Sleeping Beauty

Hi everyone, it is a while since we discussed maths paradoxes on LG and I value the people here and their way of thinking.

I hope I don't start a flaming war, I would just like for people to tell if they are more in the 1/3 camp or in the 1/2 camp.

Here it is from Wikipedia:

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake:

• If the coin comes up heads, Sleeping Beauty will be awakened and interviewed on Monday only.
  
• If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday.
  

In either case, she will be awakened on Wednesday without interview and the experiment ends.

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 Sleeping Beauty is asked: "What is your credence "Credence (statistics)") now for the proposition that the coin landed heads?"

I'm not really sure about the meaning of "credence", but I undestand that as her view on the probability of the coin distribution of heads and tails.

I'am in the 1/2 group which is said to be a minority, and I'm starting to see why some people think it is 1/3, but I don't know anyone defending this view.

Let's suppose that we run that experiment 1000 times.  Then she will be woken up 1500 times, 500 "heads" and 1000 "tails".  Thus I'm in the 1/3 camp.

I feel like the answer to this is actually not the same based on how many times this experiment has been run.

Like bill said, if you run it multiple times then for every 2 runs of experiments on average you get 3 awakenings, 2 of which happen in the case of tails so if you randomly pick one of the awakenings the chance is higher that it's one of the 2 awakenings in the tails branch. With enough experiments undertaken we know that all of these awakenings in both branches exist and so the question becomes "when did we awaken the sleeping beauty this time around" to which the answer is that it's more likely a time the coin landed tails.

However for a single experiment I feel like the above logic does not apply because really there only happened one coin landing and it's either on heads or on tails. We get 2 possible realities, in 1 of which the coin landed heads and in the other of which it landed tails. And then if you ask the sleeping beauty the same question i feel like there is a 50% chance it was on Monday after the coin landed heads, a 25% chance it was on Monday after the coin landed tails and a 25% chance it was on Tuesday after the coin landed tails.

So I have to take position in the 1/2 camp.

When Sleeping Beauty is being woken up for an interview, she knows that there is a 2/3 chance that it is Monday and 1/3 that it is Tuesday. If it is Monday, the chance of heads is 1/2; if it is Tuesday, the chance of heads is 0. This gives 2/3 * 1/2 + 1/3 * 0 = 1/3.

 When Sleeping Beauty is being woken up for an interview, she knows that there is a 2/3 chance that it is Monday and 1/3 that it is Tuesday.

This seems like an obvious application of the typical boy/girl probability problems that we’re all familiar with in introductory statistics, but I have some doubts here after thinking about it for a while as this does not completely parallel them. Could you please try to justify the statement above?

In the spirit of William Fraser's reply, in 1000 experiments she will be woken up on a Monday 1000 times and on a Tuesday 500 times.

As a side note, even if one disagrees with the probabilities for Monday and Tuesday, the answer to the original question cannot possibly be 1/2. This could only happen from my approach if the probability for Monday is 1, which it clearly isn't.

I believe that my case can be made without invoking repeated trials (as is usually the case with probability):

Let's consider at what odds she would be willing to accept a wager.  Suppose that she is told, before the experiment starts, that she can place an even bet each time she is awakened.  If she chooses to accept that bet (and picks tails) then 50% of the time she will lose $1 and 50% of the time she will gain $2.  This implies that the even bet is not fair.

Actually, the 1/2 camp isn't as easy to dismiss as that.  If we accept _syLph_'s reasoning, then she will believe that the odds that it is Monday is 75% (not 100%), so it is not ruled out quite so automatically.

Well sure, if someone told me they are gonna flip a coin and if it's heads they give me $1 once while they take $1 twice if it's tails then I surely wouldn't accept. That doesn't make the coin flip not 50/50 though.

There is also a variation (SB1000) when SB is woken 1000 times in a row if Tails and only once on Monday if Heads.

Thirders have the probability of Heads going to 1/1001 while halfers still keep 1/2...

> Actually, the 1/2 camp isn’t as easy to dismiss as that.  If we accept syLph's reasoning, then she will believe that the odds that it is Monday is 75% (not 100%), so it is not ruled out quite so automatically.

That is an important point to settle. If the interviewers don't say which day it is, but she spots the date on a screen or something and _knows_ that it must be Monday, wouldn't she have to go with 1/2?

Assuming no one has intentionally provided this information.

 Let’s consider at what odds she would be willing to accept a wager.  Suppose that she is told, before the experiment starts, that she can place an even bet each time she is awakened.

but IMO, that’s answering a completely different question because you’re intentionally giving her more chances to make a bet if the coin ends up as tails, and if you interpret it that way, then of course you’d win more frequently if you put your wager on tails. What we say that they’ll only allow her one bet (chosen at random if she awakes on multiple days) count per trial? I think this interpretation is closer to what the question is asking for.

My sentence was messed up, so let me rephrase for clarity: she gets asked to make a bet each time she wanted up, but only one bet in one of the days (chosen at random) will count for each trial.

Alright.  I've got a variant, which may or may not shed some light on this.  Suppose that she is told the procedure is as follows.  Every day she will be woken up and instructed to think about what credence she gives the coin flip being heads.

If the coin is heads and it is not Monday (i.e. she would not have been awakened in the original formulation) 5 minutes will be allowed to pass, then she will return to sleep.

Otherwise, the following will happen:

After 1 minute, she will be given a sheet of paper and told to write down her credence that it is heads.

I think it is fairly clear that during the first minute, her credence is 1/2.  After that, if she is not given a sheet of paper, her credence rises to 100%.

Thus, assuming she is given a sheet of paper, her credence must fall and cannot be 1/2.  (And, in fact, a little not-very-contentious reasoning will show that it is 1/3).

The question then becomes "is this different from the original formulation?".  If so, why?

It would quite easy to get 2/3

On the first night she would make sure her pillow case opening is on the wrong side (what ever that is)

Consider two weeks, one heads, one tails

When she wakes and finds pillow case is wrong way she fixes it and knows it's Monday so 50% it's heads, this happens twice so 1-1

When awaking with pillow case correct way it's Tuesday so it must be tails, this happens once so 2-1

But how effective are amnesia drug, maybe need to be 95% to pass safety trials

Why do you think that it’s a continuous change from 50% to 100%? Giving her the paper might as well be an immediate signal that it’s Tuesday. Unless you’re invoking a smooth probability distribution for when you’re giving her that sheet of paper (which is obviously going to be changing the question), it will be a discontinuous change from the point of the signal.

I like the point Carroll brought up. In a variation in which SB is awakened not twice but 1000 times in case of tails, do thirders really believe that the chance is 1/1001 for heads? Remember that there is only 1 experiment and only 1 coinflip ever.

Actually, according to the wikipedia page, "Extreme Sleeping Beauty" (with 1,000,001 wakings, not 1000) was formulated by Nick Bostrom, who is a thirder.

@2021, I don't think she's allowed to run shenanigans with her pillowcase.  According to the original problem:

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.

I am in the 1/2 group. The coin has a 50% change to be heads and this doesn't change because of the number of awakenings.

Let's change a little the problem. If is heads she is awakened on Monday, if is tails she is awakened on Tuesday. This makes it obvious she should always say 1/2.

This means, in the original problem, that the chance to be a Monday awakening because of heads is 50% while all other awakenings, because of tails, have together a 50% chance. In this case 25% each.

As Carol said, if she is awakened 1000 times for tails, she will have a 50%/1000 chance to be in any of them.

@William Fraser I just read and thought about your paper variant.

I agree with the chance for heads being 1/3 when the paper is given out. Imo though that is indeed not the same problem as the original. Being given the paper is information that is indicative for not being in the heads branch, because in half of the scenarios in which you awaken there that doesn't happen. However in the original problem you don't acquire any information on which branch you have entered. Basically both the heads- and the tails-branch have 50% chance allocated to them and by giving the piece of paper you take away some of the 50% that was allocated to the Tuesday awakening of the heads-branch.

Actually hold up.

 Thus, assuming she is given a sheet of paper, her credence must fall and cannot be 1/2.  (And, in fact, a little not-very-contentious reasoning will show that it is 1/3).

 I agree with the chance for heads being 1/3 when the paper is given out.

I actually think it's 1/4 chance of heads if the paper is given out. Not 1/3

Alright.  Another angle.

Let's suppose that, after she has indicated her credence, she is going to be told what the day of the week is, then asked again.  If she is informed that it is Monday, does she suddenly become very confident that heads was flipped?

Actually forget about the 1/4. I'm just mega confuzzled and not good at math, but anyway pretty sure its not the same problem.

Anyway, is that new angle for the original problem? I'm of the opinion that P(Heads | Monday) = 1/2. Being told that it is Monday in the original problem doesn't tell you anything at all. There is a 50% chance you entered the heads-branch and a 50% chance you entered the tails branch and will awaken again on Tuesday.

Hmm..... I'm not seeing the 1/4, @_syLph_.  She wakes up in 1 of 4 equally likely states, and being given a sheet of paper tells her that one of those states is impossible but tells her nothing about which of the other 3 states it is.  So wouldn't it be 1/3 if given paper?

As to whether you acquire any information, I suppose it depends on whether "not being woken up" is information.  And the equivalence of the two scenarios is dependent on whether "being woken up but not remembering anything" communicates information.

Sorry.  I forgot to refresh and missed your reply.  Yes, the new angle is for the original problem.

I don't think you can have all of these be simultaneously true:

P(Heads) = 1/2

P(Heads & Monday) = 1/2

P(Heads | Monday) = 1/2

P(Monday) = 3/4

I do think they are actually all true. I'm sure there is some mathematical contradiction in there but luckily I'm not educated enough to see it.

P(Heads) = P(Heads & Monday) should be fairly obvious.

And of course I'm arguing those are 1/2.

P(Heads | Monday) = 1/2 like I said I believe to be the case since I don't see any relevancy in the info that it being Monday provides. If the 50/50 coinflip resulted in heads then you will reach a Monday and if the 50/50 coinflip resulted in Tails then you will also have reached a Monday by the end so it doesnt serve as anything to make sense of whatever happened.

P(Monday) = 3/4 is what I wrote earlier and I still feel like that's true

I feel like this could be interesting with a dice instead of a coin where its like 5 dice results that result in one awakening and 1 dice result (lets say if a 6 is rolled) that results in 100 awakenings.

Then would the halfers from the coin problem think it's 1/6 for the 6 to be rolled while the thirders believe it to be 100/105?

I'd agree with the 100/105 proposition.

I believe that am more likely to make choices that are advantageous if I act as though I am in a universe where a 6 was very likely to have been rolled than in a universe where a 6 was unlikely.

I believe that if I am told that it is Monday, this information tells me something that affects my confidence of the distribution of the die roll.

What if sleeping beauty is interviewed before the experiment and asked how likely she thinks a 6 will be rolled? Would that still be 100/105 or is that only the case if it's within the experiment? And if it matters when the question is asked why would that be the case?

Before the experiment, it would be 1/6.  It matters because the information that she is awake under the conditions given for the experiment is relevant.

OHHHHHHHHHHH I FINALLY UNDERSTAND THE ARGUMENT NOW. ITS "I WOULD PROBABLY BE ASLEEP RIGHT NOW IF A 6 WASNT ROLLED"

SDSGAHSDGHSDKJGSDJGL

Okay tbh that actually makes sense.

Your argument makes sense, too.  That's what makes this conversation so interesting....

i'm officially still not changing position from halfer to thirder

I got an epic variant. Instead of a fair coin we use an unfair coin that always lands on the same side but SB doesn't know which side that is, she just knows the coin is unfair and always lands on the same side.

I feel like that doesn't actually change the experiment at all, except now if you do the experiment a million times there has been either exactly 1 million awakenings in case of heads or 2 million in case of tails with 0 zero chance for anything inbetween.

What do thirders think about this, does that change anything? I assume probably not.

Because it's a fair coin you only have to consider both permutations

The 3 awakens over two weeks are:-

Week 1 Heads Monday

Week 2 Tails Monday

Week 3 Tails Tuesday

Typo   week 3 > week 2 (no week 3)

@Sylph interesting variant.

How do you proceed to chose whether it is a heads/heads coin or a tails/tails coin ? I guess the distribution on how you chose the coin should impact SB decision.

Are you still a halfer in your scenario, or a double halfer ?

Someone pointed me to a nice talk by Peter Winkler about this and other probability problems, he develops interesting arguments around 00:20, https://www.youtube.com/watch?v=N3KdhrF142Y

Eh, it was supposed to just be a 50/50 thing. I mean if we don't tell SB then she really has to assume it's 1/2 for both getting the unfair coin that only lands heads and the one that only lands tails. But if I have to come up with something then i prefer something stupid that you can't apply math on like a battle between superman and a perfect copy of himself both of who hold one of the coins with the winner surviving and handing over his coin upon return, but of course nobody actually knows who won because they fought on a distant planet and they couldn't tell if original superman returned and handed over his coin or if it was the copy since they act the exact same way etc etc. Anyway, that was fun to write and I think the point is clear.

I have actually been a double halfer in the original problem and this remains the same in the new variant. My argument is basically that the coin flip is the all-deciding event that determines which reality SB ends up in and unless you provide info that gives insight as to what occured in the coinflip you have to keep assuming the 1/2 chance. Being given the information that it is Monday does not tell SB anything about whether the coin landed heads or tails as being woken up on a Monday is an event that occurs in both cases.

That being said however, I have just stumbled across someones variation here which reads the following:

Person w/ 1 minute memory span is told the following:

We will flip a coin. If it's heads, you will spend 10 hours in a green room, and spend 10 minutes in a red room. If it's tails, 10 hours red, 10 minutes green. Due to his short memory this information is written down for him.

He finds himself in a green room. Is it not reasonable for him to assume it's much more likely heads was flipped than tails?

This is essentially the same as the Monday thing, but somehow my common sense keeps screaming at me in this case that it is absolutely more likely to find himself in a heads scenario if he awakens in a green room, which would then imply that P(Heads | Monday) is indeed 2/3.

To me this is just a nice story, but in the end it's just 1 coinflip with a fair coin. So the chance is 1/2.

I think the ambiguity comes from the fact we are looking at this experiment as outsiders. And when running experiments we know that sleeping beauty is being awakened on day 1, day 2. But in our beauty queen doens't know that.

Even if the experiment would last 1000 days, for the sleeping beauty it all comes back to 1 coinflip.

I have another strange variant.

After thinking about this more I now find myself agreeing with the thirder position a lot more. In particular I can now see why there is a difference between answering the question for the result of the coinflip before the experiment and during it. The way I understand the thirders position now is that the one time you would wake up within the heads branch is a very special day because it is the only day in the experiment that you wake up, while in the tails branch waking up is just a normal occurrence. After waking up you then find it more reasonable to assume that this day you wake up on is just another day and that the coin just happened to land tails so it's completely normal you woke up, cause how likely is it that you wake up and it happens to be your birthday right? Doesn't happen a lot.

So anyway, I now have this sorta strange variant to kinda challenge that logic I just described. In this variant the experiment is the same if the coin lands heads, but if it lands tails then the SB is cloned one additional time for each day after Monday. She knows of course that the cloning happens if the coin lands tails but upon waking up she doesn't know if the cloning has happened and whether she is the original SB or not. Now instead of her waking up once on Monday and once on Tuesday, she only gets waken up on Monday and on Tuesday instead her clone gets waken up and interviewed. For the extended version for an experiment longer than 2 days a different clone gets waken up every day.

I am curious to know what thirders think the awakening SB should answer, whichever version of her that is. Like before in the case of tails there is one awakening per day and it is a common occurrence that at any day some version of the SB awakens. However, unlike before whenever SB awakens it is that one special day in the experiment that THIS version of SB herself awakens.

1/2.

It absolutely doesn't matter what day it is, or how many times she's woken up.

It matters that a coin was flipped. Each time the coin is flipped, the coin will have a 1/2 probability to land on heads or tails.

Sleeping beauty should be aware of this fact. Assuming the coin was only flipped once, and that the coin is not somehow weighted unfairly.

The coin is fair, but on average, Sleeping beauty gets asked that question twice as often when it is tails than when it is heads.

Maybe best if Sleeping beauty stays in bed until covid 19 has gone (assuming it does)

https://www.youtube.com/watch?v=Ol18JoeXlVI

Just watched this video by Veritasium on the Newcomb's Paradox. I find it very similar to the sleeping beauty problem and was reminded of it, I would assume that halfers in the sleeping beauty problems should be two-boxers in the Newcomb's Paradox. Super fun to think about this stuff. I recall that when we were discussing sleeping beauty I was flipping a lot as I began to understand the argument of the thirders better. For the Newcomb's Paradox I'm fairly sure that I will keep insisting that picking both boxes is correct.

imagine if you were running this experiment on different people, lets make it so that if tails, the subject is woken up and put back to sleep 100 times and made to forget each time in between. Now you would find that the vast majority of the time you ask this question to a subject, those who guess tails are correct. I think this gives you an idea of the relevant probabilities here.

I like them both, and I guess I have changed my mind about both, without being sure how I could have thought differently in the past (I'm glad my past opinion is written black on white in this forum).

There is a good video by a french philosopher/youtuber called Mr Phi about it: Newcomb paradoxe explained with a variant where you may drink poison... I guess syLph will not drink it.

So my personnal feeling now despite being a past convinced 2-boxer is that if you want to be rich or advice a friend on his choice like in the video, you should try to convince yourself (and your dumb friend) to be a 1-boxer until you enter the room, and maybe even after.

It is funny to ask the AIs about their opinions about SB paradox, they are quite measured accepting that there are good arguments for both views, but finally both ChatGpt and Claude put themselves in the thirder camp.

The poison version of Newcombe's paradox was interesting. However, I had trouble thinking of a reward that was commensurate with drinking poison, which made it hard to treat the problem concretely.

Things that introduce even mild risks to overall health outweigh almost any amount of money for me. Things like "World Peace" came to mind as being large enough rewards, but I couldn't think of anything on that order that could be given by the person running the experiment (whether or not it would be a physical object that fit in a briefcase). Likewise, I briefly entertained the idea "I would refuse to drink, then not even bother to open the briefcase". I think in reality, I'd actually refuse the drink, open the briefcase, then not be disappointed that it was worthless.

When he went to the 100 Euro prize and a small shock, it was much easier, and I was clearly going to take the shock. I can picture the shock that would be the right size for that.