she knows this), then she wouldn't stick to her 50% likelihood of heads and tails after seeing the wall color. When someone is asked what the likelihood is that something is true, they typically use whatever information that is available.

The real issue is whether she has any information to change her likelihood estimate. I say she does---the fact that she doesn't know whether it is Monday or Tuesday tells her something---it tells her that either:
  1. The result was heads, and today is Monday.
  2. The result was heads, and today is Tuesday.
  3. The result was tails, and today is Monday.
The fourth possibility---that today is Tuesday and the coin flip was tails---is eliminated by the information that she doesn't know whether it's Monday or Tuesday. (She would know in the fourth case).
 
  • #163
Boing3000 said:
Wrong how ? What happens that could make her change her mind about how coins works ?

If I say to you that I am going to toss a coin, you will say the probability of heads is 1/2. That's fine.

Then, I toss the coin and look at it but don't tell you. Then I ask you again. You will still say 1/2.

That's the best you can do, but your information is now out of date. I know what the coin is and, to me, your answer of 1/2 is no longer valid. Or, at least no longer up to date.

Now, I could tell you or show you what the coin is and you would be forced to change your credence from 1/2 to 0 or 1.

Alternatively, I could give you some clues about what the coin is. And, using my clues you might change your credence from 1/2 to something else, based on how big a clue I give you.

This is perhaps easier to see with a die than a coin. I could tell you it's an even number and you would change your credence that it is a 6, say, from 1/6 to 1/3 etc.

The real point of this puzzle is to work out whether there are any clues in the experiment for Beauty to pick up on.

The halfers say there are no clues, so they stick with 1/2.

The thirders, however, are much better at spotting clues, so we find these clues and change our credence to 1/3.
 
  • #164
stevendaryl said:
Once again, if instead of wiping her memory in the case of heads, suppose they wipe her memory on Monday regardless, but if the coin flip was tails, her memory is restored 10 minutes after wakening.

Then before the 10 minutes is up, do you agree that Sleeping Beauty has to consider all four of the following possibilities equally likely?
  1. It's Monday, and the result was heads.
  2. It's Monday, and the result was tails.
  3. It's Tuesday, and the result was heads.
  4. It's Tuesday, and the result was tails.
No. Case 3 never enter the possibilities.
stevendaryl said:
After the 10 minutes is up, she knows whether she is in case 4 or not. If case 4 is ruled out, then what are the updated probabilities for 1-3?
Probabilities of coins flipping does not update with knowledge. If she have to bet, she will not change her bet anyway. There is a specific wining strategy in all cases.
I think the halfers are aware that "being awake AND be Tails" is twice that of "being awake AND Heads".
Just change the question so it exclude the confusion with: "being awake AND be the SAME tail as yesterday"
 
  • #165
Boing3000 said:
No. Case 3 never enter the possibilities.

What do you mean? Are you saying that case 3 never happens, or that it has no associated likelihood?

Probabilities of coins flipping does not update with knowledge.

That is just wrong. When you acquire new information, you update the likelihood of what is true.

If heads versus tails affects what Sleeping Beauty sees, then she can update her estimate of the likelihood based on what she sees. That's the way probabilities work. If someone flips a coin, and they win a prize for heads, and they see the coin but you don't, you can update the likelihood that it is heads by whether they smile or frown. If you know that your uncle just bought a $100,000 sports car, then your estimate of the likelihood that he won the lottery would go up.

The subjective likelihood of an event in the past changes depending on what you know. Are you disagreeing with that? You think that since winning the lottery has a 1 in a million chance, that even if your uncle tells you "I just won the lottery", you won't believe him?
 
  • #166
Why can't you answer the question:

In the case where the memory restoration happens 10 minutes after Sleeping Beauty wakes up on Tuesday if the result was tails, do you agree that before the 10 minutes is up, that Sleeping Beauty would consider all 4 of the following to be possible?
  1. It's Monday, and the result was heads.
  2. It's Monday, and the result was tails.
  3. It's Tuesday, and the result was heads.
  4. It's Tuesday, and the result was tails.
 
  • #167
PeroK said:
Now, I could tell you or show you what the coin is and you would be forced to change your credence from 1/2 to 0 or 1.
No. Only me looking at the coins (both side to be sure it is fair) will turn my credence into event/certitude.

PeroK said:
Alternatively, I could give you some clues about what the coin is. And, using my clues you might change your credence from 1/2 to something else, based on how big a clue I give you.
I understand that perfectly. And that have nothing to do with coins flipping. It as everything to do with coins "flipping AND more...."

PeroK said:
The real point of this puzzle is to work out whether there are any clues in the experiment for Beauty to pick up on.
And yet the very layout of the experiment goes out of his way to drug Beauty so has to be sure "there is no clues" to be gained in the experiment. I make it clear in my first post that there is not even the need to run it. Unless I misunderstood the usage of drug. Was it suppose to make here forget only about the day, or about the whole setup explanation ?

PeroK said:
The halfers say there are no clues, so they stick with 1/2.
Indeed, the halfers says that, as well as the very definition of the problem. And they are dam right about 1/2 of Head actual counts.

PeroK said:
The thirders, however, are much better at spotting clues, so we find these clues and change our credence to 1/3.
There are not clues to spot. All data are present before hand, and there is 2/3 chances to be awake on tail.

I say: both group don't answer the same question.
 
  • #168
Boing3000 said:
Probabilities of coins flipping does not update with knowledge. If she have to bet, she will not change her bet anyway.

That doesn't make any sense. In the revised experiment (on Tuesday, if the result was tails, her memory is restored 10 minutes after she wakes), she wakes up knowing that she's in one of 4 situations:
  1. Heads and Monday
  2. Heads and Tuesday
  3. Tails and Monday
  4. Tails and Tuesday
If you asked her to bet on which situation is true, what odds would she take? I can't see any reason for her not to give all four possibilities equal likelihood.

Now, in case #4, she finds out that she is in case #4. In cases 1-3, she finds out only that she is not in case #4. Now, what likelihood would she give to each of the possibilites 1-3? How much would she be willing to bet on each?
 
  • #169
And yet the very layout of the experiment goes out of his way to drug Beauty so has to be sure "there is no clues" to be gained in the experiment.

But that's not true. When she wakes up, she finds out that she is not in situation #4 (it's Tuesday, and the result was tails). (edit: I originally said "heads")
 
  • #170
stevendaryl said:
Why can't you answer the question:

In the case where the memory restoration happens 10 minutes after Sleeping Beauty wakes up on Tuesday if the result was tails, do you agree that before the 10 minutes is up, that Sleeping Beauty would consider all 4 of the following to be possible?
  1. It's Monday, and the result was heads.
  2. It's Monday, and the result was tails.
  3. It's Tuesday, and the result was heads.
  4. It's Tuesday, and the result was tails.
Beauty has been explained all the possibilities and 3 is not one of them.
From wikipedia: " if the coin comes up heads, Beauty will be awakened and interviewed on Monday only"

stevendaryl said:
That is just wrong. When you acquire new information, you update the likelihood of what is true.
I am told that a fair coins is 1/2 chance to be head. And frankly, I tend to believe it, because it matches my experiences so far.
Then I am told that same coins have just run 1 millions head in a row. Now, I know it is unlikely, but that newly acquired information change nothing. If that coin is fair then it is still 1/2.
 
  • #171
stevendaryl said:
But that's not true. When she wakes up, she finds out that she is not in situation #4 (it's Tuesday, and the result was tails). (edit: I originally said "heads")
Maybe we haven't read the same setup. Mine if from wikipedia, and she cannot finds anything by construction of the experiment:
wiki: "Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Beauty is asked: "What is your credence now for the proposition that the coin landed heads?".
 
  • #172
Boing3000 said:
Beauty has been explained all the possibilities and 3 is not one of them.
From wikipedia: " if the coin comes up heads, Beauty will be awakened and interviewed on Monday only"

Hmm, I think I may have gotten heads and tails confused, in that case. The way I have been assuming is that
  • If it's heads, she is woken up on Monday and Tuesday
  • If it's tails, she is woken up on Monday
  • And the twist: If it's tails, of course she will wake up on Tuesday, but she will wake up knowing that it is in fact Tuesday.
I am told that a fair coins is 1/2 chance to be head. And frankly, I tend to believe it, because it matches my experiences so far.

Yes, but if I further tell you that when it lands heads, I dance a jig, and I don't dance a jig, then your likelihood changes. Likelihood of the truth of something changes based on new information. That's what probability is good for---taking into account new information.
 
  • #173
Boing3000 said:
Maybe we haven't read the same setup. Mine if from wikipedia, and she cannot finds anything by construction of the experiment:

But if she doesn't know what day it is, then that is itself information, because she would know what day it was if it were Tuesday and the coin toss had been a particular result (whichever result causes her memory not to be wiped).
 
  • #174
Boing3000 said:
No. Only me looking at the coins (both side to be sure it is fair) will turn my credence into event/certitude.

.

You are misunderstanding the point of the problem. We are not discussing here the situation with conjurers, liars or unfair coins. That is a completely different problem altogether.

In that case, none of the answers or analyses given here are valid.

If a stage magician tossed a fair coin there is a finite probability that it would come up as a playing card showing the ace of spades.
 
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  • #175
stevendaryl said:
Hmm, I think I may have gotten heads and tails confused, in that case. The way I have been assuming is that
  • If it's heads, she is woken up on Monday and Tuesday
  • If it's tails, she is woken up on Monday
  • And the twist: If it's tails, of course she will wake up on Tuesday, but she will wake up knowing that it is in fact Tuesday.
OK, but I suppose we'd better stick to the actual Beauty "problem" which is not what you describe. And btw the "twist/drug" is precisely that beauty will NOT be able to tell what day it is (or if it is the millionth time she is woken (well if there is no mirror in the room at least o:)))

stevendaryl said:
Yes, but if I further tell you that when it lands heads, I dance a jig, and I don't dance a jig, then your likelihood changes.
Not at all. Only the likely hood that you dance a jig increase (even though I am quite sure you are a happy guy on average :wink:). And the information don't concerns coins tossing, but you dancing. You make "a promise" to me (that I would assign some probability you would respect)

stevendaryl said:
Likelihood of the truth of something changes based on new information.
That's a agreeable statement but it is unlikely to be true. The precise exact (mathematical) correlation between information is what truth means for me.

stevendaryl said:
That's what probability is good for---taking into account new information.
In my book, probability is the tool to manage absence of information. Everybody knows there is no randomness in coins tossing (the lower bound of that ignorance is chaos) and only our absence of knowledge of the "exact" way the coin is tossed, force us to use probability tool to manage the truths about the situation.

New information (like dancing, or wakening) added linearly to the consequences of that original "fuzziness" don't change that original "fuzziness"
 
  • #176
PeroK said:
You are misunderstanding the point of the problem. We are not discussing here the situation with conjurers, liars or unfair coins. That is a completely different problem altogether.
I am not trying to be facetious here. I am the once trying to stick with the exact definition of the "dilema".

PeroK said:
Now, I could tell you or show you what the coin is and you would be forced to change your credence from 1/2 to 0 or 1.
I don't understand the purpose of such sentences. Telling me that a coins not tossed is 1/2 is identical to telling me the once tossed is is A or B. Observing it is some state A or B does no change credence, it deletes credence, it remove the very notion that I must use probability anymore.

I am going to op-out of this thread, because I cannot even get you to acknowledge that this problem contains the very hard assertion that Beauty cannot know what days it is (by usage of a drug).

I understand both halfers and thirders point of view. They disagree because they can't even recognize that they use different frame of reference (the lab vs Beauty)
 
  • #177
Boing3000 said:
OK, but I suppose we'd better stick to the actual Beauty "problem" which is not what you describe. And btw the "twist/drug" is precisely that beauty will NOT be able to tell what day it is (or if it is the millionth time she is woken (well if there is no mirror in the room at least o:)))

But that's not true. She knows that the combination: It is Tuesday and the coin toss was (whatever result does not result in a memory wipe) did not occur. So the fact that she doesn't know the day is itself information, which the halfers are not taking into account.

Not at all. Only the likely hood that you dance a jig increase (even though I am quite sure you are a happy guy on average :wink:). And the information don't concerns coins tossing, but you dancing. You make "a promise" to me (that I would assign some probability you would respect)

That's just wrong, but it's a confusion about the use of probability having nothing specifically to do with Sleeping Beauty.

If events are correlated, then knowing about one event gives you information about the other event. If a coin was flipped, you don't know the result, so you assign it a likelihood of 50/50. But looking at clues, you can find out more information about the result. More information means less uncertainty about the result. In the extreme case, you actually look at the coin. After that, your uncertainty about the result is zero, and you know either that it is heads or that it is tails.

There is no point in discussing Sleeping Beauty probabilities if you don't understand (or don't accept) the basics of probability theory and probabilistic inference.
 
  • #178
stevendaryl said:
But that's not true. She knows that the combination: It is Tuesday and the coin toss was (whatever result does not result in a memory wipe) did not occur. So the fact that she doesn't know the day is itself information, which the halfers are not taking into account.
Now you lost me. Yes, she knows the combination, and I already point out that going true the experiment change nothing about that knowledge.
So why bother with that memory wipe ? (which I hope you don't deny, because it is part of the problem definition)
The only reason is to project Beauty in a universe only made of Monday, while the universe of the lab is made of Monday and Tuesday.
If not, there is no reason to specify that criteria in that problem. You cannot just wave that fact away. Halfers have every rights to use that criteria to compute her probability.

stevendaryl said:
If events are correlated, then knowing about one event gives you information about the other event.
I cannot seem to be able to explain to you that 's not what I disagree with. I don't deny correlation, and I feel I am the one taking them seriously. You dancing does not change the original event probability. Establishing correlation downstream of events does not trickle up. That's all I am saying.
I don't even understand why you would bring up such bizarre correlation and passing them for information gathering. Is there not many other probabilities you start dancing ? Why "complicate" things in such a way ?
I say that seeing you dancing is an information about "coins flipped AND promise". You cannot leave out the correlation in both math and natural language, that's too many shortcuts.

In sleeping Beauty's problem the asymmetrical correlation happens BEFORE she is even put to sleep. In that lab-frame 1/3 is the probability to wake up on Monday.
The confusion in that problem is then to proceed the wipe. If you take this element as meaningful/seriously then Beauty always wake up with only Sundays memory. So it is always Monday for her.

stevendaryl said:
If a coin was flipped, you don't know the result, so you assign it a likelihood of 50/50. But looking at clues, you can find out more information about the result.
I agree 100%. But waking up isn't a clue for Beauty. She has strictly no more information to gather AFTER waking up.

stevendaryl said:
There is no point in discussing Sleeping Beauty probabilities if you don't understand (or don't accept) the basics of probability theory and probabilistic inference.
I don't accept that I don't understand them. I am one of the 3 peoples that actually agree with both answers. So we are finished here.
 
  • #179
Boing3000 said:
Now you lost me. Yes, she knows the combination, and I already point out that going true the experiment change nothing about that knowledge.

When she wakes up in the morning, the fact that she doesn't know whether it's Monday or Tuesday tells her something. It tells her that either it's Monday, or that the coin toss was (whichever one results in a memory wipe).

I agree 100%. But waking up isn't a clue for Beauty. She has strictly no more information to gather AFTER waking up.

But that's not true. She doesn't cease to exist on Tuesday in either case. It's just that if it's Tuesday, and the coin toss was (whichever result does not result in a memory wipe), then she will know what day it is. Therefore, if she doesn't know what day it is, then she learns either that it's Monday, or that it's Tuesday and her memory has been wiped.
 
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  • #180
It certainly doesn't do the halfers any good to justify their answer by rejecting the whole concept of probabilistic reasoning.

The way that probabilistic inference works is this:

If A is some random event (a coin toss, a lottery result, etc.), and B is an event whose likelihood is affected by A, then probabilistic inference works out this way:

  1. Let P(A) be the a priori likelihood of A (1/2 in the case of a coin, maybe 1 in a million for the case of winning the lottery).
  2. Let P(B|A) be the likelihood of B when A is true.
  3. Let P(B| \neg A) be the likelihood of B when A is false.
  4. Then if you see that B is true, you can compute a probability P(A|B) for the probability that A is true given that B is true. P(A|B) = \frac{P(B|A) P(A)}{P(B|A) + P(B|\neg A)}
So for example, A might be "Your uncle wins the lottery". B might be "Your uncle buys a $100,000 sports car". Suppose:
  • P(A) = \frac{1}{10^6}
  • P(B|A) = \frac{1}{2}
  • P(B|\neg A) = \frac{1}{10^6}
So it's very unlikely that your uncle would ever buy a sports car if he doesn't win the lottery, but if he does win the lottery, he has a much greater probability.

Then if you see your uncle driving in a newly purchased sports car, you can figure that there is a signification likelihood that he won the lottery:

P(A|B) = \frac{\frac{1}{2} \cdot \frac{1}{10^6}}{\frac{1}{2} \cdot \frac{1}{10^6} + \frac{1}{10^6} \cdot (1 - \frac{1}{10^6})} \approx 0.33

So the likelihood that your uncle won the lottery is updated from 1 in 10^6 to 1 in 3
 
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