Sleeping Beauty Problem

Srap Tasmaner

How do you condition on such a thing? What values do you place into Bayes' theorem? — Michael

Just scroll back up. I went through all that.

Michael

I don't think your numbers are accurate there.

But in this case we're not asking about an outsider's analysis of frequencies over many experiments, but Sleeping Beauty's when it is known that just a single experiment is being run.

Srap Tasmaner

I don't think your numbers are accurate there. — Michael

Meaning?

But in this case we're not asking about an outsider's analysis of frequency, but Sleeping Beauty's when it is known that just a single experiment is being run. — Michael

And how do you expect to apply Bayes's rule without any base rate information? SB can reason as I have described to determine what those base rates would be were the experiment repeated a number of times, and set her subjective probabilities accordingly.

Michael

And how do you expect to apply Bayes's rule without any base rate information? SB can reason as I have described to determine what those base rates would be were the experiment repeated a number of times, and set her subjective probabilities accordingly. — Srap Tasmaner

We have the base rate information.

The probability of a coin landing heads is 0.5. The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. So:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Compare with Bayes' theorem as applied to my variation:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

It should be clear that the reasoning is sound in the second case, and so too is it sound in the first case.

Srap Tasmaner

The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. — Michael

But these are useless, uninformative, ambiguous categories. What you want is the odds that this interview is a heads-type interview. If Beauty could determine her current state in the world, what type of interview she is being given, she could answer with certainty. She cannot, but she can determine the likelihood of her being in each of the two possible states, states that she cannot otherwise distinguish. And she knows that it is more likely that she is being interviewed because it was tails.

Michael

What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner

Yes, that's the left hand side of the theorem that we're trying to solve: P(Heads|Questioned).

We use the known values on the right hand side to determine it.

Srap Tasmaner

What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner

Yes, that's the left hand side of the theorem: P(Heads|Questioned). — Michael

No, that's the probability that the coin was heads given that I'm being interviewed. The trouble is on the RHS, the probability that I am being interviewed given that my coin was heads.

Take a space and partition it evenly into a heads half and a tails half. Now, in the heads half put one point, an interview; in the tails half, put many interviews.

If you choose a half space, in each you will find at least one interview. You're interviewed either way.

But if you randomly select an interview among all the interviews, is it more likely to have come from the heads half, with its one interview, or the tails, where there are more?

That's what we want to capture with the base rate of heads interviews among all interviews, not the non-emptiness of the heads half of the space.

Michael

The trouble is on the RHS, the probability that I am being interviewed given that my coin was heads. — Srap Tasmaner

Which is 1. I know that I will be interviewed if the coin lands heads.

Consider a simpler version of the experiment. If heads then I will be interviewed once. If tails then I will be interviewed once. Bayes' theorem is unproblematically:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

We don't say that because half of all questions are when tails then P(Questioned | Heads) = 0.5. That would give us this very clearly wrong calculation:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over2}*{1\over2}}\over1}\\&={1\over4}\end{aligned}$

hypericin

In our version, the base rate of heads interviews is 1 in 3. Make it 1 in 1000. (That is, 999 awakenings on tails, not 2.) Isn't it obvious that if I'm a subject in such an experiment, I know it's far more likely I'm being asked for my credence because my coin came up tails? If I'm one of 1200 subjects, I know there are 600,000 interviews, only 600 of which were for heads, while 599,400 were for tails. Equally likely that this interview is for heads as for tails? Not by a long shot. — Srap Tasmaner

This problem is like one of those optical illusions, a drawing that can be interpreted in two ways, but you can only hold one interpretation in your head at once.

It is easy to show the probability of heads is 1/3, if you allow multiple trials. I do so in my op. But the problem limits to one trial. In one trial, each of those 1000 interviews occurring is contingent on the coin flip being tails. So the probability that SB experiences all 1000 of them is 1/2, and the probability that SB is experiencing any one of them during an interview is 1/2000. While, the probability that SB is experiencing the heads interview is 1/2.

This is very easy to see if you make the coin toss unfair. Suppose heads comes up 99/100 times, and if tails, a million sleeps happen. Are you willing to believe, in a single trial, that tails is overwhelmingly likely on each awakening? Yet, if you extend to enough trials, tails is overwhelmingly likely.

hypericin

Following this logic, I wrote a python script which calculates the probability of heads given a number of trials. As I expected, the results begin at 1/2 at 1 trial, and with increasing trials converge to 1/3. This unifies the two answers in a way I haven't seen before.

Unfortunately for thirders, the problem is explicitly concerned with the 1 trial case. Halfers were right all along. As far as I'm concerned, the problem is put to rest for all time.

import numpy as np
import matplotlib.pyplot as plt

def experiment(trials):
    headsProbability = 0.0
    for heads in range(trials + 1):
        headArrangements = np.math.factorial(trials)/np.math.factorial(trials-heads)/np.math.factorial(heads)
        headsRightAnswer = heads / (heads + (trials - heads)*2.0)
        headsProbability += headArrangements / np.power(2.0, trials) * headsRightAnswer
    return headsProbability

trials = [i+1 for i in range(50)]
results = [experiment(t) for t in trials]

plt.plot(trials, results, '-')
plt.ylabel("heads probability")
plt.xlabel("trials")
plt.show()

((This is, I don't know, maybe the third time I've argued with Michael about something and then concluded he was right all along.)) — Srap Tasmaner

Sorry, Srap, it happened yet again.

↪Michael

was right.

Michael

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

…

Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment.

And given that if woken the coin landed heads iff he is 1, he ought to have a credence of P(Heads) = 1/3. — Michael

Does it not stand to reason that the probability that the coin landed heads in this example is less than the probability that the coin landed heads in the original example, given that in this example the coin landed heads iff Michael was assigned the number 1?

Do you agree that Michael’s credence that the coin landed heads is 1/3?

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

If so then it must be that Sleeping Beauty’s credence in the original example is greater than 1/3, i.e 1/2.

Michael

An interesting variation taken from here:

Four volunteers will be assigned a random number but each will undergo an experiment that is functionally equivalent to the popular version of the problem. The same sleep and amnesia drugs will be used, and each will be awoken at least once, but maybe twice, based on the same fair coin toss. Only their schedules and the question they are asked will differ, but end up being equivalent to the popular problem. On Monday and Tuesday:

#1 Will be awoken unless it is Tuesday, after Heads.

#2 Will be awoken unless it is Tuesday, after Tails.

#3 Will be awoken unless it is Monday, after Heads.

#4 Will be awoken unless it is Monday, after Tails.

Each will be asked for their credence that this is the only time they will be awoken. For #1 and #3, that means credence in Heads. For #2 and #4, it is credence in Tails. For all four, the answer has to be the same as the correct answer to the popular version of the Sleeping Beauty Problem.

On each day, we can bring the three awake volunteers together to discuss their answers. Of these three, exactly one will not be, or was not, awakened on the other day of the experiment. But none of the three can have more, or less, credence that she is that one instead of one of the others.

So with three awake volunteers, one of whom will be awakened only once, the answer is 1/3.

To be clearer with what each volunteer is considering, it is:

P(1 and Heads or 2 and Tails or 3 and Heads or 4 and Tails | Awake)

Prima facie the answer is $1\over2$ , however it is a fact that for $1\over3$ of the awake volunteers, "I will only wake once" is true.

The question, then, is whether or not A entails B:

A. "I will only wake once" is true for $1\over3$ of us
B. The probability that "I will only wake once" is true for me is $1\over3$

Thirders say it does, halfers say it doesn't.

I think this might be a Monty Hall problem. Consider a slight variation which I think is functionally equivalent to the above. All 4 are awoken on each day, are put in a room together, and then one of them is put back to sleep according to the rules. Each person left awake is then asked to consider the probability that they will be awake both days. I think it's a mistake to ignore the person who is put back to sleep, who is comparable to the door Monty opens. It was $1\over2$ before seeing someone get put to sleep, so how does seeing someone get put to sleep make it more or less likely that I will be awake both days?

So if this is equivalent to the original problem then the original problem is a Monty Hall problem as well.

sime

Always start by writing down the probability of everything :

Variables :

C = Tossed coin (Binary variable in {head, tail} )
D = Awoken Day (Binary variable in {mon, tue} )

Unknown Probability Measure :

P (C, D) = P (C | D) P(D) = P(D | C) P(C)

Constraints given on the Probability Measure :

P (C = head) = 1/2
P (D = mon | C = head) = 1
P (C = tail | D = tue) = 1

Task :

Determine the marginal distribution P (C = head ) from the above premises

Answer :

P(C = head) = 1/2 (by premise)

Why was the scientific american wasting time on this?

Michael

Why was the scientific american wasting time on this? — sime

Self-locating belief and the Sleeping Beauty problem, Elga 2000

Sleeping Beauty: reply to Elga, Lewis 2001

fdrake

As a general rule, simulating probabilities doesn't resolve disputes about which computations are appropriate. Every randomness has a generating mechanism. Addressing these paradoxes means addressing the generating mechanism.

Demonstrative example:

1) The moon is made of cheese or eggs
2) both are equally likely
3) the probability the moon is made of cheese given by the following R script:

mean(rbinom(10000,1,0.5))

So the moon is made of cheese with 50% probability.

It's just right, look at the code!

All coding something up does is let you check a calculation for it being correct, not whether it's the appropriate calculation to do.

sime

↪Michael

Thanks.

The premises of the Sleeping Beauty, at least in some popular incarnations of the problem, specify an experimental design, but they don't specify or assume how temporal self-location is related to that design. The design only specifies a few constraints relating the physical outcome of a coin toss C to the actual days of the week D that a subject is awoken. But an awoken subject's mental state M comprises yet another variable that has no specified relationships to C or D in the question's premises.

The question is literally asking, on the basis of a partial definition of (C,D) what is the marginal distribution of P(C), which it's premises already state is 1/2. So if it is intending to probe the valid philosophical question pertaining to self-location it needs to be more candid and ask

What is the probability P (C = h | M) ?

But if nothing about M and it's relationships to C and D are specified or assumed, one isn't in a position to assign conditional probabilities, and can only respond with the vacuous answer " P(C = h | M) is between 0 and 1. "

hypericin

It's just right, look at the code!

All coding something up does is let you check a calculation for it being correct, not whether it's the appropriate calculation to do. — fdrake

You seem to suggest I just arbitrarily whipped up some code and said "hey guys, code! problem solved!"

The program calculates the probability a wakening is heads, with a given number of trials.
Here are some examples:

1 trial:
Two possibilities, 0 and 1 heads
0 Heads: (does not contribute to likelihood)
1 Heads: 1/2 chance, all awakenings are heads = 1/2
Answer: 1/2

2 trials:
Three possibilities, 0, 1 and 2 heads
0 Heads: (does not contribute to likelihood)
1 Heads: 1/2 chance, 1/3 awakenings are heads = 1/6
2 Heads: 1/4 chance, all awakenings are heads = 1/4
Answer: 1/6 + 1/4 = 10/24 = ~.417

3 trials:
Four possibilities, 0, 1, 2 and 3heads
0 Heads: (does not contribute to likelihood)
1 Heads: 3/8 chance, 1/5 awakenings are heads = 3/40
2 Heads: 3/8 chance, 1/2 awakenings are heads = 3/16
3 Heads: 1/8 chance, all awakenings are heads = 1/8
Answer: 1/8 + 3/16 + 3/40 = ~.3875

If I got something wrong, it is quite an odd coincidence that the program captures both answers: 1/2 at one trial, and 1/3 at N trials.

fdrake

You seem to suggest I just arbitrarily whipped up some code and said "hey guys, code! problem solved!" — hypericin

Not what I intended, apologies for any offence caused. I whipped up some code myself to illustrate different scenarios in another thread.

hypericin

Task :

Determine the marginal distribution P (C = head ) from the above premises — sime

[/quote]

No, the question is what is the probability SB experiences an awakening with the coin being heads.

sime

No, the question is what is the probability SB experiences an awakening with the coin being heads — hypericin

Unless additional premises are included in the problem, I cannot tell the difference between how I phrased the question and your phrasing of the question.

Upon being awoken, SB is asked: "What is your credence now for the proposition that the coin landed heads?" "

She is therefore being asked "What is P(C | M) , where M is your current mental state?"

But the premises don't specify anything regarding the correlation of SB's mental state M to either the state of the coin C or the day of the week D.

So we can only interpret P (C | M ) as referring to P(C) .

Srap Tasmaner

This is a quote from Descartes posted by @Fooloso4 next door in that thread:

For a life-span can be divided into countless parts, each completely independent of the others, so that from my existing at one time it doesn’t follow that I exist at later times, unless some cause keeps me in existence – one might say that it creates me afresh at each moment.

We can say of the sleeping beauty problem, just as a way of beginning, that there are three possible future SB slices which the SB being interviewed could be.

The halfer camp seems to take the view that two of those belong to one person, and one to another, and that the correct analysis is that the coin flip partitions SB's future slices into a heads set and a tails set, just two, equal chances of being in each set.

The thirder view is that only the current slice that you might be is relevant, and there are more being-interviewed slices in the tails partition, so you're more likely one of those.

It may not be true that I am more likely to be a person who is interviewed more than once, and yet be true that this slice of me being interviewed is more likely to be a somewhat common tails slice, considering the entire pool of possible slices of me, than to be a comparatively rarer heads slice.

Srap Tasmaner

For an analogy: it's 11 am, and you're asleep in bed; how old are you? Odds are you're one of the ages where you were more often asleep in bed at 11 am, even if every year of your life has had at least one such day.

Dawnstorm

The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. — Michael

How do you get 1? How do you interpret P(Questioned)?

I get 3/4. There are two variables (Coin and Day), and two checks, for four outcomes altogether, three of which result in "questioned".

Similarly, the likelihood that the current situation is one in which we're both questioned and the coin comes up heads is 1/2.

I don't know how to do formulae, but I get the following:

1/3 = (1/2*1/2)/(3/4)

Which checks out. I mean, test it all out on the event space:

Heads and Monday = Not Questioned
Heads and Tuesday = Questioned
Tails and Monday = Questioned
Tails and Tuesday = Questioned

Likelihood to be questioned:

Heads and Monday = Not Questioned
Heads and Tuesday = Questioned
Tails and Monday = Questioned
Tails and Tuesday = Questioned

Likelihood to be questioned when Heads:

Heads and Monday = Not Questioned
Heads and Tuesday = Questioned
Tails and Monday = Questioned
Tails and Tuesday = Questioned

Likelihood to be Heads when Questioned:

~~Heads and Monday = Not Questioned~~
Heads and Tuesday = Questioned
Tails and Monday = Questioned
Tails and Tuesday = Questioned

I mean, I'm no mathematician. But this, at the very least, makes sense to me. Have I gone wrong anywhere?

Michael

How do you get 1? — Dawnstorm

Because Sleeping Beauty is certain to be questioned during the experiment.

Compare with my alternative scenario here where the probability of being questioned is 1/2.

How do you interpret P(Questioned)? — Dawnstorm

"My credence that I will be questioned."

Dawnstorm

This is messing with my head. I tried to reply but couldn't get a coherent post going.

hypericin

She is therefore being asked "What is P(C | M) , where M is your current mental state?" — sime

No, the question has nothing whatsoever with her mental state.

She is being asked, given that she is awakened, what is the probability of heads. If she is awakened 1000 times for every tails and once with heads, given enough coin flips you can see that it is overwhelmingly likely to be tails, even though the probability of heads remains 50%. This is independent of her mental state.

hypericin

The thirder view is that only the current slice that you might be is relevant, and there are more being-interviewed slices in the tails partition, so you're more likely one of those. — Srap Tasmaner

This relies on the intuition of repeating the experiment over and over. If so, then there are unconditionally more tail slices. But the coin is flipped exactly once. Therefore, even though there are more tail slices, they both exist only upon a tails flip. Therefore,

the correct analysis is that the coin flip partitions SB's future slices into a heads set and a tails set, just two, equal chances of being in each set. — Srap Tasmaner

Is the correct one.

sime

No, the question has nothing whatsoever with her mental state.

She is being asked, given that she is awakened, what is the probability of heads. If she is awakened 1000 times for every tails and once with heads, given enough coin flips you can see that it is overwhelmingly likely to be tails, even though the probability of heads remains 50%. This is independent of her mental state. — hypericin

Her "mental state" M refers to her epistemic state upon waking. In the context of repeated trials, it would include her knowledge of previous trials. But the paradox doesn't assume repeated trials. If it did in the manner of your example, then

P ( C = tails |M) = 1000/1001

In which case, her mental state has proven to be a very reliable indicator of the state of the coin.

But this isn't part of the question's premises. We are told that in the context of a single trial, that her mental state is conditioned so as to have no previous memories of waking up. This condition implies that her mental state is statistically independent to the state of the coin, i.e.

P(C | M ) = P(C) = 1/2

Here i have assumed that she is told that the physical probability of the coin landing heads is 1/2. If she isn't told this, then she should refrain from assigning any distribution to P(C | M)

Srap Tasmaner

This relies on the intuition of repeating the experiment over and over. If so, then there are unconditionally more tail slices. But the coin is flipped exactly once. Therefore, even though there are more tail slices, they both exist only upon a tails flip. — hypericin

I think it's hard.

If you wake up and ask yourself "Which of the selves I could have become while I was asleep am I?" there are two ways to take that.

You could think of the self you're seeking as a continuous self (a sort of world-line, if I have the parlance right) accumulating experiences, in which case it's reasonable to describe that self from any point along the line: in the past you're the self who's going to end up having exactly one or exactly two interviews; in the future, you're the self that had exactly one or exactly two interviews.

But you could also think of your self as this momentary self, even if it is part of a larger stream of momentary selves. Your question then is, which stream is this self-moment likely to be part of? If any such moment is more likely to be part of one stream than another, then you are more likely to be the self living out that stream than the other.

fdrake

The problem vexing the minds of experts is as follows: Sleeping Beauty agrees to participate in an experiment. On Sunday she is given a sleeping pill and falls asleep. One of the experimenters then tosses a coin. If “heads” comes up, the scientists awaken Sleeping Beauty on Monday. Afterward, they administer another sleeping pill. If “tails” comes up, they wake Sleeping Beauty up on Monday, put her back to sleep and wake her up again on Tuesday. Then they give her another sleeping pill. In both cases, they wake her up again on Wednesday, and the experiment ends.

The important thing here is that because of the sleeping drug, Sleeping Beauty has no memory of whether she was woken up before. So when she wakes up, she cannot distinguish whether it is Monday or Tuesday. The experimenters do not tell Sleeping Beauty either the outcome of the coin toss nor the day.

They ask her one question after each time she awakens, however: What is the probability that the coin shows heads?

EDIT: MISINFORMATION

Timeline:
1) Sleeping Beauty falls asleep on Sunday evening.
2) The experimenter tosses a coin. Call this random variable F.
2_1a) If F=heads, Sleeping Beauty is awoken on Monday morning.
2_1b) After 2_1a, the experimenters put Sleeping Beauty to sleep by giving her a sleeping pill.
2_2a) If F=tails, Sleeping Beauty is awoken on Monday morning.
2_2b) After 2_2a) F=tails, Sleeping Beauty is put to sleep again and awoken on Tuesday morning, then put back to sleep.
3) Sleeping beauty is awoken on Wednesday morning.

Seeping Beauty does not know the outcome of the coin toss, nor the day. She does know the logic in 2_1a), 2_1)b, 2_2a) and 2_2b). Let's just assume this occurs every week, forever.

When she wakes up she doesn't know if it's Monday, Tuesday or Wednesday. All she knows is that she has woken up. You then ask what's the probability of heads given that she's woken up? That means knowing what "Awoken on a day" means... And since she doesn't know what day it is, that's more conditional reasoning.

She's awoken every Wednesday (from 3), she's awoken every Monday (from 2_1a) and (2_2a), she's awoken on a Tuesday iff 2_2b) is triggered, which occurs when and only when 2_2a) is triggered, which is equivalent to F=tails. She also doesn't know what day it is.

I'd need to model "Awoken on a day" as a compound event of "Day" and "Awoken", Day can take values Monday, Tuesday, Wednesday. Awoken can take values "Asleep" or "Awake". If I recall correctly this formula lets you marginalise conditional distributions over another variable:

P(A|B) = P(A|B,C1)P(C1) + P(A|B,C2)P(C2)

Where C1 and C2 are disjoint events. I'm going to use A as the coinflip, B as Awoken and C as Day (so there'll be three rather than two). Now let's go onto the probabilities...

P(F=Heads|Awoken=Awake)=
P(F=Heads|Awoken=Awake, Day=Monday)P(Day=Monday) +
P(F=Heads|Awoken=Awake, Day=Tuesday)P(Day=Tuesday) +
P(F=Heads|Awoken=Awake, Day=Wednesday)P(Day=Wednesday)

P(Monday)=P(Tuesday)=P(Wednesday)=1/3 , assuming principle of indifference and the experiment runs only over these days.
EDIT: DON'T USE INDIFFERENCE HERE, WEDNESDAY'S VARIABLE DAY IS LINKED TO FLIP
P(Heads|Awake, Monday) - how many times is the coin heads if she's awoken on a Monday? She knows the experimental plan, so this should be 0.5 .
P(Heads|Awake, Tuesday) - how many times is the coin heads if she's awoken on a Tuesday? Should be 0. Since she can't wake up on a Tuesday unless F=tails.
P(Heads|Awake, Wednesday) - how many times is the coin heads if she's awoken on a Wednesday? Should be 0.5, since she's always woken up on a Wednesday.

That means we've got (1/3)*0.5+(1/3)*0.5=1/3 probability for P(Heads|Awake). Seems I'm a thirder.

EDIT: MISINFORMATION

Sleeping Beauty Problem

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