Mathematical Conundrum or Not? Number Five
The Sleeping Beauty Problem
So what is Beauty's credence that the coin landed on heads?
There is currently no agreed upon resolution to this problem. You could consider this an exercise in philosophy or probability, it should overlap both.
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, 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, 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 is asked, 'What is your belief now for the proposition that the coin landed heads?
So what is Beauty's credence that the coin landed on heads?
There is currently no agreed upon resolution to this problem. You could consider this an exercise in philosophy or probability, it should overlap both.
Comments (455)
Well let's break it down.
If the coin flip is heads then Beauty is awakened on Monday but sleeps though Tuesday.
If the coin flip is tails Beauty is awakened on Monday and awakened on Tuesday.
So we have:
H: A, S
T: A, A
So since there are three possible awakenings and only one is when the coin comes up heads, then won't that mean she has a 33% chance of it being heads?
So there's a conflict between guessing which interview this is -- and you weight by the number of possible interviews -- and guessing which day this is, right? It's not Wednesday, because you're being interviewed; if it's Monday, you're interviewed either way; so what are the odds that it's Tuesday?
Consider it this way:
P(Tails|Monday) = P(Tails|Tuesday) Right? Because they are on the same flip they have the same probability.
Now if the coin is fair then P(Heads and Monday) = P(Tails and Tuesday)
Therefore P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday).
It must add up to one, thus the 1/3.
Btw, there are other ways to look at this, Pattern-Chaser was not necessarily wrong. I only presented this side because they led with the 1/2 argument; however, they are correct in pointing out Beauty has gained no additional information. Really all she knows is what she was told before the experiment.
Yeah that's pretty clean.
I realized right after I posted that the question is almost literally what is the probability we got heads given that you have been awakened and are being interviewed? So we want the conditional P(H|A).
That's what I'm working at. Not sure what I've got so far. Nice puzzle for me because my probability skills be weak, so thanks.
P (Tails and Monday) and P (Heads and Monday) are mutually exclusive. Also, P (Tails and Monday) and P (Tails and Tuesday) are not independent. So you don't add their probabilities. Also, you left out P (Heads and Not Tuesday).
Anyway. Different experiment - much more ethical. SB comes in on Sunday and agrees to the experiment. She leaves. They flip a coin. On Monday, they place a red jellybean under an opaque cover. SB comes back and is asked whether the flip was heads or tails. On Tuesday, if the coin came up heads on Sunday, they place a green jellybean under the cover along with the red one. If the flip was heads, no jellybean is added. SB comes in and is asked again. Then on Wednesday she comes back in and they give her the jelly bean(s).
The coin is only flipped once.
Quoting T Clark
Clearly . . . .
Quoting T Clark
They are the same flip. It is pointless to argue independence. That is like saying I rolled a die and got 4 and it is not independent because I got 4. If the coin is tails she will be awakened on Monday and awakened on Tuesday, therefore they have the same probability.
Quoting T Clark
She is not awakened on Tuesday if the coin is heads.
The possible outcomes:
- - M T
H: A, S
T: A, A
If she is awakened before Wednesday, there is one awake on heads and two awake on tails. Of the possible awakens there is a 1/3 chance it is Monday and heads.
Is she told what day it currently is when she is awakened?
Here's the additive law of probability:
If events A and B are mutually exclusive (disjoint), then P(A or B) = P(A) + P(B)
In your case, P (Tails and Monday) and P (Tails and Tuesday) are not mutually exclusive, so this rule does not apply.
What is the correct answer depends on what meaning one attaches to the word 'credence'. The use of that unusual word rather than 'probability' in posing the problem is deliberate.
There's a long discussion about it here on physicsforums.
I can't remember, off-hand, whether I was a thirder or a halfer.
The trouble is that we cannot use conditional probabilities.
A conditional probability P(H|A) is the probability of event H given the probability of event A, and A will be an event that is known to be true. But the only event that Beauty knows to be true is that she has been woken at least once - because she has just been woken, so the known event is not a proper (ie smaller) subset of the entire sample space, call it S. So if H is heads and A is 'I am woken at least once' then P(H|A) = P(H|S) = P(H) = 1/2. The conditional probability is the same as the unconditional one.
To use conditional probabilities à la Bayes' Rule, we must have some information that narrows down the set of possibilities - that tells us we are in a proper subset of the sample space. But Beauty has no such information. All she knows is something she already knew before the experiment started.
I think it's to avoid that straightforward solution that the word 'credence', or sometimes 'degree of belief' is used instead of 'probability'.
I'll add that statements like P(Tails and Monday) or P(Tails | Monday) are ambiguous, because the 'Monday' could be the event 'I get woken on a Monday', for which the probability is 1, or it could mean 'today is Monday', and it's not immediately obvious how to set up the probability space so that 'today is Monday' is a well-defined event, without losing the connection to the coin toss.
P (Tails and Monday) and P (Tails and Tuesday) are the same flip. Are you really suggesting they have a different chance of occurring when they both occur on the same chance event?
P (Tails and Monday) and P (Tails and Tuesday) are the same event. You can't (legitimately) count it's probability twice.
Presented in a way that suggests we should look for a conditional, but that won't work. I could see that, but I'm still mulling it over.
Everything else seems to point to 1/3. Twice as many possible awakenings are on tails, leaving just 1/3 for heads. 2/3 of the awakenings are on Monday and half of those are heads, so 1/3 again.
But now I don't know if that's misdirection too.
As she's awake we have to dismiss Heads-Tuesday as an outcome. The only outcomes are Heads-Monday, Tails-Monday, and Tails-Tuesday. But is it right to treat these outcomes as equally likely? Perhaps not. If it it's heads then there's 100% chance that it's Monday, but if it's tails then there's a 50% chance that it's Monday. So Heads-Monday is twice as likely as Tails-Monday (and twice as likely as Tails-Tuesday).
And this is consistent with the fact that we know that, given a fair coin toss, there's a 50% chance that it landed heads. We shouldn't change our view of that just because we might be woken up twice rather than once.
I didn't count it twice. I am saying P (Tails and Monday) and P (Tails and Tuesday) have the same likelihood of occurring because they are determined by the same chance event, but for Beauty the coin flip generates three possible answers. You have to distinguish between the coin flip and Beauty determining the probability it landed on heads based on the possible times she could be awakened.
Consider two different betting games, and we assume Beauty wants to maximise her expected profit.
Game 1: Beauty places her bet before the experiment starts, paying $1, and at the end she is paid $2 if in all her interviews she guessed the coin outcome correctly, otherwise she is paid nothing.
Under this game, her expected winnings are maximised at zero, whichever she chooses. But she must decide before going to sleep the first time which side she is going to guess, because her expected profit becomes negative if it is tails and she makes one guess of heads and another of tails.
Under this game, interpreting the betting strategy as 'degree of belief', we could say her 'degree of belief', at the time of being interviewed, that the coin has landed on Tails, is 1/2.
Game 2: At each interview, Beauty bets $1 to guess what coin came up, and loses that dollar if wrong or wins $2 if right.
Under this game, Beauty's expected winnings are maximised at 50 cents if she guesses Tails.
Under this game, interpreting the betting strategy as 'degree of belief', we could say her 'degree of belief', at the time of being interviewed, that the coin has landed on Tails, is 3/4.
I find it interesting that Game 2, which seems perfectly natural interpretation to me, is consistent with a 'degree of belief' of 3/4 rather than 1/2 or 1/3.
I wonder what betting game would be consistent with a probability of 1/3. I expect there must be a pretty natural one, since most people answer either 1/2 or 1/3 to this question.
I'm off for a jog on the beach now. Maybe it will come to me.
None of it is misdirection, this problem has several possible rational answers. We are not talking about just different points of view, or different readings of the semantics, it can rationally be answered in multiple ways from the same point of view and the same interpretation of the semantics. This is because the problem is probing the relationship between knowledge and probability, in this case Beauty's knowledge.
And actually, Pattern-chaser's first answer is still a valid take. Beauty can't be sure of what day she is awakened, as she is given no new information when she is awakened, which means a 1/2 chance for Monday is a rational response. This would follow Bayesian philosophy on probability which suggest we should update our probability models when we get new information.
Yes, Beauty is aware the coin is fair, but she has also been told the details of the experiment and knows there are three possible events in which she is awakened. If awaken on Tuesday she would not know it is Tuesday; she only knows there are three possible outcomes in which she will be awakened. To Beauty, who does not know if it is Tuesday or Monday when she is awaken, Tails and Tuesdays and Tails and Monday are both valid outcomes. Beauty has to consider three possibilities and only one of them is the desired outcome.
And yet the other one was generating more discussion.
Finally got some time to come back to this and I think you can use conditional probabilities.
I have Beauty treating Monday-Tuesday as another 50-50 coin toss. Not conditional on her being awakened, mind you -- will come back to that -- just something she has no way of knowing by some other means so she forms no opinion either way.
Here's the simple table @Jeremiah posted:
[math]\small P(H \mid A) = \cfrac{P(A \mid H)P(H)}{P(A)}[/math]
[math]\small \phantom{P(H \mid A)} = \cfrac{\cfrac{1}{2} \cdot \cfrac{1}{2}}{\cfrac{3}{4}}[/math]
[math]\small \phantom{P(H \mid A)} = \cfrac{1}{3}[/math]
That makes sense to me. Am I missing something?
We can also ask, what is the chance that it's Monday, given that she's been awakened?
[math]\small P(M\mid A) = \cfrac{P(A \mid M)P(M)}{P(A)}[/math]
[math]\small \phantom{P(M\mid A)} = \cfrac{1 \cdot \cfrac{1}{2}}{\cfrac{3}{4}}[/math]
[math]\small \phantom{P(M\mid A)} = \cfrac{2}{3}[/math]
That looks right, and also makes sense. Of course it's twice as likely to be Monday given that she's been awakened, just as it's twice as likely that the coin toss came up tails, given that she's been awakened.
Am I being completely stupid about this?
That's not correct. Beauty knows that she is awake and that is relevant information.
P(Heads) = 1/2
P(Heads|Awake) = 1/3
Whether 1/2 or 1/3 is assigned depends on whether one interprets the experiment as being about a coin toss event (1/2) or an awakening event (1/3).
No. Nice analysis!
Then she knows that there's a 50% chance that it landed heads. It doesn't matter if she's only woken once if it's heads but twice if it's tails; a fair coin toss is always going to be 50%.
Quoting Jeremiah
But each outcome is not equally likely. There's a 50% chance that it's Heads-Monday, 25% chance that it's Tails-Monday, and 25% chance that it's Tails-Tuesday.
Why is P(A) 3/4? Given that she's awake (and knows it), P(A) is 1. She can dismiss P(S) as a possible outcome. And for the same reason P(A|H) should be 1, which then gives the chance that it's heads 1/2, consistent with what we actually know about coin tosses.
Using the thirder reasoning there's a 1/100 chance that it's heads, and so the £99 bet on tails is by far the best best. Using the halfer reasoning there's a 1/2 chance that it's heads, and so the £1 bet on heads is by far the best bet.
So put your money where your mouth is. What should you bet? I say bet £1 on heads.
The reason it can't be an event is that, so far as I can see, the definition of an event in the Kolmogorov framework, which is used, to the best of my knowledge, in all modern probability theory, is timeless. It cannot be relative to a particular time. So 'Beauty is woken on a Monday' and 'Beauty is woken on a Tuesday' are events because the time references they make are absolute, but 'today is Tuesday' is not an event because the reference 'today' is relative.
The same applies to 'I am awake', which is what I think you mean by the label 'A'. Because the 'am' is a relative time reference, it cannot be an Event in a probability space, and so cannot have a probability in the usual sense, that allows use of things like conditional probability rules.
To use a philosophy of time analogy, the question 'is today Tuesday', in the context of this experiment, has as much meaning in probability theory as it does when asked to a pan-dimensional being that is looking at our 4D spacetime from the outside, if McTaggart's B theory of time is the case.
Perhaps a probability space can be constructed in which 'today is Monday' or 'I am awake' is an event but so far I have not managed to construct one. I thought I had a lead, but it turned out to be a dud.
Unless we can construct a probability space in which 'today is Monday' is an event, and which also contains all the other information about the experiment, it cannot be meaningful to talk of the probability - conditional or unconditional - of the event 'today is Monday' or 'I am awake', so we cannot use conditional probabilities to calculate with them.
It is a fair coin flip, so the chance of Monday and Heads is 50% and the chance for Monday and Tails is 50%. Since on one tails flip she is awakened on two days then the chance of Tuesday and Tails is also 50%. Therefore, Beauty has a 33% chance of being awake on a heads
The chance that on Monday the coin flip was tails is 50%, but that chance that any given waking day is Monday and that the coin flip was tails is 25%.
Consider instead that if it was heads then she's woken on Monday and if it was tails then she's woken on Tuesday and Wednesday. What's the chance that she's woken on Monday? 50%. Therefore what's the chance that it was heads? 50%.
The chance that it was heads isn't reduced simply by changing which two days she's woken if it's tails. It's still going to be 50%. Or are you really suggesting that it's more likely to be tails if she's woken on the Monday rather than the Wednesday?
Maybe you could turn that around a bit:
p(Heads) = 1/2
p(Tails) = 1/2
p(Awake|Heads) = 1/3
p(Awake|Tails) = 2/3
Which strikes me as just a restatement of the conundrum.
I prefer the Monty Hall problem.
Here's a variant that is structurally similar:
No sleeping theatrics.
Examiner tosses a fair coin, and then tosses another. If the first toss was tails, she asks Beauty her credence that the first toss came up heads; if the first toss was heads she only asks for Beauty's credence if the second toss was heads as well, otherwise the round is over.
Done this way, Beauty will know that when she was not asked the first toss was heads, but she can do nothing with that knowledge. She's not asked and the round is over. What matters is that she's always asked when it was tails and asked half the time when it was heads. So her credence that it was heads should be 1/3.
Quoting Michael
I don't understand this provision. Beauty knows that she will be woken more often on tails than heads and should be allowed to show that in her betting behavior, even if she can't at the time know how many times she's been woken.
((Will try to put together a betting argument when I have time, because I haven't understood the ones presented so far. If the odds of heads really are 2-1 against, a fair book with no vig should give an expected payout of 0, and Beauty should be able to make a dutch book against anyone who thinks the odds are even. I don't understand how you and andrew are setting the odds.)
Here you might find these papers interesting.
https://www.princeton.edu/~adame/papers/sleeping/sleeping.html
http://fitelson.org/probability/lewis_sb.pdf
And a 50% chance of that she is awakened on Tuesday and Wednesday.
Therefore a 33% of Monday and Heads. What you are not considering is that from the chance mechanism 50% is allocated to two of the possible awakening as the same outcome. Not split between them, but instead, in terms of the coin flip, they are not separate events, they are the same outcome.
Just pointing out that there's only one payout, not 99 in the case of tails.
So if you were in her shoes, what would you bet? £1 on heads or £99 on tails?
So a 50% chance that it's tails and a 50% chance that it's heads.
I never disagreed with that, I don't really feel you are understanding what I am saying.
No, if there's a 50% chance that she is awakened on Tuesday and Wednesday then there's a 50% chance of Monday and heads.
Jeremiah's answer here seems correct to me. The 50/50 answer is from one who has no additional data and knows only that there is a coin flip. Sleeper here has additional data: I am awake, which reduces the case to which of the three awakenings above is this one.
But one of those awakenings is twice as likely as each of the other two, which is why the halfer answer is correct.
Right I agree with that, and have from the start.
Here's my variation:
If it's heads then she's given a red ball. If it's tails then she's given a blue and black ball.
What are the odds that she's given a red ball?
I am, you are reallocating the 50% for a tails flip across two days; however, it is not reallocated across two days, it is the two days. When Beauty is awakened there are three possible events in which she is awakened. Two of them are determined by the same outcome of the chance event, which is the coin flip, they both have a 50% chance of happening, so they are equally likely. The other possible outcome is on heads, and has a 50% chance of happening so it is equally likely as well. Since all three are equally likely, Beauty's credence that it is Heads and Monday is 33%.
They're not equally likely. Heads and Monday is 50%, because heads is 50% and heads guarantees Monday.
P(H) = P(H & M).
I don't know what you mean by adding up to 33%.
P(Heads) = 50%
P(Monday|Heads) = 100%
P(Heads & Monday) = 50%
P(Tails) = 50%
P(Monday|Tails) = 50%
P(Tails & Monday) = 25%
P(Tuesday|Tails) = 50%
P(Tails & Tuesday) = 25%
However you look at it, nothing can change the fact that the chance of heads on a fair coin toss is 50%, so any solution that says otherwise must be invalid. No other information available to Sleeping Beauty suggests that tails is more likely. The multiple (forgotten) awakenings in one case is a red herring.
Variation 2
Suppose I toss a fair coin; if it comes up heads, I ask you if it was heads or tails; if it comes up tails, I don't. When asked, you should always guess "heads".
Now move the likelihood that I'll ask in each case.
Variation 3
If the coin comes up heads, I ask you once to guess; if it comes up tails, I ask you twice.
This is different from our case because for each round you know the first question is the first question. That one's 50-50, but once I ask again, you know to answer "tails".
Now suppose there were a way to fix it so you didn't know how many times you were being asked or whether a question was a first or a second. All you know is that on tails you'll be asked twice. What do you guess?
I agree the 1/2 argument has its merits as a valid argument, but not for the reasons you are listing. Probability is not reallocated evenly between Tuesday and Wednesday since she is awaken both of those days from the same chance mechanism. From the coin flip she has a 50% chance of being awakened on Tails and Tuesday, and a 50% chance of being awakened on Tails and Monday. Not 25% each as you suggested. This means they are equally likely as Heads and Monday.
You're correct. But that's because of this:
Quoting Michael
If a bet is instead placed every time Beauty awakes, then the £99 bet on tails is the best bet.
Thirders and halfers will agree on how to bet for any given scenario. So that's not really the issue.
Quoting Jeremiah
It's self-locating information that she can update on. P(Heads|Awake) = 1/3 which Beauty already knew before the experiment began. When she awakes within the experiment, she knows she is awake so the probability of heads for her at that time updates to 1/3.
She also knew before the experiment started that P(Heads|Tails) = 0 and P(Heads|Heads) = 1. So after the experiment, when she learns the result, the probability of heads also updates, this time to either 0 or 1.
Quoting tom
The same idea really. One should update one's probabilities when given new information.
She already knows that it was a fair coin toss and that a fair coin toss has a 50% chance of landing heads. Nothing can change that.
If it's heads then I'm given a box with a red ball on Monday. If it's tails then I'm given a box with a blue ball on Monday and a box with a black ball on Tuesday.
I'm woken up and given a box. What is the chance that it's a red ball, a blue ball, or a black ball?
I say 50% chance of it being a red ball because there was a 50% chance that the coin flip was heads and heads guarantees a red ball. I say 25% chance of it being a blue ball because there was a 50% chance that the coin flip was tails and there's a 50% chance, if it was tails, that today is Monday. And then of course 25% chance of it being a black ball.
That doesn't change the fact that P(H) = P(H & M) = 50%.
And that does not change the fact that P(T) = P(T & M) = 50%
You're looking at it wrong; see the example with the balls in boxes.
However, it is not. When awakened Beauty does not know if it is Monday or Tuesday.
Quoting Andrew M
The problem with a Bayesian approach is that in order to update you need a prior to update, which in Beauty's case could be based on either the coin flip or the possible times she will be awakened. That's the issue with Bayesian probability and what makes it so controversial, it is subjective.
These cases are different because being asked (again) provides you with additional information, whereas it doesn't in the original case. You're going to be asked regardless, and you have no idea if you've been asked before.
Quoting Srap Tasmaner
There's a way to do this. If it's heads then we ask one person to guess. If it's tails then we ask two people to guess (independently). Nobody knows if they were the only person asked (because it was heads) or if they were the first or second of two (if it was tails). Given that the participants know the rules, what should they guess? The thirder response is that they should guess tails. But does that actually make any sense? I don't think it does. It was a fair coin toss, so it's equally likely to be heads or tails.
But let's imagine that we actually carry this out. We flip the coin 100 times, with it landing heads 50 times and tails 50 times. We'd have asked 150 people to guess. If they all applied the thirder reasoning and picked tails then 2/3 are guaranteed to be wrong (and the same if they all picked heads). But if they flipped a coin to make the guess for them? Assuming 75 heads and 75 tails then, what, anywhere between 1/3 and 2/3 will be right, averaging at 1/2?
The halfer reasoning gives you a better chance of being right.
Edit: But what if the thirders weight the choice, picking a ball from a bag containing two blue and one red, choosing tails if it's a blue ball? We'd expect 100 tails and 50 heads. That's anywhere between 1/3 and 2/3 being right, also averaging at 1/2.
Second edit: I think my math here is wrong.
It needs two blue balls and one red ball.
Listen I am not doing the example gambit; you people come up with all these examples and some can be representative of the problem, while others are not and at that point they become a straw-man, whether it was intentional or not. And if I argue a straw-man then I give false creditiably to it. The best response I can have here, is to suggest we stick to the main problem of interest.
I know many of you think I am just being a troll when I call straw-man, but I know where these numerous examples lead. They are fine in sort order, but people have a tendency to take them off the central path and if you follow them then you end up in La La Land and the discussion has steered way off track.
Let's just stick with the OP, then we all know we are debating the same problem and not something else.
No it doesn't. It's a blue ball on Tuesday and a black ball on Wednesday (if tails) and a red ball on Monday (if heads).
If you're given a box then there's a 50% chance that it contains a red ball, a 25% chance that it contains a blue ball, and a 25% chance that it contains a black ball.
So, if you've been woken up then there's a 50% chance that today is Monday, a 25% chance that today is Tuesday, and a 25% chance that today is Wednesday.
50% chance of heads, 50% chance of tails.
But even if it's a blue ball on both days with tails, it's still a 50% chance of being a red ball and so a 50% chance of being heads (given that a red ball is only given if heads).
There's never a reason to prefer tails over heads.
Quoting Jeremiah
The OP version is clearly confusing you, hence the need of the above variation. The point of analogies like these is to make explicit problems that are hidden in the original formulation; they're not strawmen.
Your "above variation" is confusing you and honestly the colors of the balls would not make any difference. Monday and heads is still only one out of the three possible outcomes.
I know, but it has a 50% chance of happening, which is exactly what I'm saying. 50% chance of red ball. 50% chance of Monday and heads. 50% chance of heads.
25% chance of blue ball. 25% chance of tails and Monday.
25% chance of black ball. 25% chance of tails and Tuesday.
50% chance of tails.
Would you consider the 2/3 debacle an argument against the 1/3 argument and for the 1/2 argument?
She knows it must be either Monday or (Tuesday and Tails). Do you agree that P(Heads|Awake) = 1/3?
Don't know what you mean.
The multiple person argument is really interesting and might persuade me. (One thing that occurred to me is that if there's betting, splitting the payoffs among multiple people is wrong.)
((Sadly it'll be a little while before I can come back to this.))
Sure, she knew that would be the case before even starting, but when she is awaken she has no new information on if it is Monday or Tuesday. You are not exactly describing a condition that is new to her knowledge.
Quoting Andrew M
No, I have said many times this problem has more than one rational and valid answer.
One more quick note.
The point was that being asked doesn't have to be a simple fact/non-fact, but can have a probability, and we're told that the probability of being asked is driven by the results of the coin toss.
You might still be right that this doesn't matter, I'm not sure.
Do you mean that if both people who are asked if it's tails correctly guess tails then that should only be counted as 1 success rather than 2?
Assume she's woken on Monday if it's heads or Tuesday and Wednesday if it's tails.
Do you agree that P(Monday|Awake) = 1/2?
Except in this, and in the Monty Hall problem, there is no new information.
Quoting MichaelInformation about the toss can very much change that. If she is able to actually see the result of the toss, the odds become a certainty one way or another, not 50/50. So information does change the odds, and she has information beyond the simple fact that a coin was tossed.
To assert 50/50 odds here is to use a perspective other than that of the sleeper.
There are four equal probabiltiy states (each equally has 50% chance of being visited eventually, all depending on the coin toss. Monday and Tuesday are eventual certainties):
A Monday Heads
B Monday Tails
C Tuesday Heads
D Tuesday Tails
The sleeper wakes up and knows not which of the four it is, but she has the additional knowledge (new information) that it is not C, so 33% chance of each of the other choices, and only one of those is heads. Odds are 33%
That was Lewis' argument in that paper I linked.
Starting on page 3 he lays both arguments out.
http://fitelson.org/probability/lewis_sb.pdf
Gotcha. I'm at work so I'll read later. Maybe much later since I'm enjoying fighting through this "on my own".
Another good thread.
Two of the three awakenings always happen. Only one is variable. And it depends on the coin toss. Beauty can only guess, regardless of when she is awoken, whether the coin did (or will - she is woken on Monday) come up heads.
Quoting Jeremiah
Yes, and the only time we receive new information is at the start, when the experiment is explained to us and to Beauty. No new information is provided thereafter, therefore reappraisal of probabilities will lead to the same answer we got originally. Not like the Monty Haul problem, where new information is provided.
Not exactly. If you're calculating wagers and payoffs, you'd need to add up all the gains and losses for the odds to make sense.
On this issue of whether there's new information: it's true that all she gets is that she's been awakened. She knows no other new facts, but she knows two things:
In the absence of new (certain) knowledge of what has happened, shouldn't she go with the probabilities of what has happened?
One difference between the days and the coin toss is that the days are just clues to the coin toss; it's the toss that determines whether she's awakened a second time.
If I'm making a mistake, it's somewhere around here.
I am not convinced it contradicts the 1/3, but instead just leads back to the initial problem of the inherited subjectivity in the notion that you should start with a prior belief and update that prior into a posterior belief when you get new information. This carries that subjectivity into any posterior, and as we see here those priors can look very different based on the viewer. Not really objective science.
Yes, she is. Because she is certain to be woken on Monday. So you could say "It's more likely that I am called Srap Tasmaner than that Beauty will be woken on Tuesday." Just select any certainty, and it is more probable than Beauty being woken on Tuesday. I'm not sure what you're getting at here. :chin:
A. Box with red ball
B. Box with blue ball
C. No box
D. Box with black ball
She wakes up and is given a box with a ball in it. What are the odds that it's red? You're saying 33%? I think this is wrong. She knows that if the original toss was heads then there's a 100% chance that her box contains a red ball and that if the original toss was tails then there's a 50% chance that her box contains a blue ball and a 50% chance that her box contains a black ball. From that she can infer that there's a 50% chance of a red ball, a 25% chance of a blue ball, and a 25% chance of a black ball.
There are three outcomes but one is twice as likely as each of the other two. They're not weighted equally.
The description of the problem does not make it clear that sleeping beauty knows the procedure. If she doesn't, then she has no knowledge of anything other than their asking about what is a random coin toss. The odds would be 50% then.
The 33% comes from the sleeper knowing that there will not be a waking on Tuesday if the result is heads, but there will be a waking on the other three scenarios. In this case, new information is gained (it is not Tuesday/heads), and the odds are not 50/50
I still don't see how that makes it 33%.
She knows that P(Heads) = 0.5 and that P(Monday|Heads) = 1. So she knows that P(Heads ? Monday) = 0.5.
She knows that P(Tails) = 0.5 and that P(Monday|Tails) = 0.5. So she knows that P(Tails ? Monday) = 0.25.
She knows that P(Tails) = 0.5 and that P(Tuesday|Tails) = 0.5. So she knows that P(Tails ? Tuesday) = 0.25.
Knowing that P(Heads ? Tuesday) = 0 might be new information but it doesn't entail that the other three outcomes are equally likely. In fact it's that information that allows her to know that P(Heads ? Monday) = 0.5 (rather than 0.25), and so twice as likely as either of the other two.
So, I ran 100,000 games and gave 1 point for successfully guessing heads and 0.5 points for successfully guessing tails (because you get two opportunities). It doesn't matter if you always select heads, always select tails, select tails 1/2 the time, or select tails 2/3 of the time. The average score is 0.5 in every case.
Although actually I think these analyses miss the point of the puzzle.
I don't understand this. What you're doing there is giving even money on heads and 2-1 for on tails. You can't do that.
My policy is to wager $1 on tails whenever I'm asked. You're paying even money.
If the toss is heads, I lose $1; this happens half the time, so my expected loss is $0.50.
If the toss is tails, I make $1 each time I'm asked; this scenario happens half the time, so I have an expected profit of $1.
So I make at least $0.50 on average each time I play, no matter how the toss goes.
I have made a Dutch book against you.
Suppose instead you're offering 2-1 against heads, and I still wager $1 on tails each time I'm asked.
On heads, it's the same: expected loss of $0.50.
On tails, I only make $0.50 each time, for an expected profit of $0.50.
My profit and loss cancel out, so 2-1 is a fair book, representing the true odds.
At least that's how I think it works.
How is this any different to awarding 1 point for successfully guessing heads and 0.5 points for successfully guessing tails?
1 point for each would also be a Dutch book wouldn't it?
You should be paying out $2 on heads, for a wager of $1. That's what 2-1 (against) means. On the favorite, you only pay $0.50, because the odds are 1-2 against.
Missed this.
Yes. That was exactly my argument: if you offer even money, I am guaranteed a profit.
And I seriously don't understand how you can't see the 50% argument; disagreeing with it is one thing but it has been explained to you at this point numerous different ways. Either you see it and are too stubborn to let on that you see it, or it is going right over your head.
This right here is the most succinct explanation.
The general multiplication rule, from here is:
P(A and B) = P(A) * P(B|A)
Where A is heads and B is Monday:
P(heads and Monday) = P(heads) * P(Monday|heads)
P(heads and Monday) = 0.5 * 1 = 0.5
What rule allows you to get P(heads and Monday) = 1/3? If there were one wouldn't that mean that we're using an inconsistent system?
By betting tails, I get to double what I risk only when my profit is guaranteed to double. I should not be allowed to do that.
I also understand now why @andrewk remembered that it all turns on what you mean by "credence". The weirdness here is that even though my own description has the two scenarios arising half the time each -- fair coin, after all -- I should bet "as if" the odds are really 2-1 against heads.
This is annoying, because we (Bayesians, subjective probabilists of all stripes, and the Bayes-curious like me) want to use wagering to measure degree of belief. If you can show that I must wager in a way that systematically deviates from my degree of belief, that's trouble.
So there's a good meaty philosophical issue waiting at the end of this one, which I will now ponder (and probably read Lewis). Again, nice thread @Jeremiah.
But let's apply the general multiplication rule from above:
P(Tails and Monday) = P(Tails) * P(Monday|Tails) = 0.5 * 0.5 = 0.25
P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails) = 0.5 * 0.5 = 0.25
P(Heads and Monday) = P(Heads) * P(Monday|Heads) = 0.5 * 1 = 0.5
But he says that they're all equal. So where has he gone wrong? It's in saying that P(Tails | Monday) = P(Heads | Monday).
P(Heads and Monday) = P(Monday) * P(Heads|Monday)
From the above, we know that P(Heads and Monday) = 0.5, so:
0.5 = P(Monday) * P(Heads|Monday)
What's the probability that it's Monday? It's tempting to say 2/3, but that would be wrong. From the above, we know that P(Tails and Monday) = 0.25 and that P(Heads and Monday) = 0.5. So it must be that P(Monday) = 0.75.
This then gives us:
0.5 = 0.75 * P(Heads|Monday)
Meaning that P(Heads|Monday) = 2/3. And the same reasoning gives us P(Tails|Monday) = 1/3. This is Lewis' conclusion in response to Elga.
I already posted both of their responses.
To better explain this, imagine I toss a coin. If it's heads then I give you a red ball, and if it's tails I toss again. If it's heads then I give you a red ball and if it's tails then I give you a blue ball. What's the chance that I'm given a red ball? 0.75.
In our case, the red ball is Monday and the blue ball is Tuesday.
This is Betting Game 2 from this post.
My calculation is that expecting that profit is the same as if there were a single $1 bet at even odds that the result of a coin toss will be tails and the coin had a 3/4 probability of coming up tails. To see this, note that in that case the expected profit is
$1 x 3/4 + (-$1) x 1/4 = $0.50
Hence, in interpreting the term 'degree of belief' it seems reasonable to make it the probability of tails when betting on a single flip of an unfair coin, at which one would have the same expected profit. So under this interpretation of 'degree of belief', the answer is 3/4 for tails, and hence 1/4 for heads.
Quoting Srap Tasmaner
I can't quite follow this. I think there are a couple of grammar glitches in it that make it hard to understand what it is saying. Can you please expand on it, maybe using bullet points for clarity, and explain in what way you find it equivalent to the original question?
thanks
This whole betting thing demonstrates the correct answer. It only needs to be done twice, not 100000 times, because there are only two unique cases occurring in equal probability.
I wake up and am expecting to bet a coin with even odds. I bet tails and win 2 coins (one each day) if it is tails, and lose one coin on Monday if it is heads. Sounds like a winning bet to me.
If the stakes depend on the coin toss, then I bet tails if the stakes are 0.5 coins per bet, and heads if the stakes are 1 coin in the bet. I win every time in that case.
Lewis has the same answer to me, but it looks like I got there a different way. His depends on a premise that Elga rejects whereas mine doesn't. I only apply the general multiplication rule of probability.
Sure, but it doesn't mean that tails is twice as likely to occur as heads, which is why these betting examples miss the point. All the betting examples show is that it's better to bet on whichever outcome provides more payouts, which is obvious.
This was my prior response, that I already posted before a few pages back.
Quoting Jeremiah
Is that what I've done above?
P(Tails and Monday) = P(Tails) * P(Monday|Tails) = 0.5 * 0.5 = 0.25
P(Heads and Monday) = P(Heads) * P(Monday|Heads) = 0.5 * 1 = 0.5
Therefore, P(Monday) = 0.75
It is twice as likely to occur to miss Beauty. Tails happens twice, and heads only once.
It doesn't happen twice to her. It only happens once to her, given that it was only tossed once. She just wakes up to it twice.
Nearly.
Quoting andrewk
This is why I said I couldn't understand how you were setting the odds. Doing it this way is paying off 2-1 on both heads and tails, which is incoherent. (Unless you meant both tails interviews would taken together pay off $2, but that's still incoherent.)
Quoting andrewk
Yeah that's really interesting. On the one hand, my pass at making a fair book says the odds should be 2-1 against heads. But the expected profit on betting tails clearly matches the 3-1 version.
Did I get the fair book wrong, or is there some other explanation for why they're coming out the same? (The payoff system here is so wonky, I think this could be another side effect of that, but I can't see how.)
Either way -- 2-1 or 3-1 -- offering even odds in this game loses money. We agree on that, right?
How to interpret this as credence or degree of belief, I'm not sure yet.
Quoting andrewk
I guess I could do this again -- I wouldn't mind, and will if you're really interested, but to me it's no longer relevant (if it ever was). I was still trying to figure out how the thing works when I wrote that.
I don't understand at all what you mean by P(Monday|Heads) and P(Monday|Tails).
Quoting Srap Tasmaner
I don't know betting terminology, so I may have used the wrong words. What I meant by 'wins $2' is that she gets her own dollar back, plus another dollar. Perhaps the correct betting terminology for that is 'wins $1'. My ignorance of gambling terminology is gargantuan.
Yes I think we can agree that if a bookmaker offers Beauty Game 2, which has even odds, Beauty can select a strategy under which the expected value of the bookmaker's profit is negative.
Yeah, that's right. When you win, you get your stake back and the odds represent your profit. So 2-1 against heads gives you back the $1 and pays you $2 more, total of $3. On tails, you'd get your dollar plus $0.50, total of $1.50. Losing, you lose your dollar. (I only know as much about gambling as I need to to understand philosophers who talk about probability this way!)
Quoting Michael
The betting matters, and be honest: you were happy enough to use wagering arguments when it suited you. (What's more the examples don't show what you say is obvious; they assume it and it's irrelevant here anyway.)
Why does the betting matter?
Quoting Michael
Because of this point of @andrewk's:
Quoting andrewk
Suppose Beauty is not told all the rules, the amnesia pills, etc. She's told she'll get even money -- she has to be told the odds to bet. Given enough trials -- more than is reasonable, I know, it's a story -- she'd conclude from her accumulated payoffs that the coin is unfair.
Now that's quite curious. If betting is a stand-in for hypothesis testing, it'll support a theory that we, in our omniscience, are inclined to call "false" but it will serve Beauty perfectly. Beauty will have plenty of evidence to support her theory.
It'd be interesting to flesh out competition between the two theories. If Beauty doesn't learn what's really going on, she would be incredulous that we would claim the coin is fair. Even if we explained, she might see that our theory is as adequate to the evidence gathered so far as hers, but she'd have no reason to switch to our view without new evidence, something like witnessing new trials showing the coin to be much closer to fair than she thought. That whole process is really cool.
PS: Our explanation is spectacularly more outlandish than a coin being unfair.
No, 1/3.
Both the halfer and the thirder positions are consistent. The difference stems from how the probabilities are distributed when conditioning on being awake.
Halfer:
P(Heads|Awake) = 1/2
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/2 / 3/4 = 2/3
Thirder:
P(Heads|Awake) = 1/3
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/3 / 2/3 = 1/2
One characteristic of the thirder view is that it doesn't imply an external perspective that "knows" what the probability is for Beauty's current awake state. Since Beauty has no available information distinguishing the three states from her point-of-view, she is simply indifferent about which state she is currently in, and so assigns a probability of 1/3 for each awake state.
So the thirder view can be preferred on pragmatic grounds.
In the Monty Hall problem, the host gives you information that changes the probabilities that you assign to each door. That information is new to you.
Similarly, in the Sleeping Beauty problem, awakening provides information that enables you to rule out one of the four states. However since you have no information distinguishing the remaining states, you should be indifferent about which state you are currently in.
The probability that it's Monday given the fact that it's heads/tails.
Getting to bet twice on tails doesn't mean that it's more likely to be tails. It just means she gets to bet twice on tails.
You seem to be thinking of it as:
1. Sleeping Beauty picks tails.
2. We flip a coin.
3. If it's tails, we flip again.
In that case; yes, it's more likely to be tails. This was the reasoning in my first response. But that's not happening in this case. Instead it's:
1. Sleeping Beauty picks tails.
2. We flip a coin.
3. If it's tails, we payout twice.
That doesn't mean it's more likely to be tails. It just means that tails is the better bet.
I'm questioning the validity of those probability choices. Imagine instead if it were:
Is it right to distribute the probabilities that way? I say no. We have to apply the general multiplication rule:
P(Roll 1-5 and Monday) = P(Roll 1-5) * P(Monday|Roll 1-5) = 5/6 * 1 = 5/6
Given these facts, what answer will Sleeping Beauty give?
If the toss is heads, she will awaken Monday. If she'd then say heads, she lives. If she'd say tails she'd be dead.
If the toss is tails, she will awaken first on Monday. If she then says heads, she dies. If she says tails, she'll live.
If she said tails, she will awaken again on Tuesday. If she then says heads, she dies. If she says tails, she'll live.
In the above we see there's only one event where saying tails gets you killed and two events where saying heads gets you killed. Sleeping Beauty would be smartest to state tails.
I’ll admit I was too quick to think it relevant when I thought it agreed with me. But after testing it I realised the problem.
In this case, it’s not that tails has greater likelihood odds but that tails has greater payout odds, and these seem to be equivocated.
Imagine I offered 2:1 for correctly guessing a coin toss or 10:1 for correctly guessing a dice roll. Heads is more likely than rolling a 1 but rolling a 1 is the better bet. So that tails is the better bet isn’t that it’s more likely.
We shouldn't be indifferent between the two possible states for a weighted coin. So probabilities can't be distributed on that basis. But those two states can be transformed into different states that we can be indifferent between. That is, five states that come up heads and one state that comes up tails. Then the Sleeping Beauty result is again P(Heads|Awake) = 5/7.
A simple, discrete probability space consists of two things - a sample space, which is the set of all possible outcomes, called Events, and probabilities of each Event.
It seems from what you are doing here that your sample space consists of four Events:
1. Coin landed Heads and today is Monday
2. Coin landed Heads and today is Tuesday
3. Coin landed Tails and today is Monday
4. Coin landed Tails and today is Tuesday
I think most people would agree that 2 is impossible, so it must have probability zero. The other probabilities are at first unknown and must be inferred by other relationships we have been given.
One such relationship on which I think most people would agree is that is that the probabilities of 3 and 4 must be the same.
That leaves one degree of freedom, which is the probability of 1 which, since the prob of 2 is zero, is also the probability of Heads in this probability space - which we note may not necessarily be the same as the probability space of an independent observer (who will of course say the probability of Heads is 1/2), since this is an epistemological probability space based on the knowledge state of Beauty just after being woken up.
Let the probability of 1 be p. Then the probabilities of 3 and 4 will each be (1-p)/2.
If we set p=1/2 then each of 3 and 4 have probability 1/4 and the probability of Heads is 1/2.
If we set p=1/3 then each of 3 and 4 have probability 1/3 and the probability of Heads is 1/3.
If we set p=1/4 then each of 3 and 4 have probability 3/8 and the probability of Heads is 1/4.
It seems to me that there is no indisputable way of removing the free parameter p. To do so, we need to make an assertion of probability, but it can't be derived without circularity.
We can say p=1/2, in which case we are asserting a principle that Beauty should have the same epistemological probabilities for Heads and Tails as a non-amnesified, independent observer. IN that case we conclude that the probability Beauty assigns to Heads is 1/2.
Alternatively, we can say p=1/3, in which case we are asserting a principle that all non-impossible events in the sample space should have the same probability. In that case we conclude that the probability Beauty assigns to Heads is 1/3.
Or we can say p=1/4, in which case we are basing Beauty's assigned probabilities on the expected values of winnings from a betting strategy.
I expect there are other arguments, with various degrees of convolutedness, for other values of p.
I am inclined to conclude that, if we try to use probabilities rather than betting to solve this, we are left with a degree of freedom - the value of p - that cannot be removed without making a controversial assertion about how Beauty should assign probabilities in her epistemological probability space.
In short, it seems that Beauty's degree of belief that the coin came up Heads can reasonably be whatever she wants it to be - subject to it being in the range [0,1].
Then there's 6 states, not 7. You're counting the tails state twice, which you shouldn't do. The two tails days need to share the probability that it's the tails state (1/6) giving each 1/12 which is the correct figure you get when you apply the probability rule:
P(A and B) = P(A) * P(B|A)
P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails)
P(Tails and Tuesday) = 1/6 * 1/2 = 1/12
I don't feel that this changes the situation, because although there are two events at which saying Heads can get Beauty killed, if she decides to say Heads, she will never make it to the second event, so the event of being killed at the second waking has a zero probability of occurring.
This assumes that Beauty doesn't randomly choose what to say, by tossing a coin herself after being woken. But so far that possibility has not been canvassed (and I don't think it leads anywhere productive). If the choice is not random then whatever reasoning she uses on Monday will be used again on Tuesday, since the days are indistinguishable to her, so she will give the same answer on both days.
I don't think there's any freedom. We just apply the Kolmogorov definition:
[math]P(A|B) = \frac{P(A ? B)}{P(B)}[/math]
[math]P(A ? B) = P(A|B) * P(B)[/math]
[math]P(Monday ? Heads) = P(Monday|Heads) * P(Heads)[/math]
[math]P(Heads) = 0.5[/math]
[math]P(Monday|Heads) = 1[/math]
[math]P(Monday ? Heads) = 0.5[/math]
Have I made a mistake somewhere? If not then this must be the answer.
She knows it's either Monday or Tuesday but not which day it is. I understand from previous posts you cannot apply a probability to which day it is. Nevertheless, if it's Tuesday only tails will save her. If it's Monday there's a 50% chance saying tails will save her.
I don't see how answering heads would be an equally safe bet... Also, I sucked at probabilities in high school so don't expect me to understand your answer. :razz: :rofl:
It wouldn't be an equally safe bet, but that's irrelevant.
If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.
The mistake is in going from "more likely to win if tails" to "more likely that tails".
Ok, I can see that but that seems a meta-position. I thought this was about what credence Beauty gives to the question. That makes the difference to me. Even if she knows there's a 50% chance of either, the fact that it could already be Tuesday would change my assessment if I were in her shoes.
But applying the probability rule she knows that there's only a 25% chance that it's Tuesday, whereas a 75% chance that it's Monday. It stills works out as 50% heads (as it's 25% tails and Monday).
The whole point of my answer was which was the better bet. If the odds were 2:1 instead of even (33% heads), then Beauty would make or lose no money on average by betting.
The odds of the flip are 50% from nobody's point of view. They seems to be 33% (the one point where neither is the better bet) from Beauty's POV, and they are 100% from everybody else's POV since they know the outcome of the flip during any of the wakings. It is 50% only from the external POV (not Beauty) only before the toss, which is not during any of the wakings.
If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.
Tails isn't the better bet because it's more likely but because it has a better payout.
The topic is about credence. In my example of the free lottery tickets, my credence is that heads and tails are equally likely, despite knowing that tails is the better bet.
If you tell me that you flipped a coin ten minutes ago I'm going to say that there's a 50% chance that it landed heads.
And if you tell me that I'll get £1 for guessing correctly, and that you'll let me guess twice if it's tails, I'm going to guess tails because I'll win £2 if I'm right rather than £1 if it's heads. That doesn't mean that tails is more likely. It's just the better bet.
I know what the notation means. I don't understand what these probabilities represent. Are you sure they're even defined?
Quoting MichaelYes, it is about credence. Beauty has information about the coin toss, and that alters the credence from the 50/50 credence that exists to nobody in the scenario.
Quoting MichaelYes, because I've been given no more information, so the odds remain 50%.
See my edit.
Beauty only gets one guess. If she didn't have the memory-wipe, then the situation would be as you describe it here. She would only have the information she needs on Tuesday, when the odds are 100% tails.
But Beauty is apparently being asked not if this occurrence will happen (that is obviously certain during any particular waking), but rather which of A, B, D it is. Each has a 1/3 chance if their odds are equal, and they are equal since each equally has a 50% chance of eventual occurrence. B and D do not have 25% (less than A) chance of occurrence, as you seem posit. All three have equal probability, but only A is heads.
Yes, it's the original Kolmogorov definition:
[math]P(A|B) = \frac{P(A ? B)}{P(B)}[/math]
[math]P(Monday|Heads) = \frac{P(Heads ? Monday)}{P(Heads)}[/math]
[math]P(Monday|Heads) = \frac{0.5 * 1}{0.5}[/math]
[math]P(Monday|Heads) = 1[/math]
Beauty doesn't gain relevant new information when awakening, she knew all this before hand. If we do the experiment on you, then you have a prior belief that it is 1/3, what new information would then update that? Priors need relevant new information which would allow us to update it, not just any old information that you think happened.
There's a 50% chance that both B and D will happen, but there's a 25% chance that any specific day is B.
That is were both Lewis and Elga error, in assuming that is the linchpin between which one is right.
I say nothing has changed. If it is Tuesday and Heads, she is informed of that situation, and is therefore not allowed to place a bet on the known coin toss. All of A B C D occur with equal probability, and any given waking has 25% odds of being any of the 4 (A,B,C,D) or 50% odds of eventual occurrence. But on C, she is told it is C. On the other three, she is not informed which waking it is. All she knows is that it is not C. 33% odds of being any of the other ones.
I ask now what the odds of heads is when she's informed that it is not scenario C this time? Because this is exactly the information she has been given.
Why? If the flip had a 50% chance of being heads and if heads guarantees Monday then there's a 50% chance that today is heads and Monday. You seem to just be asserting these probabilities without adequately explaining how you got there. I get to my probabilities by applying an axiom of probability.
P(A will happen) = P(B and C will happen)
to
P(A is now) = P(B is now) = P(C is now)
But I don't see how that follows.
I've done so multiple times:
Kolmogorov definition:
[math]P(A|B) = \frac{P(A ? B)}{P(B)}[/math]
[math]P(A ? B) = P(A|B) * P(B)[/math]
[math]P(Monday ? Heads) = P(Monday|Heads) * P(Heads)[/math]
[math]P(Heads) = 0.5[/math]
[math]P(Monday|Heads) = 1[/math]
[math]P(Monday ? Heads) = 0.5[/math]
There is a chance mechanism, the coin flip, it will give us one event from the population, then the ratio of the event over the total possible outcomes will give us the probability.
Possible outcomes
M1 Heads and Monday
T2 Tails and Monday
T2 Tails and Tuesday
One could stop there or, one could argue that since Beauty is asked the credence that there is a second chance mechanism in Beauty's answer. This is not updating a prior with new information, it would be a new event and the population would be dependent on Beauty's selection.
But it doesn't follow from that that each of the other three outcomes are equally likely.
How do you get from
[math]P(Tuesday?Heads) = 0[/math]
to
[math]P(Monday?Heads) = P(Monday?Tails) = P(Tuesday?Tails)[/math]
?
You are not a very honest person.
In the linked page, B is listed as unconditional probability. So yes, unconditional odds of heads is 50%, but Beauty is not working from unconditional, and I don't see how you are applying the additional information of 'it isn't Tuesday/heads' into your computation.
That's the P(Monday|Heads) = 1.
Quoting noAxioms
So what's the condition? P(Heads|Awake)? Well, let's apply the Kolmogorov definition again:
[math]P(Heads|Awake) = \frac{P(Heads ? Awake)}{P(Awake)}[/math]
[math]P(Heads|Awake) = \frac{0.5 * 1}{1} = 0.5[/math]
All four of those things were equal probability. If odds were 50 Monday heads, 25 each Monday tails and Tuesday Tails, then there would be 75% chance that it is Monday, despite the day also being the same odds as the coin toss. I say it is 66% Monday because 2 of the 3 remaining options (A and B) are Mondays.
Excuse me if I am new to the notation. I read this as the probability of it being at least one of Monday or Heads is 1, but since it might be Tuesday/Tails, this is wrong. I would think the probability of Monday or Tails is certain.
Maybe I just don't know how to read the notation.
P(Monday|Heads) means "the probability that it's Monday given the fact that it's heads". If I know that today is heads then the probability that today is Monday is 1. Whereas if I know that today is tails then the probability that today is Monday is 0.5 (because it equally could be Tuesday).
So, P(Monday|Heads) = 1 and P(Heads) = 0.5, therefore P(Monday and Heads) = 0.5.
I think you are mixing the probabilities that something will occur (or is the case, but unknown) with probabilities that something known has occurred. You seem to assign 1 to P(Awake) which is not the probability that you will be away, but rather the probability after the information about being awake has been completely (not just partially) conveyed.
Not sure if a computation of P(awake) is going to yield what we want. It seems undefined. Of course we will be awake at some point. She's not being asked if she's awake, but being awake is information nevertheless.
Sorry again, but the posts are coming faster than I can actually absorb that web link that seems to assume (reasonably) that you already know the rudiments of the notation. For instance the 'unconditional probability' hyperlinks to a page that makes no mention of the term. Not helpful.
The classical approach only works if each outcome is equally likely. As I would argue each outcome isn't equally likely it would be wrong to apply the classical approach. It would be like using the classical approach to a weighted coin toss.
[math]P(Heads|Awake) = \frac{P(Heads ? Awake)}{P(Awake)}[/math]
To get a result of [math]\frac{1}{3}[/math], [math]P(Awake)[/math] needs to be [math]\frac{3}{4}[/math]. This only happens if [math]P(Heads ? Asleep) = \frac{1}{4}[/math].
To get a result of [math]\frac{1}{2}[/math], [math]P(Awake)[/math] needs to be [math]1[/math]. This only happens if [math]P(Heads ? Asleep) = 0[/math].
Given that:
[math]P(Heads ? Asleep) = P(Heads ? Tuesday),[/math]
What is [math]P(Heads ? Tuesday)[/math]?
The host does not, that's the trick.
Quoting Andrew M
Not as the problem was described at the top of the thread. No information is given to Sleeping Beauty beyond what she was told would happen. To her each awakening is identical, and there are three of them.
There's 1/2 chance of heads and Monday happening, but that doesn't mean the chances of the Sleeping Beauty being in that situation are 1/2.
Thinking of the sailor variant, for any outsider the chances of heads or tails are 1/2, but if you're then told the sailor is your father, the chances of tails double because then he's had two children which means the odds of you being his child are higher.
If the odds were 1/2 and the Sleeping Beauty got to bet 1€ each time she was woken up, she should break even no matter what, but we notice that betting tails wins her 2€ with tails and loses her 1€ with heads.
I've gone over this:
If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.
You're conflating "more likely to win if tails" and "more likely that tails".
It's just a 2:1 payout on a 50/50 chance, but only if you bet a certain way.
But she's only given 1€ with tails. The reason she wins by guessing tails is because she's likelier to be in a situation where tails has been thrown.
Quoting Michael
I'm not conflating, I'm drawing a conclusion. She's likelier to win with tails because tails is the likelier outcome of the bet.
She's given it twice: once on Monday and once on Tuesday.
You don't even have to go through the hassle of putting her to sleep. Just tell her that if it's tails she'll get £1 on Monday and a £1 on Tuesday, and if it's heads she'll just get £1 on Monday. She's going to pick tails, because there's twice the payout.
Putting her to sleep, waking her up, and then asking her to bet doesn't make tails more likely.
A simple random sample is a subset from the population where n members of the population have a chance of being selected by random chance, such as a fair coin flip. The actual distribution of the population doesn't matter. This subset is also called an event, and probability is the ratio of the event over the total number of possible outcomes from the population.
Shouldn't matter if their odds are 25%. Then the expected value would be (1/4)*1+(1/4)*1+(1/2)*(-1)=0.
Or if the amount of people doesn't matter, why not wake up the Sleeping Beauty 0 times with heads.
For any given person there's a 50% chance that they're right, so it doesn't matter if they pick heads or tails. It's just that if it's tails and they pick heads then there's a greater number of losers and if it's tails and they picks tails then there's a greater number of winners.
But you don't say that if there's more winners under tails then tails is more likely. That's a non sequitur.
There aren't more winners because it's more likely but because you asked more people.
Of course that's true if they choose randomly whether to guess heads or tails. If they make the sensible choice of tails there're more winners because it's the likelier choice to be the correct one.
Quoting Michael
It doesn't make tails more likely to be the result of my throw, but it makes it likelier for the correct answer to be tails. That's not non sequitur, that's a textbook example of what probability means.
Quoting Michael
It's more likely because I asked more people.
No. There are more winners because you asked more people.
Quoting BlueBanana
This doesn't make any sense.
Quoting BlueBanana
All you're saying is:
Iff 100 people asked then it was tails [the rule of the game]
100 winners [the outcome]
Therefore, it was tails
But that's obvious, and not relevant.
I nowhere said the last two parts.
As for the calculations themselves, they show the odds throwing heads and it being Monday happening, not the odds of it being the current situation at the moment the question is asked. There's also 1/2 chances of throwing tails, then it being Monday, and then it being Tuesday, happening.
Basically the thing I question on them is the usage of the term "heads" and claiming its odds to be 1/2. Do you mean the odds of throwing heads or the odds of them having been thrown when the question is asked? The former leads to what I described above; right answer to the wrong question. The latter is circular reasoning.
It's not supposed to show that 75/225 != 1/3. It's supposed to show that P(Heads) != 75/225.
Quoting BlueBanana
I address the case of being asked here:
[math]P(Heads|Awake) = \frac{P(Heads ? Awake)}{P(Awake)}[/math]
The answer depends on whether P(Awake) = 1 or 3/4. The halfers have to say 1, the thirders 3/4.
I say P(Awake) = 1 because Heads ? Tuesday isn't part of the sample space.
This is more clear if we don't consider days at all and just ask her once if it's heads and twice if it's tails (with amnesia in between). The sample space is H[sub]1[/sub], T[sub]1[/sub], and T[sub]2[/sub] – and in every case she's awake.
Here's a slightly different argument.
If, when awakened, Beauty knew it was Monday, she would answer 50%; if she knew it was Tuesday, she would answer 0%. There's a 100% chance of being interviewed on Monday, but only a 50% chance of being interviewed on Tuesday, based on the fair coin toss. That means it is twice as likely that the interview is being conducted on Monday.
Thus 50% has a 2/3 chance of being the right answer (because the interview is being conducted on Monday) and 0% has a 1/3 chance of being correct (because the interview is being conducted on Tuesday).
From that we can calculate a value for my conditional expectation of heads, given that I am being interviewed:
[math]\small E(H \mid A) = \cfrac{2}{3}(50\%) + \cfrac{1}{3}(0\%)[/math]
[math]\small \phantom{E(H \mid A)} \approx 33\%[/math]
The wagering argument confirms that this is the proper degree of belief.
[math]P(Heads|Monday) = \frac{P(Heads ? Monday)}{P(Monday)}[/math]
[math]P(Heads ? Monday) = P(Heads) * P(Monday|Heads)[/math]
[math]P(Heads ? Monday) = 0.5 * 1 = 0.5[/math]
[math]P(Tails ? Monday) = P(Tails) * P(Monday|Tails)[/math]
[math]P(Tails ? Monday) = 0.5 * 0.5 = 0.25[/math]
[math]P(Monday) = P(Tails ? Monday) + P(Heads ? Monday) = 0.75[/math]
[math]P(Heads|Monday) = \frac{0.5}{0.75} = \frac{2}{3}[/math]
[math]\small E(H \mid A) = 0.75(\frac{2}{3}) + 0.25(0)[/math]
[math]\small \phantom{E(H \mid A)} = 50\%[/math]
Again, that's just a matter of a greater payout. There are two wins with every tails and only one with heads. That doesn't mean tails is more likely. It only means that if you're a winner then you're twice as likely to have won with a guess of tails.
I suppose I could ask you the same question. Heads and tails being throw are each as likely, so how does knowing that you'll wake up twice if it's tails change that?
I suppose both our answers are that common sense isn't a good measure of probability, e.g. the Monty Hall problem.
With each tails thrown, not with each tails guessed. The guesses are separate. You get to guess more times with tails, so naturally it makes sense to guess tails was thrown.
You get to guess more when tails are thrown, so if you're guessing it's likelier tails were thrown.
The probability that you get a guess is 1 in either case. The above reasoning only works if you say that if it's heads there's a 50% chance of getting to guess and if it's tails then there's a 100% chance of getting to guess. Only then can you say that tails is more likely.
Quoting Michael
It works now. It explains why guessing tails is profitable.
I think I see why this is happening, even though the odds are 2-1 against heads.
Quoting Srap Tasmaner
The $1 payoff matrix for a fair coin , betting at even money, is just
If the coin is biased 3:1 tails:heads, we can multiply to get
Betting tails you win $0.50 on average; betting heads you lose $0.50 on average.
Here's the $1 payoff matrix for our Sleeping Beauty, at even money:
It's a fair coin, so multiplying gives you:
So betting tails again gives you a profit of $0.50 on average; betting heads you break even.
So it is because your tails profit is disproportional to your tails risk.
Being more profitable isn't the same as being more likely.
What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?
This seems to be the crux of the disagreement. I say the latter. You seem to be saying the former.
It does win more times, but not because tails is more likely. It wins more times because there are two guesses for each flip of a tails compared to one for each flip of heads.
Completely different situation. There's no "eliminated" outcome. There's no difference between an outside observer and the subject.
This is just another way of saying it's more likely to win.
Quoting Michael
Because of which each guess is more likely to have been caused by tails.
What eliminated outcome?
That's irrelevant and doesn't have anything to do with the probability. We can change the scenario slightly to:
If it's heads then we wake her once. If it's tails then we wake her twice.
What's her credence that it's heads?
Then it's likelier the waking up was caused by flipping tails. As I said,
Quoting BlueBanana
So we're back to this and can ignore your "there's no 'eliminated' outcome" objection:
We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.
What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?
The coin flip with pull a subset of one from the sample space.
Our sample space contains only two possible outcomes:
H - Monday/Heads
T - Monday/Tails and Tuesday/Tails
So our event will have a 1 to 1 odds of H or T. That is a 50% chance each. That part we should all be able to agree on.
Now the second event is when Beauty determines her credence, and I am arguing that this is actually independent of the coin flip, because Beauty does not know the results of the coin flip. She can only make her determination based on what she knows about the experiment, which would mean the event from the coin flip is out of play here. It does not matter if it landed on H or T, as Beauty does not have that information.
The question is what is her credence, or what is her belief, that is independent of the actual coin flip and completely in the hands of how Beauty decides to interpret the experiment.
No, we can't. Eliminating that possibility diminishes the chances of heads being the correct guess.
Quoting Michael
3/4 but the situation is completely different. No difference between observers. Outcomes lead to same amounts of inquiries.
Now this example right here:
Quoting BlueBanana
This one's cool because it too can be calculated quite simply. 1/2 chances of tails and (1/2)*(1/2)=1/4 chances of both no question and heads.
Now what happens if you repeat the test? The Sleeping Beauty problem.
Eliminate what possibility? If it's heads we wake her once. If it's tails we wake her twice. There's nothing to eliminate.
Under Kolmogorov's framework, an event that occurs in more than one probability space can have different probabilities in different spaces. Indeed, that freedom is central to the mathematics of derivative pricing in finance, where alternative probability spaces are employed that use what are called 'risk-neutral probabilities'.
That's like saying you can't eliminate 1 from the set {1} to get { } because there's nothing in { } to eliminate.
I have no idea what you're saying, so let's start from scratch:
Mary 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, Mary 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, Mary will be awakened and interviewed once only. If the coin comes up tails, she will be awakened and interviewed twice. In either case, the experiment ends after the final interview.
Any time Mary is awakened and interviewed, she is asked, 'What is your belief now for the proposition that the coin landed heads?"
There are three possibilities; waking for the only time if heads, waking for the first time if tails, and waking for the second time if tails.
There's nothing to eliminate. This is it. So back to this:
Quoting Michael
Quoting BlueBanana
Good. So let's change it up slightly:
We flip a coin. If it's heads then we wake Mary once. If it's tails then we wake her twice.
What is the probability that it's her first awakening? 2/3, because two of the three outcomes are first awakenings? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)?
We can assume without loss of generality that she decides before the experiment begins what she is going to guess when awoken.
Say she chose heads. Then when awoken she can know with certainty that it is not Tuesday because if it were, the coin must have landed Tails, so she would be dead, having already guessed Heads on Monday.
So if she chose Heads she will never get to Tuesday.
A variant that one might take in seeking to avoid this objection is where the execution occurs on Wednesday, so she will still be woken and asked to guess on Tuesday even if she guessed heads on Monday. But on examination we see that this doesn't change the preferences because she knows on waking that if it is Tuesday and her chosen strategy was Heads then she is already condemned to death so there is nothing at stake in today's answer.
But Beauty is aware of how the experiment is to be conducted.
If I know that you flipped a coin and that if it's heads you'll give me £1 and if it's tails you'll give me £2 then I know that there's a 50% chance that I will get £2.
If Beauty knows that you flipped a coin and that if it's heads you'll wake her once and if it's tails you'll wake her twice then she knows that there's a 50% chance that she'll be woken twice.
Because that's where we get to after we eliminate the one possibility from the aforementioned case.
Quoting Michael
You forgot to weigh the calculations for the potential two wakings. Run the test 100 times, she'll wake up the first time 100 times out of 150.
What possibility have we eliminated?
Where has Tuesday come from? It wasn't mentioned at all in the experiment I described here.
Then you might as well say that we've eliminated the trivial scenario of heads + heads in this experiment. Heads + heads is utterly irrelevant and doesn't change the probability at all. The probability of heads is 0.5 + (0.5 * 0.5) = 0.75.
Heads + two awakenings is utterly irrelevant and doesn't change the probability at all. The probability of first awakening is 0.5 + (0.5 * 0.5) = 0.75.
Because that's the original scenario where it's already eliminated.
Quoting Michael
Completely possible to do so.
Quoting Michael
You mean modifying that scenario so that it includes heads+heads doesn't change the probabilities in it?
I have a new original scenario.
Quoting BlueBanana
Here you said "Completely different situation. There's no 'eliminated' outcome." as if to explain why the probability is 0.75 and not 2/3.
I'm saying that there is an eliminated outcome – heads + heads – and yet the probability is still 0.75. So it doesn't matter if you talk about an eliminated outcome or not. The probability is 0.75 whether you consider it or don't.
With the outcome added there are certain odds, and when that is modified by eliminating that one possibility the odds are changed in a way that somehow makes sense. It's just another way to look at the problem, like how in the Sleeping Beauty problem the chances are 2/3 whether one thinks about the eliminated extra possibility or not. Thinking about it doesn't change the situation but coming to the same conclusion through eliminating it can be use to confirm the solution.
But as we disagree about what conclusion is reached by looking at the problem that way I don't think we can achieve an agreement through it.
No, it doesn't. If you toss a coin once and then again if it's tails then there's a 0.75 chance that you'll finish with a heads. It doesn't matter if you consider heads + heads or heads + tails or tails + tails + heads or any number of trivially eliminated outcomes. It's always just going to be 0.75.
You're really confusing me now. Perhaps you could explain what you meant by this objection:
Quoting Michael
Quoting BlueBanana
In the coin flip example to get into any outcome you need to take away some outcome from which to get to the new outcome. Like heads is twice as likely as tails + heads, but heads + tails removes heads because you have to proceed with a new flip.
There is a 50% chance that Beauty will be interviewed once, and a 50% chance that she will be interviewed twice, determined by the toss of a fair coin.
Because the second interview, if it happens, takes place on Tuesday, the chance of being interviewed on Tuesday is also 50%.
There are two orthogonal partitions of the space, one by Heads/Tails, and one by Monday/Tuesday. Each divides the space into subspaces of equal probability. Thus
[math]\small H \cap M = H \cap M^C = H^C \cap M = H^C \cap M^C = \cfrac{1}{4}[/math].
If you now restrict the total space to occasions when Beauty might be awakened, according to the rules of the experiment, you lose [math]\small H \cap M^C[/math], and now you have
[math]\small H \cap M = H^C \cap M = H^C \cap M^C = \cfrac{1}{3}[/math].
It is this identity that Beauty will rely on, given that she has been awakened.
I'm not sure if you're just disputing the thirder position, or disputing my characterization of the thirder position for the 5/6 heads-weighted coin. Assuming the latter, there are twelve states, including awake and asleep over the two days, each with 1/12 probability. Seven of those states are awake states.
Conditionalizing on being awake, the probabilities for the halfer and thirder positions are:
Halfer (5/6 heads-weighted coin):
Thirder (5/6 heads-weighted coin):
The halfer distributes all the Tue/Heads state probability to Mon/Heads. That is, 5/12 + 5/12 = 5/6. The thirder distributes the Tue/Heads state probability evenly to each awake state. That is, (5/12 / 7) + 1/12 = 1/7.
For the thirder:
P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails) = 2/7 * 1/2 = 1/7
Edit:
Oops, the above result is conditionalized on being awake, per the above thirder table. So actually P(Tails and Tuesday) = 1/6 * 1/2 = 1/12 as Michael correctly noted. The unconditional probabilities are:
P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 5/12 / 7/12 = 5/7.
Beauty (as a thirder) knew at the time before the experiment that P(Heads) = 1/2. She also knew at the time before the experiment that P(Heads|Awake) = 1/3. When she does awake in the experiment and learns this, that self-locating information enables her to update P(Heads) to be 1/3.
Note that Beauty (as a halfer) does the same thing when she is told that it is Monday. For a halfer, P(Heads|Monday) = 2/3 so, on attaining that self-locating information, she updates P(Heads) to be 2/3.
How so? Isn't the host telling you a specific door number (that doesn't contain the prize) information?
Quoting tom
The event of waking provides information about which states one can eliminate (not merely the conditional probabilities, which were previously known). BTW, do you assign 1/2 or 1/3 to P(Heads|Awake)?
I didn't follow this part of the thread, so sorry for the late reply.
Beauty isn't assigning 1/3 from a principle of indifference. (I made the same mischaracterization earlier,)
Monday splits 50:50 because the coin is fair; Tuesday happens or not on a toss of the same fair coin, so each quadrant is 25%, by stipulation.
Beauty conditionalizes on being awakened, so the values change to
[math]\small \cfrac{\cfrac{1}{4}}{\cfrac{3}{4}}=\cfrac{1}{3}[/math].
With that preface, perhaps I misunderstood your post here. In the fifth line of the reasoning, you assert, seemingly as an axiom, that P(Heads)=0.5. While - as indicated above - I sympathise with that position, it does look like assuming one's conclusion, if the object of the exercise is to determine what probability Beauty assigns to the coin having landed Heads.
If we do not assume our conclusion, we need to make P(Heads) an unknown, which I called p, and see if we can assert other, less controversial relationships that enable us to deduce that. But I can't see how we can do that without assuming something that is essentially equivalent to P(Heads)=0.5.
To me it looks irresolvable, which is why I like to turn to betting games to make the question concrete.
She is never told it is Monday, each awaking is the same, there is no hint as to which day it is; temporally she is uncertain of her location.
So what "self-locating information" allows her to reallocate credibility? You need to be specific. If I am working on a Bayesian analysis, and I get up to go to the bathroom, siting on the toilet doesn't count as new relevant information. Just her waking up is not good enough, she has to actually gain new information.
Well, it certainly seems like you are being given information, but you're not. The host, because he has total knowledge of the situation, can always open an empty door. There will always be an empty door. So, the problem is identical to choosing 1 door or 2 doors.
You, as the contestant, know for certain that one of the other two doors is empty, once the door is opened, you still know for certain that one of the doors is empty. OK, you now now which one is empty, and that IS information of sorts, but is it relevant information?
Anyway, I've never encountered anyone who agrees with me on this.
Quoting Andrew M
In the original description of the problem, Sleeping Beauty wasn't told what day it is, so all waking events are identical. She knows there is a probability space of 3 independent events, and that 2 of them are associated with tails.
All she knows is that she is awake, and that is twice as likely to be associated with tails.
I think there are two issues.
First, getting conditionalizing on being awake right. It's clearest perhaps to imagine the coin being tossed after the first interview. Could that coin toss have come up heads? Did it have a 50% chance of doing so? Yes and yes, but Beauty will not be asked about it if it did.
Second, should Beauty conditionalize on being awakened, since she already knew she would be and cannot know how many times she is awakened. I think it only makes sense to consider the odds, in the absence of knowledge. What's more, if you don't, you're in the position of saying, "I'm right about the odds and can prove it by consistently losing money on tosses of this fair coin."
I think it's clearer with this:
[math]P(Heads|Awake) = \frac{P(Heads ? Awake)}{P(Awake)}[/math]
Our objective is to determine P(Heads|Awake), and so it's not necessarily begging the question to assert that P(Heads) = 0.5.
It really depends on what we understand by the (unconditional) probability that it's heads. I understand it as being the probability that a fair coin toss lands heads.
That is because you are wrong. Knowing which door is empty is the new information which calls for a reallocation of credibility. You know now one of the the possible empty doors. Yes there is still an empty door but before there were two empty doors and now you know at least one of them. That is a real reason to reallocate credibility.
That is not how conditionals work. If it were, there would never be a point to conditional probability. You'd just always use 1, and P(A|B) would just be a funny way of writing P(A). In how much of the total space is she awakened? Your way is just wrong.
P(Heads|Awake) = P(Heads) * P(Awake|Heads) / P(Awake)
If she applies this before the experiment then she knows that P(Heads|Awake) = 0.5 * 1 / 1 = 0.5.
So what is P(Awake)?
3/4.
It is possible to have a heads on Tuesday, same as always. It's within the total space, just not the conditional space.
Why count heads on Tuesday?
What if I were to change the experiment so that it ends on Tuesday if it's heads rather than on Wednesday? In the original we don't count Heads + Wednesday (or Tails + Wednesday) so in my variation we don't count Heads + Tuesday.
Or what if we wake her on Monday if it's heads and wake her on Tuesday and Wednesday if it's tails, ending the experiment on Thursday? Is P(Awake) = 3/6?
This is why I suggested the alternative here where there's no mention of days at all; just one awakening if heads and two awakenings if tails. The sleep day is a red herring.
How does this new information alter the fact that one door has a probability of 1/3, and 2 doors has a probability of 2/3?
If you have to pick one of three doors then the probability of being right is 1/3.
If you have to pick one of two doors then the probability of being right is 1/2.
Later I might write a script to test the Monty Hall problem. I believe prior experiments have supported the hypothesis.
If you were given the choice of 1 or 2 doors, which would you choose? I hope you would choose 2. What about the extra information that bears sh*t in the woods? I suspect you would still choose 2 doors. What about if you were told that if you choose 2 doors, the host will open an empty door as an irrelevant dramatic flourish? It makes no difference.
The host can always open an empty door, an act as relevant to the choice as the behaviour of bears in the woods. It's a choice between 1 or 2 doors, that's all there is to it.
If whether each happens is determined by the toss of a fair coin, they're all equal and 3/6 is right. (If the subspaces had unequal probabilities, you might still be picking half of them numerically but they wouldn't add up to half the space.)
So we do my variation where there are no sleep days. The experiment is just; we ask her once if it's heads and twice if it's tails.
What do you think this changes?
Or, better still, can we just stick to the problem at hand?
Here's the result of 100,000 games where you don't re-guess. 1/3 games are winners
Here's the result of 100,000 games where you re-guess. 1/2 games are winners
Go down towards the bottom and hit "Execute".
It changes P(Awake) to 1. Which means:
P(Heads|Awake) = 0.5 * 1 / 1 = 0.5
Dude, if she's not sleeping we wouldn't call the condition "awake". What matters is when she's asked.
Then:
P(Heads|Asked) = P(Heads) * P(Asked|Heads) / P(Asked)
P(Heads|Asked) = 0.5 * 1 / 1
P(Heads|Asked) = 0.5
You really just need to look up conditional probability again.
Are you suggesting that, in my amended case, being awake/asked shouldn't be treated as a conditional?
You scaled by 1/1. That's not the idea.
Seriously, read a tutorial and then we can talk about the philosophy. I don't know what else to say.
The probability of winning, when you choose 2 doors is 2/3. That is all there is to it. The fact that one of these doors is open is irrelevant. One door can always be opened, because one of them is certain to be empty.
Having two opportunities to choose one door (i.e. change your selection) isn't the same as having one opportunity to choose two doors (i.e. select both).
If I choose door A then I have 1/3 chance of winning. If I change my mind and choose door B then I still have 1/3 chance of winning (for the moment let's forget about Monty opening a door).
The option of selecting both A and B isn't an option in the Monty Hall problem, though, so I'm not sure why you'd bring it up.
On Sunday all four she will be put to sleep, and a fair coin will be flipped. Three of the Beauties will be wakened on Monday, according to the coin result and their cards.They will be interviewed, and put back to sleep with an amnesia-inducing drug that makes them forget that awakening.
This will be repeated on Tuesday.
In the interview, each awake volunteer will be asked for her belief that the coin result is the one named on her card.[/b]
Each volunteer knows that three volunteers are awake. Each knows that exactly one of these three has a card that names the coin result. None have any information that could make her believe that she is more,or less, likely to be the one. So what should her belief be?
One of these volunteers is experiencing the exact same problem that started this thread. The others are experiencing equivalent problems that must have the same answer.
Quoting tom
If the door you had selected has the car behind it (call this event C) then it is not relevant information.
But if your selected door does not have the car behind it then it is very relevant information, because it means the car is behind the remaining closed door that you did not select.
Since P(C)=1/3 there is a 1/3 probability of the information being irrrelevant and a 2/3 probability of it being relevant. So the probability of a win is increased by assuming the information is relevant, and using it, which means switching. The calcs are:
We have P(C) = 1/3. If you retain your selection then the probability of a win is:
1/3 x P(win | C & not switch) + 2/3 x P(Win | ~C & not switch) = 1/3 x 1 + 2/3 x 0 = 1/3.
On the other hand if you switch the probability of a win is:
1/3 x P(win | C & switch) + 2/3 x P(Win | ~C & switch) = 1/3 x 0 + 2/3 x 1 = 2/3.
We could call the new information 'conditionally relevant information', since its helpfulness is conditional on an unknown, which is whether C is the case.
Another approach, for those that don't find the above persuasive, is to say that there is no new information, but reject the dictum that one should not change one's guess in the absence of new information, replacing it with a dictum that one should not change one's strategy unless one receives new information. The best strategy is to change one's guess, after Monty has opened a door. That strategy does not change when Monty opens the door.
Consider a slightly different game in which Monty doesn't open a door but you can still switch your guess and, if you do so, you will win if the car is behind either of the two doors you did not originally select. The probabilities in this game are identical to those in the original Monty Hall problem, but in this case you clearly get no new info, since no door is opened until the final reveal-all. But it is in your interests to switch. You will have changed your guess, but not your strategy.
Imagine a stock market strategy of buying stocks of companies that have had a certain event, say a change of CEO. Say our research has showed that on average there is an upwards surge in stock price on appointment of a new CEO, that peaks on average three days after the announcement, and then often subsides. Then we could set up a hedge fund that buys stocks that have had such announcements, and sells them after three days. In most cases there will be no important new info about the company in those three days, but we will still sell the stock. We've had no new info. We've changed our stock-pick (our guess) but not our strategy.
I'm saying that for a halfer, P(Heads|Monday) = 2/3. As a separate hypothetical, if Beauty is told that it is Monday during the experiment, then she will have received self-locating information. So she would update P(Heads) to 2/3.
The thirder is making a parallel argument. Before the experiment, there are four states of 1/4 probability, where one is a sleep state. If she is in a state where she awakes, that is information that would rule out the sleep state. So she would update P(Heads) to 1/3.
P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 1/4 / 3/4 = 1/3
Yes. Your initially chosen door has 1/3 probability of containing the prize. The other two doors have a total of 2/3 probability. When the host shows you that one of those doors is empty, its probability goes to 0. So the remaining door now has 2/3 probability. So you should switch to that door and you will win the prize 2/3 of the time.
Quoting tom
Agreed.
P(Heads|Awake) = (P(Heads) * P(Awake|Heads)) / P(Awake) = (1/2 * 1/2) / 3/4 = 1/3
There are four equally probable states in the experiment, three awake states and one sleep state, so P(Awake) = 3/4 and P(Awake|Heads) = 1/2.
I'm saying that P(Heads|Monday) = 2/3 is a consequence of the halfer position. Do you agree?
BTW, that is David Lewis' (L6) which contradicts Elga's (E1) that P(Heads|Monday) = 1/2. See http://fitelson.org/probability/lewis_sb.pdf
Michael, I see you're still going on about this. Spent a bit skimming the posts since I dropped off.
Quoting MichaelYou're doing it wrong. You are mixing probabilities from different times, different points of view, or from positions with different information.
Choose a point and stick with it. P(Heads) of 50% is true only before the coin is tossed but you seem to assign that probability to Beauty's POV when in a waking period. That is question begging (as I pointed out early in the thread) since that's the answer we're looking for, not the premise. At that time, it is not 50% from her POV nor the administrators, where it is 0% or 100%.
No, and I already commented on Lewis and Elga.
Also Lewis is arguing that 2/3 is a consequence of Elga's 1/3 position. Lewis is arguing for the 1/2. Elga is the one arguing the 1/3, and both of their responses have already been posted, by myself.
Lewis argues two cases, L2, which is P(Heads)=1/2=P(Tails) and then he notes quod erat demonstrandum. Then moves on to L6 which is P with the plus. Read the article to find out why it has a plus.
Yes, to expand, P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 1/4 / 3/4 = 1/3.
What that equation is doing is redistributing probability from the sleep states to the awake states in proportion to each awake state's probability. In the case of Sleeping Beauty, that is equivalent to being indifferent about which awake state she is currently in, since each awake state has the same 1/4 probability. That is, when conditionalizing on being awake, the sleep state probability of 1/4 is evenly distributed to each awake state (1/4 + (1/4 / 3) = 1/3).
In other scenarios where the state probabilities are unequal (such as the 5/6 heads-weighted coin scenario), it is first necessary to transform the original state space to states of equal probability. Then the indifference principle similarly applies. So for the 5/6 heads-weighted scenario, we get:
P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 5/12 / 7/12 = 5/7.
It says right in the OP, " Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details." That is the first line of the problem.
I am aware why it has the plus. As the paper says, "Let P+ be her credence function just after she's told that it's Monday". From Lewis' halfer argument:
(L6) P+(Heads) = 2/3
(L6) is a consequence of the halfer position. Do you agree?
What on Earth do you think that tag question does?
I just gave you my response and I am not going to wade through your clear misunderstanding of Lewis's argument.
Sorry. Don't remember reading that. I brought up the issue before.
You're mistaken. (L6) P+(Heads) = 2/3 is a consequence of Lewis' halfer argument and it contradicts Elga's (E1) P+(Heads) = 1/2. As Lewis says:
"I reject Elga's premiss: my (L6) contradicts his (E1)."
You are not paying attention to what the plus means, where it shows up and where it doesn't show up.
You are confusing information with theatrics.
The Monty Hall problem and its analysis is identical to choosing 1 door, or choosing 2 doors. There is no more to it than that.
We're getting there. If the host instead told you that he could see that behind one of the other two doors, there was nothing. Do you think he is telling you something you don't already know for certain?
Would you still swap to choosing 2 doors, or stick with your initial choice of 1.
It's the same problem, with identical results, just without the sleight of hand.
Yes, if you switch you are effectively choosing two doors, which leads to the accepted solution that it is optimal to switch. That's what I said above.
Perhaps I misunderstood your post. It appeared to me from your statement that you've never met anybody that agrees with you on this, that you were rejecting the accepted solution. If you were, I'd be interested to know why. If not, there's nothing to disagree on.
It's 1/2 if you switch. I ran the experiments:
Here's the result of 100,000 games where you don't re-guess. 1/3 games are winners
[s]Here's the result of 100,000 games where you re-guess. 1/2 games are winners[/s]
Go down towards the bottom and hit "Execute".
Edit: Actually, the logic of the 1/2 winners was wrong. It doesn't switch doors but reselects at random. I'll redo it.
You (and @tom) were right. 2/3.
Ignore the bottom branch and treat the first two as Tails and Heads respectively each with probability 1/2. The Tails branch splits into 1st awakening and 2nd awakening (we're using my amended scenario where there's no mention of days) each with probability 1/2. The Heads branch extends to 1st awakening with probability 1. Using the multiplication rule as used in the above tree, we get Tails + 1st awakening = 1/4, Tails + 2nd awakening = 1/4, and Heads + 1st awakening = 1/2.
What I don't understand is why one would add a 2nd awakening branch to the Heads branch (and giving it a probability of 1/2?) if we know that in our experiment a flip of heads never offers a 2nd awakening. After all, we wouldn't add Door 1 or Door 2 as branches from the Door 2 branch because we know the rules prohibit such an event.
Under this code the prob of winning under no-switch is about 1/3 and 2/3 if the contestant switches.
Yes, the host's information is irrelevant if you can just switch to the other two doors instead of specifying a particular door number.
Now those first two branch probabilities don't seem particularly plausible since there are twice as many first awakening states. But the same can be said for the halfer view since there are twice as many tail states.
The probability of the awakenings is dependent on the coin flip (1st awakening is 1 if heads, 0.5 if tails), whereas the probability that a coin flip lands heads is independent. So I would say my branching makes sense whereas yours doesn't.
Or am I misunderstanding this?
Quoting Andrew M
I don't understand why the number of states matters. I roll a dice with 3 red faces, 2 blue faces, and 1 white face. I give you a ball with the colour of the face that the roll lands on. There might be three states (red, blue, or white ball), but the odds aren't 1/3 each. What matters are the probabilities that lead up to that state. A 1/2 probability leads to the red ball state, a 1/3 probability leads to the blue ball state, and a 1/6 probability leads to the white ball state.
If I run this experiment and give you one of the balls, hidden in a box, you don't just think "well, it could be one of any three colours, and there's no distinguishing feature in my current state to suggest that it's one rather than the others, so it's 1/3 chance of being red". Instead you think "well, the initial dice roll gave 1/2 odds of red, so it's 1/2 odds of red".
So why not apply the same reasoning to Sleeping Beauty? The initial coin toss has 1/2 odds of heads, so it's 1/2 odds of heads.
And to continue the analogy, let's say that I let you guess twice in the case that it's not a red ball. What's the smarter guess? Obviously to guess that it isn't red, because if you're right then you're right twice. If we record these winnings then we can see that you were right 2/3 of the time. But that doesn't mean that there was a 1/3 chance of getting a red ball. Introducing an element of amnesia in between the two guesses doesn't change that.
I believe this is a disagreement between the self-sampling assumption and the self-indication assumption.
This is why I suggested the alternative experiment where we don't talk about days at all and just say that if it's heads then we'll wake her once (and then end the experiment) and if it's tails then we'll wake her twice (and then end the experiment). There aren't four equally probable states in the experiment.
So we either say that P(Awake) = 1 (and/)or we say that being awake doesn't provide Beauty with any information that allows her to alter the initial credence that P(Heads) = 0.5.
This is not what the question is about. The question is whether there's 1/3 chance of being questioned as a result of getting a red ball.
Quoting Michael
There's still the fourth state of it being the time you'd be awakened the second time and heads was thrown.
What are the odds you're a part of the experiment because heads was thrown?
Also, the time of the awakening(s) is random.
What about if I’m asked in either case but don’t know if I’m first or second or third (and so on) in the case of tails? There might be multiple states for tails but I’d still say my credence should be 0.5.
That's not how it works, it needs to be a random person chosen for the situation to be comparable. The Sleeping Beauties at different scenarios are comparable to separate persons because of the amnesia, except arguably the Monday versions, who however don't have the knowledge it's Monday, unlike you, who always has the knowledge you are you.
If there's a 1/4 chance that I'm being asked on Tuesday then there's a 1/4 chance that if I'm Sleeping Beauty being interviewed right now then today is Tuesday.
Notice how you've decided (correctly) not to consider Heads + Tuesday and argue that in my example there's a 1/3 chance of it being heads. Why the inconsistency? Is the fallacy of considering Heads + Tuesday as a state in the experiment more evident in my example than the Sleeping Beauty example?
Although the main point I've been arguing here is that your example isn't at all comparable. In Sleeping Beauty's case there's a probability of 1 that she will be asked if the coin lands heads and a probability of 1 that she will be asked if the coin lands tails, whereas in your case there's a probability of 1/7.5 billion that I will be asked if the coin lands heads and a probability of 7.49 billion/7.5 billion that I will be asked if the coin lands tails. That makes all the difference and is why the cases are incomparable.
It is a possible outcome; she just won't be asked about it.
What's more, Wednesday & Heads and Wednesday & Tails are both part of the sample space. In those cases, she'll be awakened but not asked.
We can only ask her about her credence if she's awake; and we can't ask her if we're not asking her. But she can have a credence whether she's awake and whether we ask.
Think about this -- the variation we haven't discussed -- why does Beauty's credence in HEADS increase by 1/6 when she's told it's Monday? For Elga, it's an increase from 1/3 to 1/2 (and he's right), but for Lewis it's from 1/2 to 2/3 (and he's wrong). Where does this magic 1/6 come from?
It comes from the sample space, which has 6 equiprobable possibilities.
We can ask what Beauty's credence that a fair coin will land HEADS:
Looked at this way, it's a whole lot of fuss over nothing. The whole point of all the epistemic this-and-that, the centered and uncentered worlds business, all that, is just to get us to pretend we don't know things we know, on the theory that Beauty could not possibly know them, that she is in some astonishing epistemic quandary.
Beauty just needs to know how to count. Her credences are all consistent given the various conditions we might impose. It's pure semantics really to say that her credence changes at all.
We can imagine that on HEADS, we wake up Beauty Tuesday and send her on her way. That's no different from putting her back to sleep after a Tuesday interview and waking her up Wednesday. She will not know the difference between sleeping two days and only one. So that would be a sample space of 5 instead of 6 -- the experiment can conclude on two different days, depending on the coin, so that's not the explanation for the magic 1/6.
It's also the same as not counting Tuesday & Heads.
And Tuesday first week of April in the year 3030 is also in the sample space, I suppose.
The sample space is either head and tails, or awakened states.
1/6 is just 1/2 - 1/3, of no interest in itself, and we don't really need to throw in the Wednesday states or worry about whether the experiment concludes only on Wednesday or on either of Tuesday or Wednesday. Some denominators change. None of it matters. It's still just counting.
There's no conflict between believing P(HEADS) = 1/2 unconditionally and believing P(HEADS | you're asking) = 1/3. That's all there is to it. If asked, she should say "1/3". I don't think there's anything like her credence changing going on it at all. It only looks that way because she's answering the question "What is your credence?" If you don't ask, it's one thing; if you do, it's another. Just the usual silliness.
Because unlike in the Sleeping Beauty problem, in your example if there's x probability of going through any day you can draw the conclusion there's x probability of it being that day. The heads + Tuesday option is still there, but because of the rules of your variation the odds of going though any outcomes with a specific coin flip need to add up to 50%.
1/2+1/6 = 2/3
P+(Heads) = P(Heads)+1/6
1/6 is the differences between P+(Heads) and P(Heads)
I think it is just a statement noting the difference, to show they are not the same thing.
That is P(HEADS | you told me it's Monday) = P(HEADS | you asked me) + 1/6
For Elga, that's 1/2 = 1/3 + 1/6
Here's how I got the idea to include Wednesday, and it's still a good argument, I just dropped it because it all comes down to counting anyway.
Lewis accepts that being told it's Monday, and therefore not Tuesday, is relevant new information. (It tells you you're not in the future!) He does not accept that being asked for your credence is relevant new information, but it is: it tells you it's not Wednesday. If he accepts one, he ought to accept the other. Game over.
I had this whole analysis worked out about Beauty's credence upon being awakened Monday or Tuesday or Wednesday before being interviewed -- awakened on Wednesday, she won't know the experiment's over until they tell her. You can redo the weighted average credence I did before (and which is the same argument Elga makes) including Wednesday:
[math]\small \cfrac{1}{3}(50\%)+\cfrac{1}{3}(0\%)+\cfrac{1}{3}(50\%)\approx 33\%[/math]
I still find that convincing. Maybe it's even more convincing than
[math]\small \cfrac{2}{3}(50\%)+\cfrac{1}{3}(0\%)\approx 33\%[/math]
because it bakes in why being asked the question is informative, and why Lewis ought to be willing to conditionalize on H1 ? T1 ? T2.
I already know this. It is a statement showing P+(Heads) does not equal P(Heads). Elga's argument is that her credence in H, because she knew she would be awaken on Monday, should shift when she is awakened on Monday. Lewis is merely showing that the two are not the same thing.
https://www.princeton.edu/~adame/papers/sleeping/sleeping.pdf
When you are first awakened, you are here:
[math]\small P(HEADS) = \cfrac{1}{3}(50\%)+\cfrac{1}{3}(0\%)+\cfrac{1}{3}(50\%)\approx 33\%[/math]
The [math]\small \cfrac{1}{3}[/math]'s there are the days -- you don't know if it's Monday or Tuesday or Wednesday, but you know you've been awakened.
(Do we really have to say you already knew this was going to happen? That you were certain? What if you're awakened by a bloody research assistant surrounded by rubble? Do you assume the experiment is still running? What's become of this certainty about the future Elga attributes to you?)
As soon as you're asked for your credence, you switch to
[math]\small P(HEADS) = \cfrac{2}{3}(50\%)+\cfrac{1}{3}(0\%)+\cfrac{0}{3}(50\%)\approx 33\%[/math]
and now you're weighting by possible interviews per day.
Quoting Jeremiah
Well, yeah. She's awakened but not interviewed. According to the setup this does happen.
Sorry, I actually misread a few words in your post, however. . .
Wednesday is irrelevant. The interview is what triggers the response, before then you are merely speculating. Also the probability if it being Wednesday would drop to 0% as soon as the interview happened, and her response to the interview is what we are concerned with. And one more point, Wednesday is not new information. Everything we know, she knows.
We are already there and no one has been put to sleep yet.
Here is the output:
[1] 5037 - Number of Tails
[1] 4963 - Number of Heads
[1] 4963 - Number of Monday and Heads
[1] 5037 - Number of Monday and Tails
[1] 5037 - Number of Tuesday and Tails
You can see that the 1/3 argument does in fact lead to a 33% split and the 1/2 argument does lead to a 50% split; however, you should go with the 1/2 argument, as 4963/10000 is better odds than 4963/15037, therefore you have a better chance of being correct.
The simplest way to block the Wednesday argument is to change the experiment: they send you on your way immediately after your last interview. It's cleaner. The wake you up on Monday, flip a coin, and ask for your credence. If it was heads, they send you on your way. If it was tails, you get the amnesia sleep, then they wake you up on Tuesday, ask for your credence and send you on your way. No letting you sleep through Tuesday, on heads, no waking up Wednesday not knowing if you'll be interviewed.
Now when you wake up, all you have is the original weighted average:
[math]\small P(HEADS) = \cfrac{2}{3}(50\%)+\cfrac{1}{3}(0\%)\approx 33\%[/math]
I don't know if anything here counts as "updating".
Quoting Jeremiah
The problem is that wagering confirms the odds are 2-1, which, duh, there are 2 tails interviews for every heads interview. If it's all about getting to give the right answer most often, there's no way to go but tails.
She is asked about if the coin landed on heads, we can't go tails. The short and dry of it is all, is that a coin flip is still a coin flip. I posted my simulation up above, if you are considering the 1/2 argument then the odds are not 2:1 in favor of tails, they are 1:1 and those are better odds.
Oh right. Give the 50% answer you'd give on Monday, because there are more Monday interviews. I remember thinking about that a while ago -- you get to be right 2/3 of the time.
This no good though. If you know you're wrong 1/3 of the time, your credence is really the 33%.
I understand Elga's argument. I understand the wagering argument. Do you understand Lewis's argument? I don't. He tries to get you to accept P+(HEADS)=2/3 because of something about prophecy and backwards-in-time causation and accepting credences you wouldn't normally. I've read it several times but I don't know what he's on about.
He is trying to show Elga's argument leads to a contradiction, the same contradiction you posted here: https://thephilosophyforum.com/discussion/comment/184884
But I have already said several times, I think they are both wrong.
Huh. Didn't realize my first argument might be a contradiction.
I'll slog through the Lewis some more. He also notes that you can't jump straight to indifference about which of the three interviews is happening -- although you can argue for them being equiprobable -- and I didn't quite get that either.
If your position is that it's just a prior and you can pick whatever you want (thus Bayesianism is intellectually bankrupt), then the section at the end of Lewis's article is relevant right? It was all this stuff about believing now a credence you know you should believe in the future.
Quoting Jeremiah
I think "What do you mean?" is a good answer. "If you weren't asking me in this bizarre manipulative way, I'd say 50%; since you are I'll say 33%."
That was never my point. My point is that you don't update priors with priors. Priors can be subjective but the "new information" needs to be objective. Priors on the other hand need solid reasoning to be used, and I have said many time both are valid positions; as such, you could justify either one as a prior.
You are too fast for my ninja edits, so you end catching my typos. I don't want to review any more Lewis tonight, maybe tomorrow I'll take another look at the end.
Hey look at that. He saw the Wednesday argument and slipped in a defeater!
True, but the quarterer would agree (first awakening is 1 if heads, 1/3 if tails). However the probability of being Monday or Tuesday is also independent and equally probable. The quarterer would argue that those probabilities should carry through to the first and second awakening probabilities in the experiment since no new information had been acquired on awaking.
Quoting Michael
The nature of the experiment means that the odds are different for the awakened Beauty. If she is told that it is Monday, the odds are different again and this is the case for both halfers and thirders. In this latter case, it seems that your analogies apply equally to halfers.
Quoting Michael
Saying that P(Awake) = 1 is fine since we can calculate the other probabilities accordingly. But I think the four equally probable states clarifies the mathematical relationship between the independent observer viewpoint (where the odds are familiar and intuitive) and Sleeping Beauty's viewpoint.
If it's 1 then P(Heads|Awake) = 0.5 * 1 / 1 = 0.5.
Quoting Andrew M
Then perhaps you could explain how it works with my variation where Beauty is woken on either Monday or Tuesday if tails, but not both. Do we still consider it as four equally probable states and so come to the same conclusion that P(Heads|Awake) = 0.5 * 0.5 / 0.75 = 1/3?
If it's 1 then P(Heads|Awake) = 0.5 * 1 / 1 = 0.5.
Quoting Andrew M
Then perhaps you could explain how it works with my variation where Beauty is woken on either Monday or Tuesday if tails, but not both. Do we still consider it as four equally probable states and so come to the same conclusion that P(Heads|Awake) = 0.5 * 0.5 / 0.75 = 1/3?
My own take on it is that our states would be set up with the probability of being awaken on that day as so:
And I know that in Sleeping Beauty's case the observer's states would be set up as so:
But although Sleeping Beauty knows that she will be woken on both Monday and Tuesday if tails, she doesn't know that today is Monday or is Tuesday, and so from her perspective she has to reason that if the coin flip was tails then the probability that today is Monday is 0.5, giving her the following:
And this is exactly Elga's reasoning:
So her reasoning should be:
1. P(Tails) = 0.5
2. Tails ? P(Monday) = 0.5
3. P(Tails ? Monday) = 0.25
This is correct when waking up just once, so why not also when possibly waking up twice?
Because when you are woken up more than twice more awakenings are caused by throwing tails.
In scenario one there are two awakenings for every tail thrown and one for every head thrown. It's a fair coin, and so the proportion of tail-awakenings to head-awakenings is 2:1.
In scenario two there is one awakening for every tail thrown and one for every head thrown. It's a weighted coin that favours tails, and so the proportion of tail-awakenings to head-awakenings is 2:1.
These are very different scenarios, despite the same result. I don't think it makes sense to simply use the proportion of tails-awakenings to head-awakenings to consider the likelihood that the toss was tails.
Consider a scenario where if it's heads then Amy and either Bob or Charlie is asked, and if it's tails then all three are asked. How do we determine the likelihood that the toss was tails? Do we simply say it's 3:2 in favour of tails? I say no. We have to look at each individual and ask first "what is the likelihood of a coin toss landing heads?" and then "what is the likelihood that I'll be asked if the coin toss lands heads?". For Amy, Bob, and Charlie, the answer to the first question is "1/2", for Amy the answer to the second question is "1", and for Bob and Charlie the answer to the second question is "1/2". So for Amy it's a 1/2 chance of heads and for Bob and Charlie it's 1/4 chance of heads.
Speculation on Beauty's possible speculation was never a good foundation for reallocation.
Elga's argument is: If that is so then when Beauty considers the actual question that is when temporal location is relevant and at that time Wednesday is already off the table, as the asking of the question eliminates it. And speculation that Beauty was considering Wednesday as a possibility before the interview is speculation and is outside the relevant temporal location which is in consideration.
If you don't buy that argument, then you still have not explained how this supposed new information is relevant to the uncertainty of temporal location when considering if it is Tuesday or Monday. That uncertainty still remains, and probability is the measure of uncertainty. Wednesday is certainly off the table, at this point.
Done a little more sniffing around, and thirders frequently argue there's information here. Elga doesn't.
As SB, you are asked for your degree of belief that a random event has occurred or will occur.
If I flip a fair coin and ask you for your degree of belief that it landed heads, you'll answer 50%.
Suppose instead I say I'm going to tell you how it landed. What is your degree of belief that I'm going to tell you it landed heads? It will again be 50%. They're usually identical.
Now try this with SB: instead of asking for your degree of belief, I'm going to tell you how the coin toss landed. What is your degree of belief that I will tell you it landed heads? Is it 50%?
We thirders think halfers are looking at the wrong event. Just because you're asked how the coin landed, doesn't mean that's the event you have to look at to give the best answer.
(I've also got a variation where I roll a fair die after the coin toss, and ask or tell you twice as frequently on tails. Same deal: what's the random event? Is it just the coin toss?)
Why should I even consider the Bayesian approach in the first place?
Yes?
Is it?
I think the halfer intuition is that a coin toss is a coin toss -- doesn't matter if you're asked once on heads and twice on tails.
But consider this. What is your expectation that I'll tell you it was heads, given that it was heads? 100%. What's your expectation that I'll tell you it was tails, given that it was tails? 100%. Does that mean they're equally likely? To answer that question, you have to ask this question: if I randomly select an outcome-telling from all the heads-tellings and all the tails-tellings, are selecting a heads-telling and selecting a tails-telling equally likely? Not if there are twice as many tails-tellings.
Both conditionals are certainties, but one is still more likely than the other, in this specific sense.
If that means the "subjective" interpretation of probability, it's just what the question is about.
Maybe it ends up showing that "degree of belief" or "subjective probability" is an incoherent concept and we all become frequentists.
Sure, but that's not how we'd actually consider the probabilities. We can try it right now. I'll flip a coin. If it's heads I'll tell you that it's heads. If it's tails I'll tell you twice that it's tails.
Do you actually think that there's a 1/3 chance that I'll tell you it was heads (and a 2/3 chance that I'll tell you it was tails)? Or do you think there's a 1/2 chance that I'll tell you it was heads (and a 1/2 chance that I'll tell you twice that it's tails)? I say the latter.
There's a 50% chance you'll tell me "at all" that it's heads, and same for tails. But there's more than a 50% chance that a random selection from the tellings you've done will be a tails telling.
And I think the former is the proper way to talk about the probability that it was heads, not the latter. We can then distinguish between the case of telling someone twice that it's tails and the case of telling someone once but using a weighted coin that favours tails.
I think this is just SSA vs SIA.
Sleeping Beauty is a pretty unusual situation though.
Some of us think it merits switching to counting occasions instead of counting classes of occasions. There are two ways to do it. YMMV.
On our side there are confirming arguments from wagering and weighted expectations. On the halfer side I only see the "no new information" argument.
The problem with SB is that the outcomes are like a 2:1 biased coin, but the payouts (as @andrewk pointed out) are like a 3:1. If we ignore wagering, could SB tell the difference between the official rules and a variant with a single interview and a biased coin? If she can't, is that an argument in favor of one position or the other?
From the other side, wagering will tell my SB that it's not a biased coin but a bizarre interview scheme. Will a halfer SB be able to tell the difference?
P(Awake) = 1 is true when Beauty is awake in the experiment. So, under that condition, P(Heads|Awake) = P(Heads) = 1/3. (See argument below.)
Quoting Michael
No. In your variation, P(Heads|Awake) = 1/2. Your variation requires a second coin toss on tails to determine which day Beauty is woken. The unconditional probabilities are:
P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 1/4 / 1/2 = 1/2
P(Tails|Awake) = P(Tails and Awake) / P(Tails) = 1/4 / 1/2 = 1/2
P(Tails-Heads2|Awake) = P(Tails-Heads2 and Awake) / P(Awake) = 1/8 / 1/2 = 1/4
P(Tails-Tails2|Awake) = P(Tails-Tails2 and Awake) / P(Awake) = 1/8 / 1/2 = 1/4
So conditionalizing on being awake:
P(Heads) = P(Tails) = 1/2
P(Monday|Tails) = P(Monday and Tails) / P(Tails) = 1/4 / 1/2 = 1/2
P(Tails and Monday) = 1/4
Which is the same conclusion that you reached.
Quoting Michael
Because the probability from the sleep states flow to proportionally more tails states. Here's the corresponding working for the original Sleeping Beauty problem. The unconditional probabilities are:
P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 1/4 / 3/4 = 1/3
P(Tails|Awake) = P(Tails and Awake) / P(Awake) = 1/2 / 3/4 = 2/3
P(Tails and Monday|Awake) = P(Tails and Monday and Awake) / P(Awake) = 1/4 / 3/4 = 1/3
So conditionalizing on being awake:
It occurred to me that there may be a parallel between this puzzle and Nick Bostrom's simulation hypothesis. We are analogous to Beauty, and the coin coming up tails is analogous to beings in some universe developing the ability to perform simulations so intricate that consciousness arises in the simulands, call it p.
Say the average number of conscious simulands created in a universe in which those simulations are developed is N. Let the probability of conscious life arising in a universe be q and M be the number of conscious beings arising in such a universe.
We then wonder what is the probability that we are in a simulation. I think Bostrom argues that it is
pN / (pN + qM)
which he thinks would be close to 1 because N would likely be much bigger than M because in such a world, using computers or their equivalent it would be easy to conduct enormous numbers of simulands. That corresponds to the Thirder argument in the Beauty problem because it says that every consciousness, whether simulated or not, is equally likely to be the one I am experiencing, so each one has probability 1/(pN + qM).
In contrast, the Halfer position says that the probability of being a particular simuland is p/N, and the probability of being a particular non-simulated consciousness is q/M. So the probability of being a simulated consciousness is
N(p/N) / (N(p/N) + q(q/M)) = p / (p+q)
That will be much lower that Bostrom's estimate because it is not affected by N being much bigger than M. It makes no difference how many simulands the simulators create.
Hence, the Halfer position provides support to those that don't like Bostrom's suggestion that they are probably a simuland.
What do you think?
Ignore the coin toss completely. The intention of the problem is that Beauty cannot know whether this is her first or second interview. If we count that as a toss-up, then
[math]\small P(HEADS)=\cfrac{1}{2}(50\%)+\cfrac{1}{2}(0\%)[/math]
[math]\small \phantom {P(HEADS)}=25\%[/math]
That is, Beauty would expect a wager at even money to pay out as if there were a single interview and the coin was biased 3:1 tails:heads. And it does.
If I understand you, you are presenting a quarterer scenario where the probabilities conditioned on being awake are:
But, if so, how would the experiment or wagering be conducted to make it work?
Mainly so we'd get to use that word.
This is all stuff we've said before -- this comment summarizes the mechanism by which standard thirder wagering pays out 3:1, as @andrewk pointed out, instead of 2:1.
You could also think of it as revenge against the halfer position, which draws the table this way:
Halfers, reasoning from the coin toss, allow Monday-Heads to "swallow" Tuesday-Heads.
Reasoning from the interview instead, why can't we do the same?
The post I referenced had a mistake!
($2 bets below for simplicity, since the coin is fair.)
Before I gave the SB payoffs at even money as
and noted that heads will break even while tails makes a profit. That's wrong. The right table is obviously
because you bet heads incorrectly twice when the coin lands tails.
Thus the SB 2:1 table would be:
and everyone breaks even. 2:1 are the true odds.
For a reminder, the single toss for a 3:1 biased coin ($2 bet for consistency):
Same as the SB results: heads loses $1, and tails earns $1.
And, no, obviously SB doesn't break even on 3:1 bets:
At odds greater than 2:1, heads will be the better bet.
Sleeping Beauty remains its own thing: the odds really are 2:1, but the payoffs are 3:1.
(Disclaimer: I think this is the most natural way to imagine wagering, but you can come up with schemes that will support the halfer position too. They look tendentious to me, but it's arguable.)
Yes, that all makes sense.
Quoting Srap Tasmaner
Indeed.
So it's interesting that one can set up wagers for halfer, thirder or quarterer outcomes. But it seems to me that probability is not simply about what wagers one can set up and what their outcomes are since everyone should rationally agree about that. Instead, it's about what the probabilities of the states are when conditionalizing on being awake. And that just results in the thirder position.
As to the question of what new relevant information (if any) arises when Beauty awakes that justifies an update on P(Heads), I think it's really just a change of context. When referring to the coin toss outcome from the perspective of an independent observer, P(Heads) = 1/2. However what is relevant to Beauty is P(Heads|Awake) = 1/3. When she awakes, P(Awake) = 1 and so P(Heads) = 1/3.
So seen this way, no new relevant information has been learned on awaking. Instead, P(Heads) is indexed to a context. Which context is relevant depends on the perspective one is taking - the perspective of the independent observer or the awakened Beauty.
The relevant context is the awakened Beauty after the interview, as she is asked the question when awakened; it is right in the OP, which would mean you don't consider periods of sleep in the sample space. They effectively do not exist to Beauty. I mean you are arguing against yourself here.
Our task is to formulate Beauty's response to the interview; that means the relevant temporal locations, are not when she is awakened, but immediately after the interview. By the nature of the experiment the interview itself conveys information, which eliminates all days not contained in the three awakened days of the experiment. This means by the time Beauty is ready to response she already knows the relevant possible awakened states.
Beauty can either consider the probability based on the three possible awakenings that include an interview, or she could consider the coin flip, because regardless of any other considerations, a coin flip is still just a coin flip. Those are the only two relevant sample spaces.
You can slice up probability up whichever way you like, and argue all types of samples space that pay no mind to relevant temporal location or Occam's razor; however, when you get right down to it: it is still just a coin flip. Pushing numbers around on a page will not change that.
And when I see the word "just" used as it is here, I assume someone is trying to manipulate my intuitions. If ever a coin flip wasn't "just" a coin a flip, surely it's Sleeping Beauty.
I'm for trying to tackle the problem as posed. I don't think we should assume, for instance, that Beauty is informed by being awakened. But I'm also for examining the problem statement carefully to see if information is being smuggled to Beauty.
Assume you need information to update your prior; either Beauty doesn't update (and the problem has manipulated our intuitions to suggest she has) or she has received information (and the problem makes it appear she hasn't) or some third option (like a really complicated credence function).
We don't know which. Maybe the problem is just underdetermined or ambiguous (but is pretending it can be answered). We have to look. Eliminate the blind alleys. Maybe when we're done there will be a choice we can't eliminate, maybe not.
Or perhaps that is just innocent word usage. Just is kind of a hard word to avoid, there are not a lot of options that fill its role so concisely.
Quoting Srap Tasmaner
Why? There was one coin, it was flipped once, it was fair and only had two sides. Seems to have all the characteristics of being just a coin flip.
This is a common failing of human perception, people tend to overthink things until they can no longer see the simple. It is still just a coin flip; let's not think our way out of that practical aspect. Probability is not about pushing numbers around on a page, it is about making a reliable model.
Quoting Srap Tasmaner
I didn't say she was being informed by being awakened, I said she was being informed by the interview.
Here's the picture halfers actually use:
[IMG]http://i.imgur.com/0fqxqKd.jpg[/IMG]
And I think they use the same sort of weighted expectation I keep posting, only theirs looks like this:
[math]\small P(HEADS)=\cfrac{3}{4}(67\%)+\cfrac{1}{4}(0\%)[/math]
[math]\small \phantom {P(HEADS)}=50\%[/math]
There are a couple of curiosities here:
I find these proportions strange. Lewis ends up here and shrugs. I'm not sure what to make of it, but this is a far cry from the way I think halfers want to think of their position, that it's just a coin flip with some meaningless frosting on it. Whichever position we take, something about it is counterintuitive.
There is another sample space you are not even considering.
{H,T}
Meaning?
Really?
Also make sure we don't confuse proportion with the sample space.
I know what you meant by "{H,T}". I was asking what your point is.
You know the math better than I do, so if you have something to say, I'm going to listen.
I'm willing to do the work to figure things out on my own, but what you have in mind is not one of those things.
Can we just get back to SB now?
I have been talking about SB this entire time.
First a few terms:
You have a sample space, that is the total possible outcomes, then you have the event, that is the subset randomly selected from that sample space, then you have frequency of occurrences, which is just the proportion of the outcomes.
Lewis is very much arguing it is a simple coin flip with a sample space of {H,T}. Elga does not disagree with that but rather that when Beauty is awakened and interviewed then her temporal location becomes relevant. Elga then argues, that due to this the sample space then becomes {H1, T1, T2}; the periods in which she is awakened and interviewed.
When you see Lewis assign 1/4 to T1 and 1/4 to T2 then what are looking at is the frequency in which is he assigning to those two possible outcomes from the 1/2 in the event of tails. He saying that since being awakened on Monday and Tails and Tuesday and Tails is under the same event then you are equally likely to be in either one upon being awakened and interviewed. However, that does not mean they are pulled from a separate sample space. You have to understand that Monday Tails and Tuesday Tails are pulled from the same event of the coin landing on tails, which is 1/2.
To justify the 1/3 argument you need to give a good reason why the sample space should be from the periods of her being awakened and interviewed over the sample space of two sides of the coin.
This is your sample space {H,T}
In the event of T, which is 50% then you have a new sample space {M,Tu} where M and Tu have an equal chance of occurring. In the event of H your new sample space is just {M}.
However, that never removes the 50% of T or the 50% of H, those sill remain regardless.
You can follow conditional probability from there, but whatever you do, no amount of number pushing will change the 50% T and the 50% H. The only way to do that is to redefine the sample space, which is Elga's argument.
I don't think anybody will argue that her belief that a 3 rolled should be 1/6. She has no evidence that could make it anything else. But "3 rolled" and "it is Wednesday" represent the same concept: if a 3 rolled, it must be Wednesday, and if anything except a 3 rolled, it can't be Wednesday.
The point is, that "it is Wednesday" is a perfectly valid proposition to evaluate. Like "a 3 rolled," its probability is 1/6.
Demonstration of concept #2: Same as #1, but two dice are rolled until the result is not doubles. Beauty is wakened twice, and the same amnesia drug is used in between wakings. She is asked about the same two beliefs.
Note that "a 3 rolled" can refer to either die, so her belief on Sunday is 10/30=1/3 (remember, the six possible ways doubles could roll are eliminated). When awake during the week, she gains no information that can affect that, so her belief remains 1/3.
"It is Wednesday" is still a valid proposition. Jeremiah, this does not mean that Beauty has "temporal awareness while asleep." Only the period of time from when she woke, until when is put to sleep again, exists in her awareness. But that entire period exists within a single day; a day that has a constant name, even if she does not know it. So she can represent that awareness with a probability for each possible name. And since she has no evidence that any name is more or less likely than "Wednesday," her belief must be 1/6 in Wednesday.
It is is not surprising that this is half of her belief in a 3, since she wakes twice. In the combinations where a 3 rolled, she is awake on Wednesday half of the days.
Demonstration of concept #3: Same as #2, but roll only once and accept doubles. In that case, she will be wakened only once.
Her temporal awareness is still that it can be only one day, within a set of six equally-likely days. Her belief that it is Wednesday is still 1/6. (Whether a 3 rolled is not quite the same issue as "Heads" in the original problem, so I won't obfuscate the point by discussing it.
+++++
The halfer argument is based entirely on treating Monday and Tuesday as the same day in Beauty's awareness. They are not; Beauty cannot distinguish them through her senses, but she knows that one name has applied since she woke up, and the other name has not. She can treat that name with probability.
Each of the four combinations "Heads+Monday, Heads+Tuesday, Tails+Monday, Tails+Tuesday" is equally likely to represent a random point in time during the experiment. The probability is 1/4 each.
Yes, Tails+Tuesday" still can happen, even if Beauty has no awareness of it when it does. The point is that it can exist, she knows it can exist, and she knows she won;y be awake to see it. From that, she can update her belief in Heads to 1/3.
I'm suggesting that we should start with a background sample space that includes all possible combinations of days and coin toss outcomes and then assign probabilities according to a principle of indifference. That reflects our prior intuitions that we should be indifferent to an unknown coin toss outcome for a fair coin and also indifferent to an unknown day of the week (or any two instants in time) ceteris paribus. This is the probability distribution for the resulting background sample space:
The next step is to transform it to a second sample space that reflects the Sleeping Beauty scenario by conditionalizing on when Beauty is awake. That excludes the Tuesday/Heads state and gives us the thirder distribution, as follows:
So {H1, T1, T2} is the relevant sample space for Beauty. But it has been derived from a background sample space that distributes all possible coin toss and time interval outcomes according to an indifference principle.
And that's it. That is an independent motivation for the thirder distribution based on the consistent application of a plausible principle.
The halfers use the same principle of indifference, but apply it preferentially to the coin toss outcomes over the day outcomes. Which skews their resulting probability distribution for Beauty.
Yes, it would be good to hear some halfer reasoning for the P(Heads|Monday) = 2/3 consequence.
Quoting Srap Tasmaner
I agree. That's why I think we should seek to derive the distribution from assumptions that are independently plausible. As I argue, the halfer is treating indifference to coin toss outcomes preferentially to indifference over day outcomes. Whereas the thirder is treating them on an equal footing (which is a further application of the indifference principle). (The hypothetical quarterer is the symmetric complement to the halfer, preferring indifference to day outcomes to indifference over coin toss outcomes.)
The double interview is not a single event, for the simple reason that Beauty makes two decisions.
I think this problem arose out of earlier problems and chitchat about decision-making given imperfect memory.
I keep thinking that if it had developed on its own, it would be a time travel puzzle. "Tuesday" appears here essentially as "another Monday". You have no way of knowing for sure it's the first Monday or the last you will experience, etc. etc.
You forgot Heads and Wednesday, Heads and Thursday, Heads and Friday. . . . . and on forever.
Then do the same with Tails.
And in the process don't forget the terms on the days Beauty will be awakened AND interviewed (AKA the sample space), was defined before the experiment started.
I was assuming the two possible interview days in the experiment. But since the probabilities are all equal, it doesn't matter how large the background sample space is (at least in a finite universe). On conditionalization, all the probability will still be distributed equally between H1, T1 and T2.
Quoting Jeremiah
Yes. So do you see any problem with my approach in principle?
Oh really? So you think they are the same thing.
If your distribution is the same as Elga's then they should have the same center.
Let's find out, by comparing the means.
Egla's distribution: (1/3+1/3+1/3)/3=1/3
Your distribution: (1/3+1/3+1/3+0)/4= 1/4
So in your distribution, on average Beauty will get one out of every four attempts correct, because oddly enough you are actually suggesting zero is in that distribution of possible outcomes. When working with sets zero is not the same thing as null.
Let's look at this from the ground up.
A sample space is the set of all possible outcomes of a random process.
An event is a subset of that sample space.
Let E be the event and let S be the sample space.
Then the Equally Likely Probability Formula is:
P(E) = the number of outcomes in E/ the number of outcomes in S or P(E) = N(E)/N(S)
(N(R) is just the number of elements in R)
Now just to clear it up, in set theory {A,B,C} is not equal to {A,B,C,0}. Let {A,B,C} bet set 1, and let {A,B,C,0} bet set 2.
Consider,
We randomly select one element from each sample space, then our possible outcomes are:
For set 1: A or B or C
For set 2: A or B or C or 0
That means for set 2 there is a one in four chance of 0 being selected. Put that in the context of our problem and that really does not make any sense.
Now conditional probability is the the probability of event K given that event L has already occurred.
---
Consider this argument:
Our sample space is {H,T} with P(H)=N(H)/N(S) or P(H) = 1/2.
Let H equal the set {M1} and let T equal the set {M2, Tu}
Where,
M1 equals Monday and Heads
M2 equals Monday and Tails
Tu equals Tails and Tuesday.
So now our possible events are sets. Set H has one element and set T has two elements, each with a 50% chance of being selected. Then P(M2) or P(Tu), given the event T, by our Equally Likely Probability Formula is P(M2) = 1/4 and P(Tu) = 1/4. Given Tails she could be in P(M2) or P(Tu), so 1/4 + 1/4 = 1/2 therefore P(M2) + P(Tu) = 1/2 = P(T) = P(H).
I was referring to the background sample space in my earlier post that included all combinations of day outcomes and coin toss outcomes and that are assigned the same probability. I'm saying that it doesn't matter what size the background sample space is, as long as it is finite and the elements have equal probabilities.
I'm then conditionalizing on Beauty being awake (and interviewed) to produce a second sample space with only three elements {H1, T1, T2} with 1/3 probability each. That second sample space is the relevant sample space and matches Elgas.
The second probability distribution from my earlier post should have been:
Thanks. I think I finally understand the halfer position. (The one thing I'm not completely clear on is how the Monday interview is retroactively determined to be a single or half of a double in the variant where the coin is tossed after the first interview.)
What puzzles me is why Beauty would reason this way.
My Beauty reasons this way, as I've said before:
(1) If I knew it was Monday, I'd know it could be heads or tails, even chance.
(2) If I knew it was Tuesday, I'd know it was tails.
(3) I know I'll be interviewed on Monday, but interviewed on Tuesday only half the time.
(4) Therefore my weighted expectation of heads is 2/3(1/2) + 1/3(0) = 1/3
The halfer Beauty reasons this way:
(1) If I knew I was in the single interview track, I'd know it was heads.
(2) If I knew I was in the double interview track, I'd know it was tails.
(3) I'm in the first track half the time and in the second half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2
But this is just pretend reasoning.
It's like "working out" your expectation of heads in a simple coin toss this way:
(1) If I knew it was heads, I'd know it was heads.
(2) If I knew it was tails, I'd know it was tails.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2
What's the point of that?
And indeed, Lewis's "proof" has but a single step.
(No argument in this post, just clearing my head.)
How many times does Beauty expect to be asked for her credence?
(1) If I knew it was heads, I'd know I'll be asked once.
(2) If I knew it was tails, I'd know I'll be asked twice.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(2) = 3/2.
If I'm in the single interview track, and I am half the time, I get 2/3 of my expected interviews.
If I'm in the double interview track, and I am half the time, I get 4/3 of my expected interviews.
My expectation for getting to say "heads" and be right, because I'm in the single interview track, is 2/3(50%) = 1/3.
My expectation for getting to say "tails" and be right, because I'm in the double interview track, is 4/3(50%) = 2/3.
When does anyone ever make a random selection from among only the tails interviews?
I misunderstood.
But you are going to say exactly this about M2 and Tu (or T1 and T2), so the question stands.
In the event of Tails, Beauty will be awakened on Monday and Tuesday, but due to the nature of the experiment she will not be able to tell the difference, either one is equally likely when interviewed. Hence P(M2) = P(Tu) =1/4. It is 1/4 as only a total of 50% was allotted to T.
This temporal uncertainty, is actually where the 1/3 argument is placed. If the uncertainty is about her location in time, and probability is the measure of uncertainty then shouldn't her sample space be {M, T1, T2}?
It all depended on which uncertainty Beauty decides to consider.
So what? It's not a situation that arises. Neither she not the experimenters are ever in the position of knowing that the coin landed tails but wondering what day it is. Beauty only wonders what day it is to figure out how the coin landed.
Suppose there was another coin toss to determine whether heads was the single interview or the double this time around. Then half the time heads would be 1/3 of the interviews, and half the time 2/3, so heads would on average be half the interviews and same for tails.
But that is not the case here. The interviews are not randomly distributed.
Beauty knows that when she is asked for her credence, 1/3 of the time the coin has landed (or will land?) heads and 2/3 of the time the coin has landed (or will land?) tails.
Therefore her credence that the coin has landed (or will land?) heads must be 1/3.
If my position was that she somehow magically knew it was tails, then why would I claim it has a 50% uncertainty? If she knew it was tails it would be a 100% certainty and M2 = 50% = Tu with temporal uncertainty. The 50% is the uncertainty of T or H and the 25% is the uncertainty of M2 or Tu.
The conditional probability of tails given that it is M2 or Tu is P(T|M2)= P(.5|.25) = .125/.25 = 1/2 = P(T|Tu) which is equal to P(H). Both days still have the same 50% uncertainty when considering H or T. In fact from that direction all three days have the same uncertainty when considering H or T, which is where we get the 1/3 argument.
Quoting Srap Tasmaner
Randomly just means equal probability. Figured I should clear that up. In the technical sense when talking about random, it means each element in the sample space has an equal chance of being selected. So in the event of tails, on any given consideration of the interview between M2 and Tu Beauty, has a 25% uncertainty of being in either of them. It is 25% and not 50% because of the uncertainty in it being Tails. We are stacking uncertainties.
How it comes out all depends on the considerations of the uncertainties.
This is still slightly puzzling to me.
P(H | M1) = 1, right? And this is the thing about the double interview track: both them happen if and only if the coin lands tails. From your calculation, P(T | M2 v Tu) = 1, yes? But it should be P(T | M2) = P(T | Tu) = 1, and P(T) = P(M2) = P(Tu) = 1/2. You always get both on tails. You get them one at a time, but we don't necessarily care.
That space of three possibilities, {M1, M2, Tu} has three elements each of which has an unconditional probability of 50%. Conditioned on the whole space, they'll each be 33%.
This is why I keep saying it depends on how Beauty decides to consider her uncertainties.
Remember conditional probability is the the probability of event K given that event L has already occurred. Order matters.
So what is T given M2 or Tu? 50%
Hence, .50 +.50 +.50 = 1.5 > 1. The sum of probability cannot be greater than 1 and since P(H|M1)=P(T|M2)=P(T|Tu) we reallocate the credibility to 1/3 each.
That is when Beauty is considering the uncertainty of her location in time.
Now what if Beauty considers instead the uncertainty of H or T?
Then, purely for demonstration, what is M2 given the event T?
P(M2|T) = P(.25|.5) = .125/.5 = 1/4.
However, if Beauty was considering the uncertainty of H or T, and not her location in time, the only reason to consider the conditional probability of M2 would be for completion; practically she could end at the uncertainty of H or T.
The real issue here is not that we get two different yet seemly reasonable answers; this is not about 1/2 vs 1/3. What the Sleeping Beauty Problem demonstrates is that decision affects the outcome of her solution.
What I find interesting is that this decision also is very likely an unconscious decision. Which may be why we get people who are convinced it is 1/3 and people who are convinced it is 1/2. The unconscious mind made a decision for them on how to consider the probability, a decision that consciously they were never aware of.
Something I don't remember us talking about: should Beauty, knowing the rules of the experiment, subject her expectation of a tails interview to a discount? It occurs to me that this may be the regime Lewis is describing.
Here's a physical version. You decide to test if a coin is fair by throwing a red marble in an urn on heads, and a blue marble if tails. After a bunch of flips, you'll count the marbles, expecting them to be about equal. Drawing a marble randomly will have the same distribution as the coin itself.
Suppose instead on tails you throw in two blue marbles. Then you'd expect a 2:1 ratio if the coin is fair. A randomly selected marble is now twice as likely to be blue, but each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads. Each blue does represent a tails, certainly, and only got in the urn because a tails was tossed, but there's another blue that's evidence of the same toss.
Now suppose the marbles are all white. Still true that you're twice as likely to draw a marble representing a tails toss, but you have to discount.
@Andrew M, @andrewk, @JeffJo: do you find this argument as convincing as I do?
@Jeremiah
Quoting Srap Tasmaner
Is this what the conditional probabilities we've been talking about are trying to express? I'm still not clear on how this idea is formalized.
It's as we were discussing: each marble represents an interview event. To count coin toss outcomes you only need one red marble, but two blues, to make up the entire double interview event. Each blue marble is one kind of event, but that event is half of the kind of event we want to count.
That's all we're asking Beauty about. That is, the probability that the next marble to be drawn (or the interview that is being conducted) will be associated with a heads outcome. Which is 1/3.
Quoting Srap Tasmaner
Discounting doesn't help Beauty. Suppose she is a halfer. Should she discount the next drawn marble (or current interview) by 1/2 or not? Well, she should 2/3 of the time. So 2/3 * (1/2 * 1/2) + 1/3 * 1/2 = 1/3. The thirder says that is the real probability of each state for her.
This is the wrong model. This table
is right, and here's why.
Suppose you have a machine set up like this: there's a hopper full of red marbles and a hopper with twice as many blue marbles; you push one button and it transfers a single red marble or two blue marbles to another hopper, one you can't see; you push a different button and it dispenses one of the marbles from the small hopper. What are your odds of getting a red marble? 1/2. Half the time only a single red marble goes into the small hopper and then gets dispensed in the second step. (Half the time, two blue marbles go in, and then one of those two is dispensed, so the chance of blue -- a blue, any blue, one of the two in the small hopper -- is also 1/2, despite the fact that twice as many marbles were dispensed at the stage you don't see.)
Now do it this way: you have a hopper full of white marbles; push one button and half the time a single marble is moved to the small hopper, half the time two; you push the second button and get a single marble. How many marbles are left in the small hopper? Dunno. Half the time there's still one there, and half the time there isn't.
You could randomize. Any number of marbles could be dispensed to the small hopper. Getting one tells you exactly nothing. You could have it transfer a random number of reds to the small hopper half the time and a random number of blues half the time. When you push the second button to get your marble, the chances will still be 50:50 of getting a red or a blue.
Agreed. But that scenario is equivalent to randomly waking Beauty on either Monday or Tuesday if tails, but not both days. To be analogous to the Sleeping Beauty scenario, the second blue marble has to be dispensed in a separate event (with amnesia in between). That is an additional possible state that Beauty could be in which, for the thirder, decreases the probability for the Heads and Monday (or red marble) state to 1/3.
All that means is that if Beauty is asked to guess which state she is in and she guesses Heads and Monday, then she will be correct 1/3 of the time. Similarly for Tails and Monday and Tails and Tuesday. On the thirder view, probability is about the state she is in, not the coin toss (or day) outcome itself.
The halfer view, while seemingly just representing a fair coin toss as coming up heads half the time, has the consequence that P(Heads|Monday) = 2/3 instead of 1/2. I think that is a reductio of the halfer view.
If you toss a fair coin 100 times and throw one red marble into an urn on heads and two blues on tails, you'll end up with (roughly) 50 reds and 100 blues. If you count each of the marbles as an outcome of the coin toss, without discounting the blues, you'll end up with 100 tosses having 150 outcomes, which is absurd. It's an attempt at alchemy.
One of the side effects of this attempted alchemy is that each red represents twice as much of an outcome as each blue. Yes, there is something absurd about the 2/3, but it's a result of putting in twice as many blues per toss but then taking them out one at a time, as if they were the same as the reds. You can pair off each red with two blues -- that is, taking the marbles back out of the urn the same way you put them in -- without absurdity. If you insist on taking the blues back out singly, the absurdity of the result (a marble representing half an outcome) is on you.
As for Beauty's state, try thinking of the Tuesday interview this way: I am (still) being interviewed about a tails outcome (the same one as yesterday). There was just the one coin toss, with just the one outcome. Smearing the interview across two days doesn't change that. Beauty does not know which interview this is, but she knows there will be two interviews for each tails outcome and she discounts accordingly.
Think about what's going on if she wagers. She can make a Dutch book on tosses of a fair coin at even money. That should not be possible. That's just as strong a principle as the business about no updating without new information. If it is possible, someone's performing alchemy or cheating.
Here's another way to look at it.
You could say it depends on what we take to be the event that must be predicted -- I've made that argument myself, and not long ago. Should Beauty predict the outcome of the coin toss or being asked about the coin toss?
I used to think that being asked was itself evidence, but that's only true on two conditions: (a) Beauty doesn't know the rules and is unable to figure out the true chances of heads and tails; (b) she is willing to accept the absurdity that a single coin toss has, on average, 1.5 outcomes.
It is certainly true that there are on average 1.5 interviews, but some of those interviews (the extra .5 on average) are about the same outcome. That extra blue marble left in the hopper is not another outcome; it's just the rest of the outcome you already know about from the first blue marble.
I have argued, as thirder, that there are three outcomes each of which has a 50% chance of occurring (the one heads interview and the two for tails). Taken conditional to the whole set {H1, T1, T2}, each would shift to a 33% chance.
The thing is, this 2:1 proportion of interviews is right, but remember that SB does not payout like a wager on a 2:1 biased coin. It pays out like a 3:1 coin.
It's not just the ratio of wins to losses, but the actual payouts that bothers me. If it were a genuine 2:1 deal, we'd expect a $1 wager on tails to pay out 2/3(1) - 1/3(1) = 1/3. It doesn't. It pays out 2*1/2(1) - 1/2(1) = 1/2.
That means that scaling 50:50:50 to 33:33:33 never happens. The actual payouts represent a coin that has a 50% chance of heads and a 100% chance of tails, not a coin that is 33:67. That's pretty weird.
I'd like to analyze the halfer's P(Heads|Monday) = 2/3 consequence further because I think it is key to how we see the Sleeping Beauty scenario. Just to make the consequence more stark, suppose that Beauty knows she will undergo 1000 memory-erased interviews on tails (subject to ethics approval). The halfer probability distribution is:
P(Heads and Monday) = 1/2
P(Monday) = 1/2 + 1/2000 = 1001/2000
P(Heads|Monday) = 1/2 / 1001/2000 = 1000/1001 (approx. 99.9%)
Now suppose that Beauty knows before the experiment begins that the coin will be flipped Monday evening after she goes to sleep. The experiment proceeds and, during her Monday interview, she is told that it is Monday. As a halfer, she concludes that it is a virtual certainty (99.9%) that the outcome of the yet-to-be-tossed coin will be heads!
Is that how she should understand P(Heads|Monday)? If she should discount the possible tails interviews to get the more sensible result of 1/2, then where does that show up in the halfer's math?
Quoting Srap Tasmaner
Yes, this is an important point. I think the intuitive comparison with a weighted coin is misleading since SB is just structured differently. Adding more interviews (and thus bets) on tails is not like increasing coin bias.
The thirder reasoning works against intuition as well, especially in the extreme version. It suggests that if we’re to be woken a thousand times in the case of tails then we should be almost certain that it’s tails upon waking, despite the fact that it’s an unbiased coin flip. And all because if we’re right then we’re right more often? That shouldn’t be the measure.
I think the sensible position is to just accept that a coin toss is 50:50 and so we shouldn’t have a preference, no matter how many more times we are woken if it’s tails. The probability tables are only useful if we have no further information and want to work out the probability that today is Tuesday, in which case I think the halfer’s table is more intuitive.
Indeed, I think it means that the odds here are not truly 2:1 at all.
I can't figure out how to make this into a normal wager of any kind. The closest I can come looks very similar to the problem itself: the bet is at even money, but on tails the wager is doubled. Now this is very strange. Conditional wagers are fine, but the one thing they would generally not condition on is the event whose outcome determines the outcome of the wager! Even your stake is unknown until the bet itself is resolved?! I don't know why people who find the wagering argument so attractive, as I once did, aren't more troubled by this.
Quoting Michael
This position is attractive, but I just don't understand how it works.
Your expectation of heads is 1/2 before you know what day it is, right? So that includes the possibility that it's Tuesday. If you go with the way Lewis splits up the space, that's fine:
[math]\small P(H) = \cfrac{3}{4}(67\%) + \cfrac{1}{4}(0\%) = 50\%[/math].
Learning that it's not Tuesday should increase your expectation of heads if you were taking the possibility of it being Tuesday into account, and that's how we end up with the 2/3.
Is there a principled way to just ignore Tuesday, and any subsequent Tuesdays?
Quoting Andrew M
Michael is certainly right that this is the mirror image of the thirder problem -- we each seem committed to unreasonable confidence about heads or about tails as you increase the number of Tuesdays.
I'm not convinced this is what the halfer position really implies, but the two are clearly related.
Quoting Srap Tasmaner
This is what I keep thinking about. You can add as many Tuesdays as you like to represent the tails outcome, but Beauty would only need to know about one of them to know the toss landed tails. All of the Tuesdays are clones of each other -- if you know the fact of the matter about one, you know them all.
Still working on it. Maybe in another day or so I'll have something new to say.
As Andrew M said the coin isn’t flipped until Monday evening, so if you know it’s Monday then you’re being asked about your credence that a future coin flip will land heads, and of course that’s going to be 1/2. And as you’re only woken on Tuesday if it’s tails then if you’re told it’s Tuesday then you know it didn’t land heads.
I see those equations as a formalization of the rules that we learn in familiar settings. In unfamiliar settings, the question would be whether it is the equations that are at fault (or inapplicable) or our intuitions.
Quoting Michael
The thirder is not measuring the outcome of the coin toss as an isolated event. She is instead measuring uncertainty about her location in a state space.
In a familiar setting, there is no difference between those two ways of thinking about probabilities. It does not matter whether you think of the coin as being in one of two possible states or, alternatively, you as being located in one of two possible states with respect to the coin. There are only two states in both views, so 50:50 is the agreed answer.
In the Sleeping Beauty scenario, those two methods give different answers. The first method is the halfer view which understands the coin to be in one of two states. But that leads to absurdity when conditioning on Monday. The thirder doesn't have that problem, because conditioning on Monday eliminates all but two states that she could possibly be in. So 50:50 is the natural answer as with a coin toss in an everyday setting.
In the initial Sleeping Beauty scenario, Beauty is located in one of three possible states, she just doesn't know which one. Since they are indistinguishable to her, she assigns each of them an equal probability. That assignment would make no sense if it meant that the coin's intrinsically equally likely states had somehow become weighted. But it does make sense if it means that there are three indistinguishable states that she could be located in and two of them are associated with a tails outcome.
Quoting Srap Tasmaner
I think they are, but not as a measure of the coin toss outcome. I think probability is a measure of self-locating uncertainty in a state space. It's not directly about coin outcomes, days, or even interviews at all, except in so far as they contribute to the construction of the state space.
Quoting Srap Tasmaner
The simplest wager is just to bet on being located in a single state. If you bet on being located in the Tails and Monday state, then you'll win once and lose twice (on average), just as the thirder probabilities imply. Similarly for each of the other states. Whereas if you bet on being located in a Tails state, you'll win twice and lose once. And sure enough, there are two tails states that add to 2/3. Similarly if you bet on being located in a Monday state, you'll win twice and lose once and those states indeed add to 2/3.
That makes sense to me.
I think the halfer reasoning should just be that it’s a 50:50 chance that it’s heads, whether unconditioned or conditioned to Monday. We shouldn’t be applying some formula and should just consider what we know about coin flips.
Quoting Michael
That's saying P(A | B) = P(A), and therefore A and B are independent events. Of course in one sense the coin toss and the day of the week are independent, but whether Beauty is interviewed on that day is clearly not independent of the coin toss: you yourself noted that if she's told it's Tuesday then her credence should be 0. Is that right? You have to decide whether you're using this table
or this table
Switching between them as needed is just equivocation.
Quoting Andrew M
But Beauty is asked about the coin toss itself: it has one outcome which gives rise on average to 1.5 states for Beauty. Not discriminating between Beauty's states and the contribution of the coin toss to those states is another sort of equivocation. Beauty can make this distinction, just as you could in counting 1 red marble on heads and 2 blue marbles on tails, expecting a 2:1 ratio if the coin is fair.
The thirder model relies explicitly on there being a single toss of a coin with heads and tails distributed 50:50. (And we've agreed you cannot construct an alternate model with a weighted coin.) How can Beauty take that as a premise and then be unable to reach the conclusion that the chances of heads were 1/2?
It is independent. She’s interviewed on Monday regardless, and in Andrew’s example the coin isn’t flipped until after the Monday interview.
So Monday is independent of the coin toss but ¬Monday isn't? Aren't we just using ambiguous vocabulary here?
Speaking of ambiguity, does when the coin toss happens affect Beauty's credences? Don't we want a solution that applies to a toss before the Monday interview as well as after?
The OP uses the toss-before model.
Yes, because the Monday interview happens regardless, and in this example before the toss, whereas a Tuesday interview only happens if there’s a toss of tails.
Quoting Srap Tasmaner
Perhaps. And as it’s obvious that if you know it’s Monday and the coin toss hasn’t happened that the probability that it will be heads is 1/2 then it should be obvious that if you know it’s Monday and the coin toss has happened that the probability that it was heads is 1/2.
Two urns, one with a one red marble, and one with one blue marble. One blue marble.
You toss a fair coin to determine which urn you'll be taking a marble from. One coin toss with one outcome. The unchosen urn is removed.
If we stop here, it's obvious that the chance of getting a red marble is 1/2, and the chance of getting a blue marble is the same. (In each case, there's a single marble with a 100% chance of getting chosen from the urn.) This is Monday.
Now here's the wonky part: selection from the red urn is without replacement; selection from the blue urn is with replacement. You can go just twice, as in standard SB, or you can go a thousand times. Whenever you want, there's a single blue marble available.
Looked at this way, you can see why there doesn't ever seem to be any discounting or conditioning -- not in the chances of a tails interview on Monday, not in the payoffs from wagering, not in the chances of a second interview, nowhere. Each marble inherits the full 1/2 probability of its urn because it is the only marble there. Tails doesn't show up as an event chopped up into parts, but as an event that repeats as often as you like.
The coin toss determines whether you draw the one red marble or some number of blue marbles. The number of marbles drawn has nothing to do with the chances that you're drawing red or blue though.
I'm not sure how to finish formalizing this, but I think it represents SB pretty well.
OK, that's the double-halfer view. When Beauty is told it is Monday, all the Tuesday and Tails probability is reallocated to Monday and Tails which violates conditionalization. That is:
Quoting Srap Tasmaner
Because of the initial conditioning on being awake. You also accept that conditionalization changes Beauty's probability of heads when she is told it is Monday, so probability isn't simply about the nature of fair coins. It's also about the information that an agent has that she can update on.
No, I'm in the double halfer camp now. The post right above explains my current thinking.
((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.))
Just to clarify my position a little:
According to the design of the experiment, there are two protocols:
- a single interview on heads;
- multiple interviews on tails.
Beauty is asked for her degree of belief that the coin landed heads, that is, that the experiment is following the heads protocol. Chances of that are 1/2. As it happens, being told it's Monday provides no additional information because both protocols include a Monday interview. (If she were told it's Tuesday, she'd know it was the tails protocol.)Back when @Michael said it's just the difference between being asked once and being asked repeatedly, and that this makes no difference, he was absolutely right.
Here's a slight variation. Beauty will receive payments at the end of the experiment for each state that she was either awake or asleep in, as follows:
When Beauty is awakened and interviewed, she is asked how likely it is that she is in the state that pays $1.
My answer is that she should condition on being awake and so her answer should be 1/3.
I'm just following the principal principle. If I can figure out what the objective chances are, so can Beauty, and she can set her credences accordingly.
That is of no use to her. When awakened, she doesn't know whether she is in an awake state that she should assign a probability of 1/2 to or 1/4 to. She can only condition on being awake and thus assign 1/3.
I see the problem.
Yes, I argued recently for "discounting", basically the model that Lewis presents.
Now I think that's wrong. There is no discounting. None of the 1/2's should be reduced to 1/4's. Monday is not 1/2:1/4 either.
When Beauty is asked, "What is your credence that the coin landed heads?" she knows there's a chance the experiment is using the heads protocol, in which case this is her one and only interview, and a chance that it is using the tails protocol, in which case this may be her first interview, last, or one of many, depending. By stipulation, there is no evidence she can use to distinguish one interview from another; all she has to go on is her knowledge of the experiment's design. So the right answer is that there is a 1/2 chance of heads protocol -- which is the chance of heads -- and a 1/2 chance of tails protocol. It's 1/2 everywhere, all the time. When she is told is Monday, this makes no difference. The number of interviews conducted in the tails protocol also makes no difference. They are all interviews about the same solitary outcome.
And the question of "updating" never arises because she never does.
I'll take one more stab at this. I did it with marbles above, but here's the application.
Which protocol to use is determined by the toss of a fair coin. Each protocol is an interview pool; when they awaken Beauty before Wednesday, they select an interview from the pool. Each pool has a single member.
This is where Lewis goes wrong, and where I went wrong when I first came around to the halfer position. It's natural to imagine the tails interview pool as a collection of 2 interviews or 1000 interviews or whatever, in which case you end up with each interview being "discounted", as I put it. If there are 2 tails interviews, they each have a 1/2 chance of being selected; since the pool as a whole has a 1/2 chance, they're each 1/4.
There are two problems with this view: (a) there are absurd consequences, like the 2/3 heads advantage on Monday, the likelihood of a second interview being 1/4 instead of 1/2, etc; (b) it does not represent how the experiment is conducted. Remember me frustratedly asking, back when I was a thirder, when anyone ever randomly selects between the two tails interviews? No one ever does, not even Beauty.
The key for me was to recognize that there is only a single interview in each pool, but under the heads protocol you select that interview from the pool (100% chance on heads) without replacement, but under the tails protocol you select that interview from the pool (100% chance on tails) with replacement. Thus the chance of that tails interview is 1/2, just as it was for the heads interview, and the next time you do a tails interview, its chance is once again 1/2, and it's always 1/2, as often as you go back to the tails pool and select that interview again.
(As a thirder I argued for conditioning this 50:50:50 to 33:33:33, but that's also wrong. The heads and tails interviews are never part of the same pool; it's one or the other. The wagering payoffs make it clear that this conditioning does not happen: every 1/2 stays a 1/2. The chance of a second interview is clearly 1/2, not 1/3. Now I understand how all this is possible.)
The question Beauty needs to answer is, which protocol is in force? Each has a 1/2 chance. That's it. Either she's being interviewed once about a heads or repeatedly about a tails, but the chances of heads and tails remain 1/2 regardless.
When Beauty is being interviewed, what probability should she assign to Monday and Tails? If 1/2 then Tuesday and Tails would be 0 which doesn't seem right.
No. I'm not sure how to formalize this (@fdrake help!), but I think if we want to do this as a table, it will be n+1-dimensional, where n is the number of tails interviews. We're going to multiply at each step, but that's just multiplying by 1 since each interview is a certainty. At the front, we're multiplying by 1/2 for the outcome of the toss.
(We usually write about chains of independent events "combinatorially" -- HHHH, HHHT, HHTH, etc. We could do that here: "H" and "T" toss outcomes, "h" and "t" interviews, and then we're choosing between Hh and Ttt. Each has a chance of 1/2. Other permutations are eliminated by the rules: there is no Ht, no Tht, and so on.)
There is no uncertainty in the interview pools themselves. This I have tried to express by having a single interview available to be selected. In the case of tails, that selection is with replacement, so you can repeat that same certain selection indefinitely. The only uncertainty here is in the outcome of the coin toss.
Consider that from the experimenter's point of view, there is never any doubt about which interview comes next. There is uncertainty for Beauty, of course, but again if she can figure out the objective chances, that's what she sets her credences to. Knowing which of, say, 1000 tails interviews she's in would be useless to her -- either she knows it's a tails interview or not, and she'll never care which one it is. Knowing that it's not Monday would be useful, but by stipulation she can't.
I've not been following the thread so I'll be of no help.
Here's a story where Lewis's table seems to make sense:
Beauty wonders to herself whether she's already been interviewed, and whether she'll be interviewed again. She reasons that there's a 1/2 chance of tails, and then a 1/2 chance that this is the first of two interviews, for 1/4; and there's also a 1/2 chance that this is the second of two of interviews, for another 1/4. The chance of either one of those being the case is 1/4 + 1/4 = 1/2, so she concludes that, since the chances this is not her only interview are the chances of tails, then the chances of tails are 1/2. Huzzah!
But this is pretend reasoning. She starts out knowing the chances of tails are 1/2.
I don't think the halfer view ultimately flies. I think it conflates the question of the nature of fair coins (which we all agree come up heads half the time) with the question of what an agent knows about the outcome of a particular fair coin toss.
For a straightforward example where those two answers differ, suppose that the experimenter will toss two fair coins in sequence. Alice (in Wonderland) will not know the outcome of the coin tosses until after the experiment has completed. However the rule of the experiment, which Alice knows beforehand, is that she will only be interviewed if the outcome is not a double-header. If she is interviewed, she will be asked the probability that the first coin came up heads.
When being interviewed, Alice should condition on being interviewed which results in a sample space of { HT, TH, TT } and an answer of 1/3.
I think the Sleeping Beauty experiment is analogous to this. The theater is just that she is interviewed twice if the first coin toss comes up tails and amnesia is added to make the interviews indistinguishable.
Which in Beauty's case is zilch, isn't it?
I agree about your double-header example, but don't see the similarity to SB at all. Interviewing here clearly gives you information. In SB it does not.
Beauty knows beforehand that she will be awakened and interviewed. But P(Heads|Awake) does not become relevant until she is actually awakened in the experiment, in which case P(Heads) consequently changes from 1/2 to 1/3 for her. Before and after the experiment, P(Heads) = 1/2.
Quoting Srap Tasmaner
Yes, but what is actually relevant is that the interview condition has obtained. It is on that basis that P(First coin heads|Interviewed) = 1/3 becomes relevant and P(First coin heads) consequently changes from 1/2 to 1/3 for Alice.
Both Beauty and Alice can calculate all the applicable conditionals beforehand. The only difference is that Beauty knows with certainty that the interview condition will obtain in her future whereas Alice does not. But what makes the conditional relevant is whether the condition currently obtains for the agent, not whether it constitutes new or old information.
Here are the rules:
I have tossed the coin and given you a box.
What are the chances there's a red marble in the box?
Monday is a different day than Tuesday. You are treating them as the same event from Beauty's point of view, and they are not. They are within the overall experiment, but Beauty sees only half - one day of two - of what the overall experiment encompasses.
I provided an un-refuted (and irrefutable) demonstration that the answer must be 1/3 here. But you apparently need another.
So, do everything as in the original experiment, except don't tell Beauty that she might sleep though a day. Tell her that she will be interviewed on Monday, but only on Tuesday if Tails was flipped. If it was Heads, Something Other Than An Interview (SOTAI) will happen (you can even give an example: say that last week's volunteer was taken to DisneyWorld). BUT, the stipulation that she cannot distinguish the interviews still applies; only Interview/SOTAI are distinguishable.
You were right before that some "discounting" needs to be done. You just did it incorrectly. Beauty does not see the entire experiment, she sees just one day of it. Using SOTAI demonstrates how she should do this discounting: To Beauty, there aren't just two protocols, there are four:
Monday+Tails = I am to be Interviewed,
Monday+Heads = I am to be Interviewed,
Tuesday+Tails = I am to be Interviewed, and
Tuesday+Heads = SOTAI.
On Sunday Night, she knows that each of these is equally likely to occur. But from the point of view of the overall experiment, the events that happen on different days are not disjoint; each has a probability of 1/2 that it will occur in the future. There is no inconsistency, since two will happen.
But when she is wakened, she sees only one day and not the entire experiment. She has evidence only of the present, not the future or past. So Monday+Tails and Tuesday+Tails are just as much a different presents, as are Monday+Heads and Tuesday+Heads.
If SOTAI happens, she knows the sub-protocol was based on Tuesday+Heads. So the probability of Heads is 100%. This should be a big clue to Halfers, since an increased probability in some circumstances requires that it decrease in others.
If she is interviewed, she can deduce that one of these equiprobable sub-protocols can't be the protocol responsible for today. The other three remain equiprobable, and only one includes Heads. So the probability of Heads is 1/3.
Finally, note that it does not matter what SOTAI is, just that when she is in an interview she knows that SOTAI is not happening. Her logic is based entirely on the fact that the three possible interviews are indistinguishable from each other, but distinguishable from SOTAI. So, let SOTAI be "don't wake her up."
If I condition on ~(Tuesday & HEADS), I exclude neither the heads protocol nor the tails protocol, as neither included it. This helps me not at all.
1/2.
Yeah that one was too easy. If I could have come up with a way for you not to know the difference between getting one box and getting two, I would have done that. Every now and then I think if I can find just the right analogy, I'll convince you!
I'm disappointed that I'm still struggling with this, but it's a chance to learn.
It happens! Here's how I see it. Lewis' halfer view fails because it gives an absurd result of 2/3 when conditioning on Monday. But the double-halfer view also fails because on informing Beauty that it is Monday the result of 1/2 violates conditionalization.
Whereas accepting conditionalization and reasoning from the Monday result of 1/2 leads naturally to the thirder view. So that's one argument in favor of it.
Note also that all the real action takes place after the Monday interview since the coin toss can occur that evening. So suppose the Monday interview were removed from the experiment entirely. If Beauty were interviewed at all, then she would know that the outcome is tails (i.e., she conditions on being interviewed on Tuesday).
If she can condition on being interviewed in that scenario and reallocate probability from Tuesday/Heads to the remaining state, then she can also condition on being interviewed in the standard Sleeping Beauty scenario and reallocate probability from Tuesday/Heads to the remaining three states. The condition has obtained just the same whether or not she has learned something new.
Suppose we really were asking Beauty to guess the result of the coin toss, rather than give her credence. We'll do this with 100 Beauties and tally the results.
100 tosses, presumed result of 50 heads and 50 tails, 150 interviews.
If the Beauties all guess tails all the time, they will get 100 right out of their 150 answers.
That looks like a 2/3 success rate, right? But is it?
Out of the 100 tosses, they got 50 of them wrong. Looked at this way, that's a 50% success rate.
There's an element to committed tails-guessing of over-performing in districts you're sure to win, if you see what I mean. That extra interview, the one that tips that the result was tails, it happens in the tails track. If you're guessing tails, you've already guessed it. You're not "extra right" about tails just because you get to be right twice about the same event.
Some of the math puzzles me. Figuring out how to formalize it puzzles me. It's not a situation there's an off-the-shelf model for. Getting the math to work in a satisfying way is a chance to learn. None of that makes me at all uncertain about 1/2 being the right answer though.
What you are trying to do is like trying to get a sum of 10 or more on two six-sided dice. If you look at only one, and see that it is not a 5 or a 6, can you conclude that the sum can't be 10 or more? After all, you have to have a 5 or a 6 to get that large of a sum, and you don't see one.
Just like my example excluded the possibility that the unseen die is a 6, you excluded part of the HEADS protocol when you conditioned on ~(Tuesday & HEADS). Specifically, the part where SOTAI happens. Whether or not Beauty is awake to see it, it is still a part of the HEADS protocol and you are treating it as if it is not. You are inconsistent because you insist you can't separate the two parts of the TAILS protocol the same way.
The "help" I am trying to offer, is to get you to see that you have to separate both protocols into individual days. And you are right, it will be of no help to you if you refuse to see this, just like you won't address my "four volunteers" proof that the answer is 1/3.
+++++
In probability, an outcome is a description of a result. A set of such descriptions with the property that every possible result of an experiment fits exactly one of the descriptions is called a sample space of the experiment.
There can be more than one sample space, depending on what you are interested in describing. Possible sample spaces for rolling my two dice include 36 outcomes (if every ordered combination is considered), or 11 (if just the sum is considered), or 2 (if all you care about is whether the sum is, or isn't, 10 or more). But note that it is never wrong to use a larger sample space than the minimum required to distinguish what is of interest to you: the 36-element sample space describes the experiment where you are trying to get 10 or more, and in fact is easier to use. The most common mistake made by beginners in probability may be using the wrong sample space, and assuming the outcomes are equiprobable just because it is a sample space. ("I have two children, and there is a 1/3 chance that I have two boys because the sample space is 0, 1, or 2 boys!")
An event is not the same as an outcome, it is a set of outcomes. The two are easily confused. The difference is that your schema for providing descriptions can, depending on the event, separate it into subsets that are also events. By definition, an outcome can't be separated that way unless you change the schema.
So if your schema is to look at the sum, a 10 is a 10 whether the combination is (4,6) or (5,5). But that schema isn't useful if all you see is one die: "I see a 4" doesn't tell you anything about which "sum" description is appropriate. You have to change to a schema that describes the possible companions of the 4 you see.
In your approach to Sleeping Beauty, you are considering Monday&TAILS and Tuesday&TAILS to be inseperable parts of the same outcome. Probably because they are both part of what you call "the TAILS protocol." That is, you consider Monday&TAILS, Tuesday&TAILS, and just TAILS to be different names for the same outcome. This is a point of view that is only valid from outside the experiment; the lab techs, or Beauty on Sunday night.
What you are ignoring, is that when she is inside the experiment, even though she doesn't know which day it is, she does know that that it is not currently both days. So TAILS (or TAILS protocol) is not an outcome. She can separate it into the distinct outcomes Monday&TAILS and Tuesday&TAILS, and know that only one applies to the current moment. And the fact that you do make this distinction for HEADS requires you to do it with TAILS. (And even if you think it should not be necessary to do so, you can't be wrong by doing it.)
The two-day protocol is irrelevant to Beauty, because she is inside the experiment and so participating in only one day. Her sample space is the set of four possible single-day protocols: {Monday&TAILS, Tuesday&TAILS, Monday&HEADS, Tuesday&HEADS}. Each has a prior probability of 1/4 to apply to a random single day in the experiment, which is all that Beauty knows is happening. But because she kn0ows SOTAI is happening on that single day, she ca rule out one of those outcomes.
This is why I suggested the alternative where we forget about days and just say that if it's heads then she's woken once and if it's tails then she's woken twice. Our sample space is just {Heads + only awakening, Tails + first awakening, Tails + second awakening}.
(a) No it isn't. From the OP:
I personally think it's slightly cleaner to describe the experiment as immediately sending her home at the conclusion of her final interview, whether that's Monday or Tuesday -- it reduces the temptation to argue about Tuesday and Wednesday -- but this is a standard presentation and it works just fine.
There's also little difference if you modify the experiment as @Michael suggested and push the tails interviews back a day to Tuesday & Wednesday, and give Monday entirely to heads. Or do them all at different times of day on Monday, as some internet poster (on one of the LW threads, I think) has suggested.
The only thing that matters is one for heads and two for tails.
(b) If it were part of the heads protocol, by eliminating it, you would be eliminating heads as an outcome. Simply being interviewed would tell you the coin landed tails.
If that seems like a tendentious interpretation, consider what happens as you increase the number of tails interviews: whatever the ratio, that's your odds it was tails. Do a thousand tails interviews, and it's a near certainty -- according to thirders -- that a fair coin lands tails.
Quoting JeffJo
I was speaking as Beauty there. I appreciate your input very much -- my point was that Beauty reasoning in this way makes no progress.
Quoting JeffJo
I'll look at it. After 18 days now in this thread, I had grown weary of alternative presentations that require analysis to figure out if they're even equivalent to SB. But I'll look at it.
Because the SB scenario doubles tails outcomes, it is difficult, or at least unnatural, to express your confidence about the outcome through wagering. For instance, suppose the coin is biased 2:1 in favor of heads. You have inside information and are thrilled to be given even money odds. Then we get these results:
100 tosses, 67 heads, 33 tails, 133 interviews.
Betting heads consistently on a biased coin at even money, you break even. WTF?
The sucker who was betting tails, who didn't know the coin was biased? He breaks even too.
ADDED:
Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6?
Yes that's the nature of the experiment. There are two ways of looking at it.
In my view, probability is a measure of the state that the agent is in. Unconditionally, there is a 1/2 chance that Beauty will be in a state associated with heads.
However when conditioning on being awake and interviewed, there is a 1/3 chance that Beauty will be in a state associated with heads.
So the issue between double-halfers and thirders is whether conditioning is valid here.
Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.
However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not?
Normally you would just give 1 M&M. But it is ultimately a question about what sample space and rules are appropriate in the circumstances.
My point is that it can't matter what SOTAI is. There is a protocol for Tuesday&HEADS, and in an interview Beauty knows that it is not the protocol that is currently in progress.
What matters is that there is a protocol on both days for both HEADS and TAILS. And that one of these four protocols is inconsistent with Beauty being interviewed. You keep treating the fact that she sleeps through a day as if that makes the day nonexistent,or that it is not something the lab techs have to have included in their protocol.
?????
In general, I dislike the use of the word "eliminate." People forget that it means "An outcome which was possible has been shown to be incompatible with the current information state."
On Sunday Night, Beauty knows that her information state during the experiment will be limited to a single day's experiences. She knows that there are four possible such states:
If she is interviewed, she knows that one has been shown to be incompatible with her current information state.
Yep. Get two thousand volunteers. Order them randomly from #1 to #2,000. House #1 thru #1,0000 in the HEADS wing of your lab, and #1,001 thru #2,000 in the TAILS wing. Then flip your fair a coin.
On each of the next 1,000 days, wake all of the volunteers in the wing that corresponds to the coin result, and one - the one whose number corresponds to the day - from the other wing. Ask each of the 1,001 awake volunteers for her confidence that her wing is not the one that corresponds to the coin that was flipped.
Each of these women is in an experiment that is identical - except for the labels you put on coins and days - to what you just described. Each knows that 1,000 awake women came from one wing, and only 1 from the other.
Yes, it is a fair coin flip. But Beauty is not asked about its flip in an information vacuum. She knows that there is a 1/N chance that she would have been interviewed today under one result, but a 100% chance under the other. Her confidence in the first result must be 1/(1+N).
Here's the thing: it sure does look like the design of the experiment involves conditioning heads on ~Tuesday, so you get (1/4)/(1/2) = 1/2 for heads -- heads ends up by definition (heads & Monday). I'm tempted to say that since this is baked into the design of the experiment, this bit of conditioning has the status of background knowledge, more or less. At any rate, I consider it an open question whether this bit of restricting the space can or should be treated differently from the conditioning that Beauty might do in considering her personal situation.
Lots more to say, but first I want to ask you two about another quickie alternative experiment, which we might have done before, I've lost track.
I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.
My question is this: do you think this is equivalent to SB? And why or why not?
Sorry, I'm not getting your experiment, or its equivalence to SB.
One thing I'm generally uncertain about is how strongly to lean on "what day today is" being random. There are some things we can say about their equivalence for Beauty, but Elga and Lewis are both pretty cautious about that. I don't think we can just throw a big principle of indifference at "what day today is" and be done.
Here's how I converted from thirderism to halferism. Heads interviews are red marbles, tails interviews are blues.
If you select a marble from an urn with 1 red and 2 blues, sure, chances of getting the red are 1/3. But that is not Beauty's situation. Instead we have two urns, red in one and blue in the other. A coin is tossed to determine which urn to select from. It doesn't even matter how many marbles are in each; your chances of getting red are 1/2. Beauty cannot tell the difference between one interview on heads and any number of interviews on tails, but she knows that each procedure has a 1/2 chance of being followed.
Yes, in the sense that one should condition on being interviewed.
Quoting Srap Tasmaner
He's simply saying that if you interview different people for each permutation instead of just one person, then the thirder-style (1/(1+N)) result follows. Which is equivalent to the SB scenario.
In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track.
I really like this argument. I meant to ask about it myself -- I saw a variation of it on StackExchange a few days ago while I was digging around for other approaches -- but I forgot.
Maybe an even cleaner version is for the additional rule to be: if and only if it's a (Tue & Heads) interview, you will be told at the start that it's a (Tue & Heads) interview; then when Beauty is not told this, she infers ¬(Tue & Heads). Now when you delete the (Tue & Heads) interview, absolutely nothing else changes. You could even run standard Sleeping Beauty by having the experimenter lie, tell her it's Informative-SB, but then never do the (Tue & Heads) interviews. This looks like the perfect way to solve SB by treating it as a special case of something more obviously solvable.
Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end.
I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success. Michael (this is my 8 year old, not TPF's Michael) gets 1 M&M and a fatherly lecture on how to measure the success of predictions.
Thanks for hanging so long, @Andrew M (weird, the Andy in my story is my 10 year old, not you). Think I learned some things. Going to take a long break now from Sleeping Beauty.
Still thinking about how to properly score this thing.
The Lewis table is what you get if you try to compensate for SB's structure by treating the coin itself as biased 2:1 heads:tails. You start with this table
drop (H & Tue) and conditionalize on P(H1 ? T1 ? T2) = 2/3 to get
So it's true that the Lewis does represent an attempt to "discount" the overabundance of tails, but it does it in the wrong place. You can't mess with the coin.
The only thing to do, to get a better measure of success rate, is to score the results at 2:1, so that you can get a payoff table like this:
I'm not sure this is sophisticated enough though. What if instead of going all heads/tails, you use a mixed strategy? The payout events (W and L) have the right ratio, but the payout values are still screwy.
Not sure I even need to worry about mixed strategies here though. The coin being fair gives a lower bound to failure of 50% and an upper bound to success of 50%.
Thought I was done, but I'm going to keep thinking about the scoring problem.
One more point.
If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty.
Statistics is about expectations. Statistics uses Probability Theory to calculate expectations. In conventional experiments, we find that the value of the expectation is the value of the probability. The concepts of "probability" and "expectation" are very different.
The Sleeping Beauty Problem bends conventionality, and makes statistics inapplicable, by making the number of trials depend on the result for which you are trying to assess a probability. So arguments about frequency and expectation are meaningless until the "bent" issues are resolved.
Specifically: people disagree about whether there are two, three, or four disjoint events that comprise Beauty's sample space. On Sunday Night, the answer is clearly "two" since how the concept "today" applies at that time is different than when she is awakened during the experiment.
Halfers want the number to be "three" inside the experiment. They treat it like Tuesday doesn't happen if Heads is rolled. The problem with the rationale they use, it that it only supports "two," which nobody thinks applies within the experiment. The inconsistency in their argument is that Monday is different than Tuesday if TAILS was flipped, but not if HEADS was flipped. In that case, Tuesday does not exist.
You reject the existence of Tuesday yourself, when you say things like:
That is not what the condition is. It is {AWAKE}, which is can also be written as ~{HEADS&Tuesday} or {HEADS&Monday,TAILS&Monday,TAILS&Tuesday}.
I've presented an alternative problem that unequivocally demonstrates the solution. You give no reason for "not getting .. its equivalence to SB." It is my opinion that you "do not get it" because doing so would make you change your answer. To "get it," all you need do is let Beauty be wakened on HEADS+Tuesday, but do SOTAI.
"Random" is not a property of what you are looking at in an experiment. It is a property of what you know about it, but can't see. Either because the experiment hasn't happened yet, or it has but you can't see what happened.
Here's an example that I've used in this thread, that directly addresses your uncertainty: Forget the coin, and the two possible awakenings. Put Beauty to sleep, and roll a six-sided die. Call the result of the roll R. Leave her asleep on five days during the following week, waking her only after R nights (so R=1 means Monday, R=2 means Tuesday, etc.). When she is awake, ask her for her confidence that today is Wednesday.
Is the answer 1, meaning "what day today is, is not random," or is the answer 1/6, meaning "what day today is, is random" ?
The reason it is random, is because Beauty has no evidence of what day it could be. Does this sound anything like the Sleeping Beauty Problem to you?
Thanks. We've moved on here, but I appreciate your thoughts. You've addressed a lot of the gaps in my understanding of this stuff, and I especially appreciate you taking the time to do that.
Your scheme for removing the time element is pretty cool, and I'm going to spend some more time looking at it.
Welcome back!
Quoting Srap Tasmaner
I regard them as two distinct epistemic perspectives. One is Beauty's on Sunday (or Wednesday) who doesn't condition on being interviewed and the other is Beauty's on Monday or Tuesday who does.
Quoting Srap Tasmaner
Yes. You could say the setup is that there is one of two experiments which is randomly selected. In one experiment, only one random person gets interviewed. In the other, everyone does. You find yourself getting interviewed. So what is the probability that you're in the single interview experiment? 1/(1+N) where N is the number of people.
As a result, it is structurally unlike a biased coin or regular betting scenario. There is no one-to-one mapping when the coin toss outcome itself has a different number of interviews (or agents) associated with it. Instead the sample space must properly account for those degrees of freedom with probability being the measure of which state an agent is currently in.
It was worth the wandering just for the return. I'm still kind of stunned by the elegance of the argument that convinced me. The symmetry of it. In standard SB, on (H & Tue) our Beauty receives no information at all, is not even conscious; in Informative-SB, she receives nearly all the possible information. We gather all of it together into one box -- and then close the lid. Just like that, transforming one into the other.
It's quite beautiful. And a fine reminder to look for the general problem of which the one at hand is only a special case.
The issue is, does she receive information on ~(H & Tue)? And the answer can't depend on how, or even if, she would receive information on (H & Tue).
So, instead of the (H & Tue) protocol being "let her sleep," make it "take her to DisneyWorld." Just before she finds out where she is going, the Law of Total Probability says:
Here,
So we can now say that:
If Y=1/2, as you believe, that means X=1. This is a contradiction, since it means she cannot be taken to DisneyWorld. Whether or not you accept that Y=1/3, the fact that there is a chance of going to DisneyWorld proves that Y<1/2.