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The sleeping beauty problem

Off-topic: Share\(\rm\LaTeX\) has had a facelift, I was told by the folks behind it
Sean Carroll's blog posts are getting stupider and stupider. The latest one is about the
Quantum Sleeping Beauty and the Multiverse
He believes that the blog post is about quantum mechanics or a "quantum version" of a silly problem that philosophers like to answer incorrectly. Of course, there is no quantum mechanics in his musings whatsoever. He hasn't even attempted to think quantum mechanically in his whole life and all his thinking has always been purely classical. In this case, he is thinking about the probabilities in a classical Universe with many worlds and a sleeping forgetful lady in it.

Miss Czechia 2012 Tereza Fajksová in Paris. How many legs does the sleeping beauty in a Škoda car have? I find the obsession of similar (Czech) women with getting the suntan excessive.

Even though it's purely classical, Carroll – along with some stupid philosophers – gives a completely wrong answer to a very trivial problem. What is it?

The problem is the following:
On Sunday, a blonde is told that they would flip a coin once (and never again). Depending on what the coin shows, they will play with her in one way or another.

The games will include no sex but she will be interviewed on Monday, and maybe – if the coin came up heads – also on Tuesday. In the case of the "heads", she will be made forgotten about the Monday interview before they interview her on Tuesday.

In advance, they tell her that it is a fair coin that comes up tails with probability 50%.

During the interview(s), she is asked: “What is the probability you would assign that the coin came up tails?”
What should the babe say? Well, it's an intelligent blonde who was previously told that the probability for the coin to come up tails was 50%. So she answers that she would assign the probability 50% that the coin came up tails. If they told her that M-theory predicts an 11-dimensional spacetime and asked her what dimension she would assign to the M-theoretical spacetime, this unusually smart woman would be able to say "eleven", too!

A smart blonde. Unlike Sean "Dumber Than a Blonde" Carroll whose answer is 33.3%. No kidding! ;-) And he even defends this silly wrong answer by references to his crackpot papers about many worlds. I just can't believe it. This is a puzzle I would correctly solve when I was 7 years old. Why is it so hard for Sean Carroll who is an adult?

The probability that a coin comes up tails is\[

P_{\rm tails} = \frac{N_{\rm tails}}{N_{\rm flips}}

\] i.e. the ratio of coin flips (in repetitions of the same situation) that come up tails and the number of all flips. It's clearly 50% in the problem we were explained above. We were told that it is so. That's how the problem was defined. To answer correctly, we just need to repeat one sentence we were explicitly told!

Sean Carroll believes that the probability for the coin to come up tails is 33.3% because if the coin comes up heads, she is interviewed twice – on Monday and on Tuesday – and not just once, on Monday, which is the case when it comes up tails.

It is very obvious where his retarded childish mistake comes from. He computes the probability that "the coin comes up tails" as\[

P_{\rm tails,Carroll} = \frac{N_\text{interviews with her after tails}}{N_\text{total interviews with her in all days and all scenarios}}

\] Because the number of interviews with the babe is 2 times higher in the case that the coin comes up heads relatively to the case when it comes up tails, he only gets \(P=1/3\).

But the probability that a coin comes up tails has nothing whatsoever to do with some number of some interviews. It doesn't matter whether a woman talks about "heads" twice as often than about "tails". The probability of "tails" is still equal to 50%. Whether her memory about the previous interviews was erased or not cannot influence the answer, either. The behavior of a coin doesn't depend on what someone does with a woman's brain.

The ratio that leads to the "result" \(P=1/3\) isn't even a "probability" because the "units" that are being added in the numerator as well as the denominator aren't even "outcomes in mutually different repetitions of the history". Instead, this ratio is the expectation value of the number of interviews after tails, \[

\langle N_\text{interview after tails}\rangle.

\]. Note that there is a variable in between the brackets so it's not a probability. Probabilities are expectation values of projection operators, something whose possible (eigen)values are just zero or one.

If she is told that the coin comes up tails with probability 50% and if she doesn't forget about this rule, she must be able to say that the probability that the coin comes up tails is 50%! The probability that a randomly chosen interview occurred after "tails" may be 33.3% but that's an a priori different question and a possibly different probability than the question about the probability that the coin came up tails. Incidentally, the probability that a random interview occurred after "tails" is still equal to 50%! The result "heads" may lead to more words and interviews but its probability is still 50%.

The coin was flipped "Once In a Lifetime". It doesn't matter whether the "Talking Heads" are "talking" about the coin more often if the coin comes up "heads": the probability for the flip to come up heads is still just 50%. You may find yourself in another part of the world, behind a large automobile, or in a beautiful house. Or in a room for interviews. But the probability of "heads" is the same as it ever was, the same as it ever was, 50%. Lyrics.

For Sean Carroll, however, it's too difficult. What a moron. It's not surprising that he has no chance to understand quantum mechanics and the concept of probability that is intrinsically incorporated in the foundations of quantum mechanics if he isn't able to solve this trivial problem for the schoolkids.

Yes, Carroll's mistake is very similar (and perhaps identical) to Walter Wagner's "calculation" of the probability that the LHC would destroy the world. On the Daily Show, the high school teacher Wagner would say that the LHC may either destroy the world, or fail to destroy the world, so the probability is 50%. Similarly, Carroll thinks that the woman may be either interviewed after "tails" or Monday, or after "heads" on Monday, or after "heads" on Tuesday, so the probability of "tails" is one-third. Note that the key both to Wagner's and Carroll's idiocy is that they attribute the same probability to each of the list of several options that have absolutely no reason to be equally likely (and, in Carroll's case, the three events even fail to be mutually exclusive).

I guess that many of the "physicists" who talk about the "many worlds" would do the same mistakes as Wagner and Carroll. It's insane that Carroll – and arguably many others – are pretending to do cutting-edge science by presenting completely wrong solutions to some problems that intelligent enough schoolkids can solve correctly.

Additional remarks on the sleeping beauty

If you need a six-page, 100+ citations paper by a Princeton philosopher saying the same things as me, \(P_{\rm tails}=1/2\) and she is getting no relevant information when woken up (L1), because both hypotheses imply that she would have the experience of being woken up (at least once), so there is no justification to adjust the prior probabilities \(1/2\)-\(1/2\), see e.g. David Lewis' article. Lewis died shortly after he published that paper in 2001.

In the comments below, people say lots of right and wrong things. A typical extra error of the "thirders" ("truthers"?) – perhaps an error equivalent to all other errors they are making – is that they do not take sampling bias into account while calculating credences.

I also say that a single "share" in a financial scenario is worth $33.3 but the average values of costs and benefits match and are both equal to $50, as expected from the probability \(P_{\rm tails}=1/2\).

Most of the papers about the problem also discuss the value of \(P_+(HEADS)\), the credence that the coin is "heads" right after she is told on Monday that it is Monday – and let's assume that the rules now include the comment that she is always told it is Monday on Monday evening. The original "thirder" Adam Elga would say that \(P_+(HEADS)=1/2\). It's calculated as \(P(HEADS|MONDAY)\) by dividing his overall previous (wrong) probability for all Monday outcomes, \(2/3\) of the wakeups, to two equal parts. However, the correct answer is obviously \(P_+(HEADS)=1/3\) exactly because in this case, we are incorporating new information (about "Monday") by Bayesian inference. The "heads" hypothesis only predicts that she would be told it's Monday with probability \(1/2\), unlike the "tails" hypothesis that predicts the Monday announcement at 100%. That's where the ratio "two" of the probabilities (suppressing the "heads" hypothesis that didn't predict "Monday" too certainly) comes from via the Bayes rule. Equivalently, the pie is divided to \(1/2\) for "tails Monday", \(1/4\) to "heads Monday", and \(1/4\) to "heads Tuesday". Among the Monday piece of pie, \(1/2+1/4\), "heads Monday" represents one-third of the piece, which is why it's the conditional subjective probability \(P(HEADS|MONDAY)=1/3\). (Note that Lewis' version of the sleeping beauty has "tails" and "heads" interchanged.)

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reader Truth said...

Hi Lubos, I agree with you that Sean Carroll says lots of wrong things on his blog, which is very bad for the dissemination of science.
However, I couldn't tell if this was conclusion in his "quantum version" of the problem. I think he might have ended up concluding that the odds are 50/50? Am I wrong?

reader Luboš Motl said...

Yes, I think that you are wrong. Look for this quote:

Our derivation of the Born Rule is actually based on the idea of self-locating uncertainty, so adding a bit more to it is no problem at all. We show that, if you accept the ESP [the silly "principle" in the Carroll et al. paper], you are immediately led to the “thirder” [P = 33%] position, as originally advocated by Elga.

reader JollyJoker said...

Replied on Carroll's page. One thing that comes to mind is whether you could set up some "counterfactual bomb detector" style scenario where she can guess with a better than 50-50 chance of success what the coin toss was due to quantum effects? That would of course require that more than one path exists in "non-many" worlds ;)

reader Luboš Motl said...

LOL. The coin was *sad* to produce heads with 50% probability, so getting a different result for the same quantity isn't a great achievement in physics or engineering. Instead, it is a self-evident mathematical contradiction. I don't believe we will be even talking about this here.

reader andrew said...

Hello. Suppose that SB's credence is always one-half, as you say it should be. A bookmaker (without more knowledge than SB) can expect to make money from SB by offering her bets that rationally she will accept - she can be "Dutch-booked."

If SB can be Duch-booked, her credences violate the axioms of probability. She cannot be Dutch-booked if her credence is one-third. That suggests her credence ought to be one-third.

There are many subtleties - the problem is not trivial and Sean Caroll is not silly for answering one-third.

reader Luboš Motl said...

No, on the contrary, there are no subtleties - at least no subtleties that an intelligent 10-year-old boy would be unable to solve. Carroll definitely *is* stupid, and so are you and everyone who proposes any different answer than 1/2 based on any definition of the probability or any argument.

If she is promised $100 for every correct guess about the coin she may give on an interview, of course that it's better for her to guess "heads" because she wins $200 in total because in that case, she is interviewed twice.

If they want to offer her a "balanced" offer so that she will be ambiguous whether she guesses "heads" or "tails", they have to offer her $50 for each correct guess if the correct answer is "heads", and $100 for a correct guess if the correct answer is "tails". In that case, she will get 1 x $100 in the case that it's "tails", and 2x $50 = $100 if it's "heads". The correct ratio of probabilities is equal to the inverse ratio of the *total* bounties she wins in the two possible outcomes, and it is obvious that the ratio of probabilities is therefore 1:1, with Dutch book, Bayesian, frequentist, or any other way to describe probabilities. They never really differ.

reader andrew said...

A bookmaker could be a fellow participant in the experiment, sleeping the same room as SB, offering her bets when they are both awake.

If SB's credence is always one-half, the BM can expect to make money from her. The BM knows no more than SB.

If a BM can expect to make money from bets that, with your credences, you ought to accept, and the BM knows no more than you, it is not silly to suggest that your credences are faulty.

reader Dan said...

There is a response below Carrolls blog by one "Joe Polchinski". Is it really him? I hope not. It would make me sad to think he really lost it so badly. :-(

reader Uncle Al said...
Sean et al. are not necessarily stupid - they could be deluded, unlike Mythbusters' "Wheel of Mythfortune."

Many Worlds demands coins land exactly on their edge. Given at least a small infinity of those worlds, the summed probability of a coin landing either heads or tails is then asymptotic to zero. Observation is defective.

reader Luboš Motl said...

This addition of the people who know exactly the same as the SB clearly cannot make any difference to the right reasoning whatsoever, can't you see that?

If the woman or any other participant of the experiment knows that there are 2 interviews after heads and 1 interview only after tails, they will know that the number of interviews - and possible opportunities to win something, if they're included in the interviews - will be 2 times greater (the first part of this sentence is a tautology because I am just fucking repeating what I have already said). This factor of two has to be taken into account when computing the correct probabilities. The ratio of probabilities isn't equal to the inverse ratio of money one wins after one correct guess on an interview. It is equal to the inverse ratio of the total money that one wins in the case of one outcome of the coin flip. So it doesn't matter at all how the "bounty in the case of heads" is divided between the two interviews in that case.

reader Luboš Motl said...

Holy cow, Joe has lost it entirely.

reader Dilaton said...

A sad loss ... :-(

reader JP said...

The case that the blond was just woken up is not independent of the result of the coin toss. There is extra information. With this information the probability of tails is no more 50%, no?

What if the game is this: If coin comes up tails, you are killed tonight. Otherwise you wake up happily tomorrow. Now you are asked, when you wake up next time, "what is the probability that the coin came up tails yesterday?". Will you answer 50%?

reader Mikael said...

Dear Lubos,
I think Carroll is trying to apply the concept of probability to postdictions instead of predections. For example if I through a coin while you couldn't see it and I told you afterwards that it was heads and you trusted me completely you would assign also the probability one to the outcome heads. (Probably it is a little less than one because I might have become insane or malicious.) You could also think that I want to deceive you and assign a probability close to zero to heads. Now somebody who has recorded the outcome could offer you a bet. If it was good enough you would probably take it. Not sure if the concept of probabilities applies but it seems like a reasonable question to me.

reader yiannis said...

Of course she should answer 50%. Even if she had to answer twice, the coin was flipped only once with probability 1/2. People that start adding things on their own, about how much money she would win depending on the number of her interviews they just reply to different questions.

reader DN said...

Suppose the SB has this share on offer when waking up:

"This share pays 100$ on Wednesday if the coin came up tails."

Then she should pay up to 33.3$, but no more, to make a profit (or break even). If she chose to pay 50$, she would (obviously) lose money in the long run.

If we change the rules of the game to the SB being immediately killed in case of heads, then of course she should pay up to 100$ for the share because she then knows for sure that the coin came up tails, given that she was asked about it at all.

It is not enough to only focus on the odds of the coin toss before the experiment since the SB has additional info (ie. "I am being asked about this knowing that these and these effects took place after the coin toss.")

reader John McVirgo said...

Sean Carroll worded the question as:

“What is the probability you would assign that the coin came up tails?”

Whereas I think he should have said:

"“What is the probability you would assign that the coin currently shows tails?”

For the first question I would say 1/2, for the second 1/3.

I would therefore say both Lubos and Sean are right, but the phrasing of the question is the real problem.

reader jon said...

John, your re-phrasing does not change the problem. The repeated head interviews all share 1/2 of the probability space. The easiest way to see this is to imagine the heads and tails are drawn from an urn with one head and one tail. The two head interviews share the same head draw. So they each get 1/4 probability, or 1/2 total. The one tail interview gets the other 1/2.

reader Luboš Motl said...

Amen to that.

reader Luboš Motl said...

No, there is no extra information whatever. The blonde was waken up and the only thing she knows are the rules of the game and the fact that she was just waken up. These things unambiguously imply that a coin was previously flipped with probabilities 50%, 50% to come up heads, tails.

reader Luboš Motl said...

No, the answer is 1/2 to both. They don't differ at all. If the "tails" predicted - at least with some probability - that what the woman sees and observes right now could "fail to happen", then she would be getting new information at the moment of the interview. But she is not getting any new information whatever, and neither is anyone else - except for those who may know it's Tuesday, and indeed, it is Tuesday today. ;-)

reader Luboš Motl said...

Huh? She should obviously pay $50 for that share. In half of the cases (weeks), she would get nothing bad, in the other half, she would get $100 back, so in average, she would get her $50 back so it's a fair price. I have no clue how you can fail to see that.

reader DN said...

She is buying two (worthless) shares in half the cases, and one share worth 100$ in the other half. (The share bought on Monday stays with the broker.)

So, over two weeks (tail and head respectively), she buys 3 shares, with one of them being worth 100$.

So she cannot pay more than 33$.

I thought of this question as a prediction market when you posted it; I can't see how you can justify paying 50$ when you are in her shoes?

reader Don said...


reader Don said...

"It’s exactly the kind of wacky thought experiment that philosophers just eat up." That's because they don't know math. What company that guy keeps. Wacky is right.

reader Luboš Motl said...

Sorry, I have already discussed this point as well. If your rules about the shares are that you offer her to buy 2 shares during the week in the case of heads, one has to sum up the costs as well as benefits from the whole week. It makes no impact on the probability of heads and tails.

The credence of tails is Ptails=1/2 and along with the price Price=$33, they obey the "fair" costs-and-benefit equations for the expectation value of the total costs and benefits.

The expectation value of the benefits is

Ptails * $100 + (1-Ptails) * $0 = 1/2 * $100 = $50.

The expectation value of the costs for the stock(s) is

Ptails * Price + (1-Ptails) * 2 * Price = 1/2 * $33 + 1/2 * 2 * $33 = $50.

Feel free to verify that it works, or ask an average clever 8-year-old kid to help you. The credence is P=1/2. It was important to realize that the costs of the stock are 2 times higher in the heads week. If she is told about the "two wakeups after heads rule", she knows that as well as we do.

reader DN said...

Of course she pays twice in case of heads (and she has forgotten about the Monday purchase on Tuesday). How else would I arrive at 33$.

Your calculations show perfectly that 33$ is the break even point, with expected benefit and cost then each equaling 50$.

Another case:

If tail, then she is killed 50% of the times, the other times she is offered the share ("interviewed").

If heads, same as before (two interviews).

Now the break even point is 20$, with expected cost and benefit each 25$.

You can make all kind of games like this; the rules obviously influence the correct betting strategy for a participant in the game (her "credence").

reader Luboš Motl said...

Sorry, Mikael, it makes no sense.

The very point of probability - of any kind, in any of its formalization - is to determine the fraction of events that obey a certain condition in the case that the same initial state is repeated many times.

One can't define any "mirror probability" where we would repeat the final state because we don't have the control over the future: the final state is always less determined than the initial state.

Moreover, he's not just reverting the past and future. He's also totally unjustifiably assuming that the probabilities of "Monday interview with tails", "Monday interview with heads", and "Tuesday interview with heads" are all equal, and therefore equal to 1/3 each. But that's just not the case. There exists absolutely no justification of this wrong result.

The property "Monday or Tuesday" is independent of the property "heads and tails", so if we consider the conditional probability that a coin shows heads or tails, and that it's Monday or Tuesday, and we ask about the probability of all these four combinations (Mon-heads, Mon-tails, Tue-heads, Tue-tails), each of them clearly has the probability 25%. It doesn't matter a tiny bit that a woman is not waken up on Tuesday if the coin is tails. The probability of that is still 25% even if she is not around.

reader Luboš Motl said...

You can invent lots of games with different results but the particular "sleeping beauty game" discussed here has the right answer for credence P = 1/2. Do you finally agree?!

reader Am I Who said...

Let's reformulate the problem as this.
There are three participants in the game. The participants are sitting in their own room so they can't see other participants and can't know whether they were interviewed or not. Of course, they are informed of the rule of the game.
If tail comes out, then one of the participant is chosen and interviewed. If head comes out, then two of the participants are chosen and interviewed. The coin is tossed only once.
Now you are one of the participant and you are asked about the outcome. What will you be your assignment?
As you can see, the problem is completely equivalent and now the probability is the simple conditional probability.
P(You are interviewed| tail) = (1/3 * 1/2)/ (1/2) = 1/3
P(You are interviewed|head) = (2/3 * 1/2) /(1/2) = 2/3
Plain simple. Can you point out the error in this logic?

reader an avid fan said...

Lubos, as philosopher Nick Boström, who argues ( or jokes?) that the probability we don't live inside a computer simulation is near zero has also written about the Sleeping Beauty problem in his book "Anthropic Bias: Observation Selection Effects in Science and Philosophy" - the text is online

The book also addresses the Dooms Day -argument:

From Wikipedia, the free encyclopedia

The Doomsday argument (DA) is a probabilistic argument that claims to predict the number of future members of the human species given only an estimate of the total number of humans born so far. Simply put, it says that supposing the humans alive today are in a random place in the whole human history timeline, chances are we are about halfway through it.

It was first proposed in an explicit way by the astrophysicist Brandon Carter in 1983, from which it is sometimes called the Carter catastrophe ... etc

Boström says the he thinks the argument "ultimately fails" but there haven't yet been valid refutations, so why not bust it next ...

reader Luboš Motl said...

LOL, thanks for your kind words. Yes, I know these musings - e.g. Brian Greene's The Hidden Reality book that I was a translator of is full of these things.

It's been discussed on this blog about 10 times but I forgot what was the best optimized blog post dedicated to the doomsday fallacy.

Maybe something else.

reader Am I Who said...

Let's reformulate the problem as this.
There are three participants in the game. They are sitting in their own room, so they can't see each other and they don't know whether others are interviewed or not. They are informed of the rule of the game.
And one coin is tossed once and for all.
If tail comes out, one participant is chosen and interviewed. If head comes out, two participants are chosen and interviewed.
You are one of the participant and you were asked about the outcome of the coin toss.
You can see the problem is completely equivalently formulated. Now you can use a simple conditional probability.
P(You are interviewed|head) = (2/3 * 1/2)/(1/2) = 2/3
P(You are interviewed|tail) = (1/3 * 1/2) /(1/2) = 1/3
Can you find any mistake in the logic?

reader Luboš Motl said...

Sure, I can easily point out the error in your logic. Your error starts when you say that your "problem is completely equivalent" to the sleeping beauty problem.

It's not equivalent at all. In your case, the conditional probabilities that you are interviewed given the heads vs tails theory are

P(You are interviewed| tail) = (1/3 * 1/2)/ (1/2) = 1/3
P(You are interviewed|head) = (2/3 * 1/2) /(1/2) = 2/3

reader Am I Who said...

Why aren't they equivalent? Each sleeping beauty can be regarded as one participant.

reader Marcel van Velzen said...

In “Physicists Should Stop Saying Silly Things about Philosophy”
( there are 225 comments and our “philosopher” hasn’t answered a single comment. Joe Polchinski writes a silly comment and within 15 minutes he answers! Only because it’s good for his career. I tell you, Sean Carroll is a sneaky person! I would advise you to completely ignore him, like I do, but on the other hand your explanations of the correct way of thinking about these problems are invaluable.

reader Am I Who said...

I see your point.
To make the two problem equivalent, I need to do an additional trick.
Trinity. Three participants are actually the same person. Then we will get 50:50. I agree with you.

reader Luboš Motl said...

Thanks a lot, happy about every small advance for the mankind like this one. ;-)

reader Luboš Motl said...

I sort of agree with your explanation of the asymmetric reactions of Carroll towards different readers of his blog...

reader Luboš Motl said...

It's funny that some philosophers look like grown up children who revenge to their math teachers and others who were not satisfied with the philosophers' answers to math questions when the philosophers were kids.

An exchange I remember as if it were yesterday: we were fifth grader, just switched to a special mathematical class, and I already had some name due to some victories in mathematical olympiad (Pythagoriad). But we would learn about probabilities, and among hundreds of similar things, the teacher asked what was the probability that you get 6-6 if you roll two dice.

Now, a classmate, a nice guy and a deep historian, would answer that it was 1/21 because there are exactly 21 possible outcomes of the two dice - 11, 12, 13..... 56, 66. Note that he only counted the ordered pairs of numbers, so 56=65. There are 21 options so it has to be 1/21.

Of course, the teacher would tell him that it was 1/36. The outcomes 56 and 65 may be grouped into one group but if you do so, the probability of this group - that the sum is eleven - is twice higher than it is for 6-6. You may just do it experimentally if you don't believe it! ;-) The guy got totally pissed off, insisting that it was a teacher's demagogy for the rest of the class.

He would become a historian but I suspect that some of these lousy schoolkids in mathematics deliberately become philosophers and they publish ludicrous papers with wrong solutions to such basic puzzles - so that they may say: You see, my mathematics teacher? I am now a big philosopher and may publish papers claiming that you were wrong! ;-)

It's them who is still wrong but they don't care.

reader Marcel van Velzen said...

True, but I would nevertheless advise your readers to comment as “Ed Witten” or something :-)

reader Luboš Motl said...

LOL, or maybe Thomas Rosenbaum. ;-)

reader Shannon said...

Thank you for this excellent post, Lubos. All the comments are great too.

reader Luboš Motl said...

Merci to you, Shannon, for your interest and kind words!

reader Shannon said...

And you are the most interesting adult I know ;-)

reader Giotis said...

Polchinski said:

"I thought that your 1/3 sounded correct, but when I heard that certain other blogs were vigorously denouncing your answer I became certain that you must be correct."

I'm not sure but if he means you Lubos I think such statement is highly provocative and unexpected. I'm begining to think that he is not Polchinski but an impostor.

reader Alex said...

It's always 50%. You either win or you lose.

reader John Archer said...

Dear Luboš,

I'm currently working on a quantum-mechanical version of the unexpected hanging paradox. It involves a radioactive pussy cat, a heavy crossbeam, a phial of cyanide, a length of piano wire and Tony Bliar.

The thing is, I can get the result I want without the cat and the phial of cyanide but I need to be able to pin it on the cat.

Can you help me out, or should I ask Sean Carroll?

Hey, maybe we could pin it on him instead! Two birds with one stone. Whaddya say? :)

reader Alex said...

hallucinogenic instead of the cyanide should do it. It would be a mystery then. Was it the cat or was it suicide?

reader SteveBrooklineMA said...

Like "Monty Hall" and "Tuesday Boy", this question causes problems because the English it is written in is a little vague and subject to interpretation. How do you model an "amnesia potion" mathematically? Clearly, if Beauty always guesses "tails" she will be right about 1/3rd of the time out of a large number of trials. Some will interpret or model the "amnesia potion" as a kind of de-correlator, or scrambler, allowing each Beauty interview to be considered as an independent event, and making it "seem to Beauty" (whatever that means) that the probability of the coin coming up tails is 1/3.

Take a large number of trials: THHTTHHHHTHHHHHHTHHTTHH ... where the H's always come in pairs, then scramble the letters with a permutation chosen uniformly at random, and the result will look just like trials from a (1/3,2/3) coin.

reader Luboš Motl said...

Yup, I say that torturing of animals is bad, so you might be right that it's a better idea to ask Sean Carroll. ;-)

reader Luboš Motl said...

There are many reasons why I would bet against your possibility...

reader Luboš Motl said...

Dear Steve, there is no problem with English.

When she says "tails", she will be right 1/3 of the times when she opens mouth, but that doesn't mean that the "tails" is the state of the coin 1/3 of the *time* (or weeks).

reader SteveBrooklineMA said...

Yes, no doubt, we know it is a fair coin. But some are interpreting Sean's question as asking how the coin would appear to our scramble-brained Beauty.

If we didn't interview Beauty at all when the coin comes up tails, it would appear to Beauty that the coin always comes up heads. It's a little silly, but I think that's how people are interpreting Sean's question.

reader Luboš Motl said...

Dear Steve, by construction, she isn't told about the state of the coin at all.

If you consider a completely different thought experiment in which she is shown the coin after memory erasure, she will always see one state of the coin only, so she will clearly be unable to make any statistical deductions out of this observation - except for excluding the hypothesis that the coin always gives the opposite result than the result she just saw.

reader SteveBrooklineMA said...

I agree. If Beauty were able to keep count with a black=tails and white=heads marble dropped into a mixing bag, she could count the marbles in the bag after each interview and eventually think the coin was about 1/3 - 2/3. As written though, with N=1 forever, what you have written seems right to me.

If she knows the experiment beforehand, and she only has the N=1 experience, then p=1/2 obviously, just as you have written 1000 times now.

reader Luboš Motl said...

OK, but in your other experiment without memory erasures where she collect marbles, the 1/3 - 2/3 ratio of the marbles wouldn't change anything about the fact that the coin flip gives 1/2 - 1/2, would it? I can't believe that you are really suggesting something like that.

reader Rehbock said...

Carroll's blog question and constraints were poorly stated. The answer depends on whether the probability is of the toss or of prior tosses leading to being where she is in the game. It is not controversial as Carroll suggests. It has no meaning to QM or MWI except that one must think carefully about what states are asked about.
If Polchinski has to look this old riddle up on Wikipedia and decides on basis that Motl must be wrong on the proposition that 1/2 is always the probability for the question asked then he has been wrapped up too much in strings too long.

reader SteveBrooklineMA said...

Of course I'm not suggesting that Lubos! The coin comes up heads half the time. But if we play games with poor Beauty, and show her only some of the data, or some of the data more than once, we can make it seem to her that the coin is biased. Beauty's biased-coin conclusion wouldn't reflect reality though, or cause the coin to become biased!

reader Gordon said...

When I saw Sean "solving" yet another QM "problem", I thought

Ohhh, nooo...

Oxbridge philosophers sketch:

reader Luboš Motl said...

Dear Robert, thanks, I obviously agree with something you say, but I still disagree with your claim that the probability depends on whether it's the probability of the toss or prior tosses. All of these probabilities are equal to 50 percent!

reader Luboš Motl said...

Dear Steve, it depends how stupid the woman is but in my opinion, she would have to be extraordinarily stupid if she could duped into thinking that the coin is unfair in this way.

One could only be fooled into thinking that a coin is unfair if she could actually see the coin being *tossed* many times and the ratio of heads and tails would be very different from 1/2. But you can't do that.

If you show a static coin on the desk to a blonde 1,000 times, no one moves the coin in between, and the blonde always sees heads - obviously - will she conclude that the coin is biased or that it always shows heads? If she does conclude that, she should probably be stored in a sanatorium.

reader Uncle Al said...

Go ahead, do the experiment. A coin does not land on its edge. Many Worlds proves observation is defective - with one exception,

reader JollyJoker said...

After reading Polchinskis post, I at least feel I understand the thirder position somewhat. It's the ambiguity in "what probability would you assign". It's not necessarily asking about what the _real_ probability is and with the betting going on as outlined, the thirder position does give a fair bet. Of course, this is simply due to betting twice if it comes up heads; SB could exploit that knowledge by always guessing heads and winning 2/3 of the bets.

I do feel this is a completely contrived interpretation of the words "would you assign".

reader Rehbock said...

Now it was I that was sloppy. We agree that the toss is always 1/2.
There are four paths chosen by two sided coin and one path leaves her asleep. If asked what the odds are among four tosses that one is heads it is 1/2. If shown that one of these four is heads, I have three left to choose from and the odds are 1/3 that I will locate the other head among the three remaining choices.

reader JollyJoker said...

I just realized Polchinski wants SB to pick the only odds she cannot ever exploit for profit... As the problem is stated, "what probability would you pick", Polchinskis contrived betting scenario has her pick the only objectively stupid answer.

reader JP said...

SB is woken twice, if coin comes up heads (or was it tails), otherwise just once. Therefore being just woken up (given the amnesia scenario) is evidence about coin toss result. Simple as that.

reader Luboš Motl said...

No, there is no evidence in that because both hypotheses predict exactly the same thing - that she will experience an interview.

reader RAF III said...

Lubos - In a previous thread I compared the philosophers quantum conundrums to magic tricks. Reading your post and the comments here I was reminded of a story told by the logician Raymond Smullyan (in one of his popular books).
As a young man Smullyan performed a magic act for small groups, family parties, etc.. During one such performance he noticed an older gentleman who reacted to each trick, not with delight, but with an annoyed scowl. Thinking that his tricks were too elementary, Smullyan began to perform ever more elaborate and difficult feats of prestidigitation culminating in the most extraordinary illusion he could produce. When he looked to the old fellow for approval he was surprised to see him, red faced and furious, slam his fist on the arm of the chair and exclaim 'It's a trick!'.
I trust that no further elaboration is neccessary. Cheers!!!

reader scooby said...

Amazingly you have stumbled onto the topic of next week SB blog post:

reader Luboš Motl said...

Haha, funny, and it has probably happened many times.

reader Luboš Motl said...


reader ny-ktahn said...


reader Luboš Motl said...

No, DN, pretty much everything you wrote down is self-evidently wrong. It doesn't matter whether we talk about the coin on the table or its last toss. The state, heads or tails, is exactly the same.

In the sleeping beauty problem, she is getting *zero* information when she's waken up.

Your other example from the bar brings nothing new at all. It's been covered at 5 other places of this thread and most people got it.

She may quantify the credence that she's doing N-th interview or (N+1)st interview after heads, and those values of N may be equally likely.

But they are still dividing a predetermined fixed "pie" of probability, 1/2 for heads and 1/2 for tails. So if she has 2 interviews after heads and 1 interview after tails, and she observes that she's being interviewed, then the probability is 1/2 that it is the first and only after-tails interview, 1/4 that it is the first after-heads interview, and 1/4 that it is the second and last after-heads interview.

If you screw these numbers 1/2, 1/4, 1/4 and replace them by 1/3, 1/3, 1/3, it's because you are doing the very same retarded mistake as Walter Wagner who said that the probability that the LHC would destroy the world was 50% because there were two options and 100%/2 = 50%. But they are not equally likely - and the three combinations of coinstate/rank_of_interview are not equally likely, either!

reader RAF III said...

It's a trick!!

reader DN said...

This is getting a bit silly. When prompted, you even verified yourself that the "fair price" for a 100$ tails share is 33$ here:

Anyway, let me try and counter faithfully, firs with regards to the 1/2, 1/4, 1/4 argument.

It is correct that these are the odds for our current state, GIVEN that we know that we are 50% sure we are in a tails state.

And we know this once the coin toss takes place. But once specific events begin to unfold, depending on the outcome of the coin toss, and we see those events, we have to update our odds, and this premise of yours no longer holds.
(The trivial counterexample to your position has already been given, namely when we are immediately killed in case of tails.)

Think again about the situation with the neutral observer observing a game (among thousands in a big hall), under the rules of a 4-1 tails-heads interview ratio and equal duration of the interviews.

1) If he arrives at a random point in time and an interview is taking place, there's going to be tails 80% of the time at that table. And if he were to guess at the state, he would have to guess 1/5, 1/5, 1/5, 1/5, 1/5 (not being told about when the game started).

2). If he sees the initial coin toss and stays at the table and an interview eventually takes place, there's going to be tails 50% of the time.

[Here, coincidentally, the observer doesn't have to update his odds since a first interview is always going to take place and he knows it is the first interview.]

3) If the rules are "4 interviews when tails" and "1 interview half the times when heads (else no interview)", and the observer sees the coin toss and stays at the table and eventually sees an interview, then it's going to be 66% tails.

[This is an example of the observer updating his odds due to observed events.]

4) The "only" (not quite if course) way you are going to get 1/2, 1/8, 1/8, 1/8, 1/8 for the current state is by ADDITIONALLY telling the recently arrived neutral observer that there is now 50% chance for heads and 50% chance for tails at this table where he is seeing an interview. (You can think of a number of suitable chosen tables being picked from the hall of 1000 tables, and one of these being drawn.)

But exactly this information (50/50) is not available to the SB when interviewed, indeed by virtue of being interviewed by a non-neutral interviewer. As soon as she is being interviewed by the non-neutral interviewer whose behavior is influenced by the outcome of the coin toss, she will have to update her odds. The issue is NOT per se that she can't distinguish between the various possible interview states, it's that the 50/50 tails/heads guess is no longer reasonable once she has knowledge of events influenced by the coin toss.

She's in a situation similar to 1 above, and not in a situation similar to 4.

Another example:

If tails, you are immediately killed 50% of the time and you live the other 50% of the time.

If heads, you live 100% of the time.

The next day, you are offered a 100$ tails share before being told whether tails and heads happened.

How much would you pay?

If you pay more than 33$, you're going to lose money (talking in terms of expected value of course).

Lubos, it seems to me that you are making a mistake similar to the Monty Hall, where one refuses to update his odds in the face of new events unfolding (here, the SB being interviewed by a partial interviewer)?

reader Michael said...

Most "thirders" are so dense that any admitted insight on their part makes you feel excited and swell with pride. Like Feynman said: teaching is useless, except in the few cases where it is not needed.

Anyways, thanks for all of your very lucid and consistent explanations. I enjoyed them.

reader dreamfeed said...

No, this is a wrong way of thinking. If you're interested in the probability of coin flips, then they are half heads, half tails, and which day it is is irrelevant. If you are interested in the probability space consisting of the SB's observations, then a randomly chosen observation has a 1/3 chance of being monday / heads, 1/3 chance of being monday/ tails, etc. There's no space where you get 1/2, 1/4, 1/4.

reader JP said...

This whole problem seems so trivial it is hard to see what is the disagreement. Of course the coin is not biased, and an outside observer sees it coming heads and tails as often. Also SB herself sees this, if she checks the coin (assuming it is stored) e.g. the next sunday. But _at the moment of an interview_, SB's subjective probability, given the information she has (and information she has not) is clearly 2/3 for heads. This can even be seen 'frequentistically'. If the experiment is repeated 1000 times, the sleeping beaty has about 1000 heads-interviews, and 500 tails-interviews, as you acknowledged. If she does not guess at the interview time that coin came tails, she is wrong more often than she is correct. What weird definition of probability is one using if it is not a definition that helps one guess correctly in questions like this?

reader Angry Jesus said...

Monty Hall problem is similar. But this is exactly why QM is mathematical. This linguistic crap is left for the birds(and bird brained). Analytic Philosophers should be thrown out of the Academy. Analytic philosophers are no more than horse's asses ironing out brain teasers that are solved when you figure out the lazy linguistic construction that causes the debate. This is crap that would last 5 minutes at a MENSA meaning.

reader Angry Jesus said...

MENSA "meeting." Sorry, so angry.

reader John McVirgo said...

Thanks for the links. First I'll take a look at the original Adam Elga paper that argues for 1/3, over 230 citations ;)

reader JollyJoker said...

Frankly, I don't know enough about QM to understand whether there are situations in which the "thirder" view makes sense. This should be Tommaso Dorigo's specialty and I'd be pretty happy if he takes some time off his Greek island vacation to explain things.

reader JollyJoker said...

A friend asked the only really pertinent question here when I tried to explain the issue: "What is SB trying to achieve?". Depending on what she wants to do, the answer is completely obvious, even to people who disagree on the answer to the original problem. The only way to get the "thirder" opinion is to get SB to answer a question which is different from "What is the probability the coin toss was tails" which is defined as an axiom here.

“What is the probability you would assign that the coin came up tails?” is dependent on what she wants to do and she could answer anything. If she doesn't care she could answer 500 or LOL. If she wants to give the objectively correct probability she would answer 50%. If she wants to play a game where she guesses at the correct answer every time she is asked the question and wants to maximize the odds of being correct every time, she would answer 1/3. In this case she's inventing random crap that wasn't in the original question, but since the problem didn't specify any objective for her, she can answer absolutely anything and that's fine.

reader Matthew said...

"Firstly, the fact that research has not ruled out a hypothesis does not mean the hypothesis necessarily has any validity."

"One may show that a hypothesis has no validity without ruling it out?"

I stopped reading after that. If you don't have the comprehension to tell the difference between not having proof that something is true and having proof that something is false then you're not worth taking seriously.

reader Luboš Motl said...

Everyone who has read at least 2 sentences about the definition of the sleeping beauty problem knows that the sleeping beauty never makes any *observations* of the coin at all (the problem asks about the subjective probability - credence - that the coin she doesn't see is showing tails), so your comments about the "probability space consisting of observations" are completely irrelevant for this problem because there are none.

If she were observing the coin, which she doesn't, the fraction of her observations that would be "tails" could still not be interpreted as the probability that it actually *is* "tails" because the body of observations demonstrably (and, according to the rules of this problem, it is known to the sleeping beauty as well) suffers from the sampling bias (observations of heads are overrepresented). See this article

if you fucking don't know what is sampling bias. If she makes many observations and converts the fraction of "heads" observations to the probability of heads, removing the sampling bias, she will correctly conclude that the probability of heads is 1/2 and the probability of tails is 1/2 (within the statistical error), much like every person who is not a moron.

reader Luboš Motl said...

No, this is not getting a little silly. It is getting profoundly moronic.

As everyone who can read knows, the comment where I calculated a price to be $33, I also explained by detailed formulae that this price follows from - or implies, depending on what we assume - that the probability of tails is P=1/2 and the expectation value of costs and benefits match and are equal to $50.

I won't waste extra words by repeating it because imbeciles like you don't have any chance to understand the concept of probability, anyway. This was your last comment on my blog.

reader Luboš Motl said...

I agree that Dana1981's quote is idiotic if that's what you said.

What I added was the comment that one can't rule out a theory without ruling it out. If you don't agree with *this* sentence, please see your psychiatrist.

reader Luboš Motl said...

Neither subjective nor "objective" probability of "tails" is ever equal to 1/3 or 2/3 anywhere in this experiment.

reader jim z said...

If I may ask a question of all of the knowledgeable people here.

If I have a lottery ticket, where the lottery drawing is six successive draws from a pool of 50 numbers, and before the drawing my chance of having the six winning numbers is vanishing small.

If the first number drawn is the same as the first number on my ticket, my chance of winning is greater at that point in time. Ditto if the second number drawn is the same as the second number on my ticket. Etc, up to before when the last number is drawn. If my ticket has the same first five numbers as the first five numbers that were drawn, at that point I have a one-chance-in-45 of winning the big loot.

When the last number drawn is not the same number as is the last number on my lottery ticket, I have zero chance of winning the big loot.

So the odds of winning have changed, even though nothing has changed, nothing at all has actually changed, during the drawing. The "Monty Hall choice" is totally bogus. Switching doors makes no actual difference. The Everett multi-doors-world-line is just a bogus copy of a TV game show.

reader Luboš Motl said...

Sorry, I completely disagree. The probability for a coin to come up "tails" is an intrinsic property of the coin so its value surely cannot depend on "what the sleeping beauty wants to achieve".

The objective value of the probability, P=1/2, may also be studied by observations. With many observations, one may measure the probability more accurately. It is possible even if we know that the "heads" observations will be 2 times overrepresented. If that's so, it's called the sampling bias. The sleeping beauty is told about this sampling bias so she knows about it, so she must take it into account when calculating subjective probabilities, and because she's getting no new information when she's woken up, she keeps the prior probability P=1/2 for the tails.

If she is told the day of the week once she is woken up, then "Tuesday" makes her 100% certain it was "heads". When she's told "Monday", then she gets nontrivial information about heads/tails, too. She adjusts the prior 1/2 - 1/2 probabilities to 2/3 tails, 1/3 heads, because among the probabilities for "coin state plus day of the week" arrangements, Monday tails has 2/3 of thepiece and Monday heads has 1/3 of the piece.

reader jim z said...


What I meant is that nothing about the state of me, or the state of my ticket changes during the drawing. Before the drawing the numbers have a probability to be winners, and after the drawing they have no probability of being winners, but they are the same numbers, before and after.

reader Luboš Motl said...

Dear Jim, right, the probability that you have a winning ticket is changing with time. If you have a promising ticket, it may go up for a while, but it may still drop to zero when you see that the last number is wrong.

It isn't really true in the lottery case that "nothing is changing during the drawing". The value of the ticket surely *is* changing. If you had 5 correct numbers, then just before the 6th number is drawn, I would surely buy that ticket for a much higher price than what you paid when it was just an average ticket. This changing price reflects the changing degree of agreement between the "ticket's numbers" and "winning numbers". This degree of agreement is changing, in this case, because what the "winning numbers" are is continuously changing when the numbers are being drawn one by one.

reader jim z said...

Numbers with the same probability of winning any future lottery drawing, but with no possibility of winning any past lotteries.

reader jim z said...

For the "monty hall" problem:

I have a 1/3 chance of picking the winning door, at the start.

Monty Hall has a 1/1 (100%) chance of showing me that one of the two door that I didn't pick is a looser. That changes nothing. Any door that I pick, he always shows me a looser.

No matter what, after he shows me a loosing door, I have a 1/2 (50%) winning chance. That only means that he always can throw away one looser. Always.
Every choice goes from 1/3 to 1/2. That's the show.

He throws away useless knowledge. But it is only useless knowledge to him, because he knows which door the prize is actually behind.


For the Sleeping Beauty 'problem', I agree exactly with the unchanging 1/2 coin flip outcome. What else could it be?

But on the 'monty hall' problem, it can't be that you win more when you change your choice. No?

reader jim z said...


I understand the changing value of knowledge over time, in the lottery ticket. I was only saying that nothing intrinsic to the lottery ticket changed the outcome during the process.

The changing probabilities of a 'prediction' of an outcome don't change the probabilities that the outcome would have occured, or will occur, from start to finish. No?

reader jim z said...

Never mind about the Monty Hall problem, I figured it out.

reader yiannis said...

No. You are not right. In the beginning you have 1/3 chance of winning, while you also have 2/3 chance of loosing. When he opens one door, you have to switch your choice, cause you increase (double) your chance. You don´t have 1/3 chance at the end but 2/3.

reader John Archer said...

Jim Z,

"The "Monty Hall choice" is totally bogus. Switching doors makes no actual difference."

Take an extreme case but otherwise keep the rules the same:

There are a million doors. You pick one then Monty opens 999,998 others.

Do you switch? :)

reader KP said...

" If asked what the odds are among four tosses that two is heads it is 1/2."
Actually I think the chances are 6/16.

reader Matthew said...

He didn't say anything about ruling something out without having ruled something out. He said it doesn't need to be ruled out without good evidence. A subtlety that your misparaphrasing of what he said indicates you were unable to understand.

On top of that, he claims in a response post to this one* that when he tried to correct your errors you deleted his posts and blocked him. Are you truly so afraid of having your ideas challenged?