The new ATLAS preprint with this potentially interesting finding is called

Search for supersymmetry in events containing a same-flavour opposite-sign dilepton pair, jets, and large missing transverse momentum in \(\sqrt{s}=8\TeV\) \(pp\)-collisions with the ATLAS detectorSo look at it: we want final states that contain \(\ell^+\ell^-\), a lepton pair – it's the leptons that apparently follow from a decay of the Z-bosons that will matter; jets; and large missing transverse momentum (assumed to be composed of the lightest superpartners such as the lightest neutralinos – possibly particles of dark matter). The Standard Model didn't do too well.

One shouldn't overstate the deviations from the Standard Model. Most of the channels are OK. For example, a particular channel in the ATLAS data refused to confirm a recently announced CMS' 2.6-sigma excess – although the kinematic constraints were slightly different in the two experiments.

However, while rejecting this CMS excess, ATLAS saw an even larger one. It was in the events in which the lepton pair looked like one from a Z-boson of the correct mass (no neutrinos are produced in that). The key table is table 7 which describes SR-Z (the signal region with a Z-boson):\[

\begin{array}{|c|c|c|c|}

\hline {\rm Channel} & e^+e^- & \mu^+\mu^- & \ell^+\ell^- combo\\

\hline {\rm expect} &4.2\pm 1.6& 6.4\pm 2.2 & 10.6\pm 3.2\\

\hline {\rm observed} & 16&13&29 \\

\hline

\end{array}

\] The excess is mostly in the electron pair channel, 3.0 sigma again, but even in the dimuon channel, there is some excess, 1.7 sigma. When combined, the dilepton channels are back to 3 sigma. One expects \(10\pm 3\) and gets \(29\) events, not bad! I vaguely remember some other recent excess that – surprisingly or disappointingly – appeared mostly in the electron channel only but I can't figure out where it was. Do you know where we saw it?

As far as I understand, CMS hasn't released its own data on the on-Z signal region where ATLAS is seeing the excess yet.

If you want to believe that this excess is from new physics, it may be a gluino or a squark that is decaying. One exclusion plot – Figure 13b – weakly suggests that the exclusion curve is being repelled from a point with a \(1\TeV\) gluino and a \(600\GeV\) lightest neutralino which may be "the truth" for those who want to be real optimists.

(On Monday, March 16th, this paper by Barenboim et al. will appear that will make pretty much the same conclusion as I did 4 days earlier. A SUSY scenario of "general gauge mediation" is the simplest thing that is needed. The gravitino is light and it's the LSP. Fundamental higher-energy scales are at hundreds of TeV. Only this paragraph was added after March 12th.)

Your humble correspondent doesn't really understand whether they assume these strongly interacting superpartners to be pair-produced and if they do, why the other member of the pair doesn't destroy the identity of the final state. But maybe it's giving the jets and the missing transverse energy is only required to "be there" and its magnitude isn't important. Can you help me?

**New Higgs CMS excess**

CMS will release a paper tomorrow looking for heavier friends of the now familiar \(125\GeV\) Higgs boson. There is a 2.56-2.64 sigma excess in the mass range \(700\)-\(800\GeV\). The newer papers released recently seem to have many more excesses than 1+ year ago – which supports the conjecture that whenever they see something "not quite mundane", they (both ATLAS and CMS) are delaying the publication, an approach that may be reasonable but also a potentially dangerous bias we should be aware of (and force us to deduce lessons from "comparative literature" – like the statistical evaluation "how many excesses have been seen anywhere" – much more carefully than otherwise).

Incidentally, a few hours ago, there was a CERN webcast kickstarting the new season, the 2015 run. You may watch the recorded press conference here. Be ready for lots of French and German accents. I believe that the content isn't important enough to be analyzed.

Assume that there is betting and if she guess correctly she wins some money. If you are using a betting argument, the total stake when coin lands heads or tails should be same. This way you one easily sees that the correct answer should be 1/2. If you bet same money each time she awakens, in effect you bet more money when it lands tails. You should bet 1/3 $ each time she awakens when coin lands tails, if you bet 1 $ when she awakens when it lands heads.

ReplyDeleteHi again Lubos,

ReplyDeletefine post, with such nonsense out there I am starting to understand why you need to be so "tough".

I don't understand how some people who even are hired scientists cannot get that you are right on this. Maybe this explanation will help.

Imagine a hundred sleeping beauties, and a hundred beds. I paint half of them red, and leave half of them white. The hundred beauties are now put to sleep, distributed on the beds at random, and the ones on the white beds are woken up once and asked which color bed they think they lie on, while the ones on the red beds are woken up twice (or a million or a billion times, it doesn't matter for this part of the question) and asked about the probability of the color of their bed. Because there are 50 beds of each kind it is pretty clear that every beauty will say that there are a 50 percent chance they are lying on a red bed, and 50 percent that they are lying on a white bed.

If you regard probability as a subjective evaluation of the probability of certainty, then the withholding of information is a condition that must be taken into account, not a cheat. The correct probability under this definition is then 1/3. And whining about the cheating is like whining about what things in QM are really like, instead of just accepting the probabilities of the observables are all we can know.

ReplyDeleteOf course, by a frequentist interpretation, you are correct. Modern mathematics is apparently based on rigorous proofs but the conclusions still seem to depend upon what one chooses to assume. This has left me unsure about what kind of knowledge mathematics is.

PS There is a typo in Bayes' theorem above I think. Also, it is definitely wrong to imply there's anything Marxist about DeLong, who is a committed anti-Marxist.

This sleeping beauty thing also reminds me of the Monty Hall problem. 3 doors, behind two of them you find a sheep and behind the last you find a prize. You are asked to pick one at the beginning, then the game host will open another door revealing a sheep (he always chooses a door that reveals a sheep) and the participant can now keep the first door he chose or switch to the other option. What should he do? Assume the prize was really behind the first door, and play out the three different scenarios where he chooses number one, number two or number three at first. So if he chooses to keep his first choice he only wins when his first choice was door number 1, that is 1/3 of the time. If however he chose door 2 or 3 at first and switches (in which case door 3 and 2 respectively is revealed with a sheep behind) he wins 2/3 times, so therefore it is more sensible to switch.

ReplyDeleteAnother way to make it easier to see is to choose a door among 1 million, and then the game host opens all the other doors (always revealing sheep, he knows where the prize is) and now it is obvious that it is better to switch.

In the tv-game deal or no deal, the participant is asked to choose a suitcase among 20 containing different money amounts from almost nothing up to a million. After this first choice he is asked to gradually choose one at a time from the remaining ones revealing their content. Say he continues the game until he only has his first choice and one remaining suitcase among the remaining ones, and the amounts one dollar and one million dollars are left. Should he switch?

In this case it doesn't matter because he chose all the suitcases himself and choosing which suitcases to eliminate is no different from choosing the ones you want to keep(as he did with the first one). So it makes no difference. However for unknown psychological reasons I find it much cooler to keep the first choice ;-).

Yes. I think the 1/3-er crowd find this confusing because they sense intuitively (and correctly) that if this was a game where the win condition was to get >50% correct guesses over repeated runs of the experiment, then obviously a winning strategy for beauty is to predetermine to always guess tails.

ReplyDeleteBut this is due simply to the fact that a correct guess of tails is effectively rewarded with "2 points", whereas a correct guess of heads is only awarded "1 point". We can imagine an isomorphic game without the distraction of the repeated awakenings + memory erasure where you repeatedly toss a coin and I simply guess tails all the time and collect my 2/3 (points-weighted) overall score.

Obviously on a single run of the experiment with only one coin toss, whether she consistently guesses heads or tails beauty has a 50% chance to win and there is no winning strategy, and over repeated large-N runs half the coin tosses will have come up heads and the other half tails.

The one thirders seem almost to imagine that were beauty to be woken 99 times in the case of tails, that this would somehow retrocausally bias the actual coin tosses to be 99% tails!

Exactly, Michael, you are using a similar emphasis as Bob Walters, see the Markov text

ReplyDeletehttp://motls.blogspot.com/2014/08/sleeping-beauty-thirders-rudimentary.html?m=1

The key in your analogous description of the situation is that the total number of beds - the denominator used for the probabilities - is fixed to constant at the beginning.

Only with this choice, one may interpret the fractions as probabilities. In effect, this is equivalent to saying that we are computing - and we must be computing - probabilities of evolution to a certain state during the week but *from* the known initial state.

The possibilities are mutually exclusive only if the "tree of evolution" branches. But if one follows a single branch, like beauty with tails on Monday and Tuesday, those really share the same number of "beds" to start with, the same probability, and if one wants to decide about the probability of coin_state+day, the probabilities are "pouring from one day to another" but just in between the points on the same branch, not to other branches, as long as something on the branch is alive.

Dear Steven, the bets that the bookmaker offers on Monday and on Tuesday are *different* bets with *different* odds for heads and tails (50-50 vs 0-100), and if he pretends that they're the same while he can influence whether the client is woken up on one day or another, he may manipulate the betting to his advantage.

ReplyDeleteIf she is offered these bets in the "thirder" way, she can of course figure out that she should demand 3 times the price of the lottery ticket by which she bets on "heads". But this sentence is not equivalent to the statement that the probability of "heads" at the point when she is woken up is 1/3 because the probabilities are only linked to the corresponding ratios from the betting if the bookmaker can't manipulate with the asymmetric knowledge.

I don't see how there's any room for confusion about this problem except that people *think* they have a precise understanding of some terms but actually don't. There's no need to dodge the bookmaker scenario. Simply describe in advance what betting tickets will be offered (a neutral constable will enforce this contract). If we're making odds for a betting ticket to be offered either once in event of H or twice for T, this should lead naturally to Thirder indifferent-odds (H ticket paying off 2:1). It's up to you if you choose to call that contractually-agreed payoff ratio the 'proper subjective credence'. I do like your argument that even if you're going to be woken (w/ memory erasure) thousands of times if you're in Guantanamo you shouldn't be subjectively near-certain that you're being tortured in Guantanamo :)

ReplyDeleteDo you agree, Michael, that the same logic that gives 1/2 for the sleeping beauty is the logic that gives 2/3 for the Monty Hall problem? This numerical difference may be confusing - because the results are kind of reverted in the two problems.

ReplyDeleteThanks for the answer Lubos.

ReplyDeleteI will look at the other article :-)

The same logic, hmm, I agree that the answer is 1/2 for the sleeping beauty as I also mentioned in my comment below about the colored beds. I am the same Michael as him. But maybe you know that and mean something else?

ReplyDeleteThere were a few more but the Markov one was a bit more focused on formalism - and based on a different man who carefully studied it - so it may be more interesting than to read my other, similar texts.

ReplyDeleteApologies, I don't understand why you seem anxious here.

ReplyDeleteDid I count Michael at one moment and Michael at another moment as two different people?

That would be the fallacy that the "thirders" are doing to the sleeping beauty - they effectively pretend that the tails-sleeping beauty when woken up on Tuesday is a different lady than the tails madam who woke up a day earlier.

Exactly - to interpret the fractions of winning tickets as probabilities, one has to reward each successful guess with 1 point, while the correct "tails guess" is rewarded by 2 points in the counting of the thirders.

ReplyDeleteA way to fix it is to guarantee that e.g. when it's tails, she is sold a half-price ticket on Monday and half-price ticket on Tuesday so that in total, she buys 1 full ticket in the case of tails just like she does in the case of heads.

With this fix, she may assume that the probabilities for heads and tails are the correct 50% and she will get her money back if the winning tickets pay 2 times what she bet. In effect, the only deviation from the simple head-tails bet is that in the tails week, the same ticket is divided to two halves.

Your 99-day extremization of the wrong calculation ad absurdum is almost equivalent to my previous blog post about the sleeping beauty in Guantanamo Bay:

http://motls.blogspot.com/2014/08/sleeping-beauty-in-guantanamo-bay.html?m=1

Okay, sorry.

ReplyDeleteThe probabilities change when the beauty is given some new information about the day, as you explained, and the probabilities also gets updated in a similar way in the Monty Hall problem when the participant learns something new.

I agree with Luboš that this is a good different example. The problem I have is easily illustrated with it:

ReplyDeleteThe sleeping beauties all know that they will not be woken in 1/4 of the cases. So only 75 beauties are effectively asked whether they are among the 50 on one side or the twenty five on the other side of bed. far as I can tell the beauties are given the information that they are making a selection among three equally likely outcomes because they know 25 of 100 beauties are never asked.

Hey awesome twist and a much more dramatic and clear explanation! :-)

ReplyDeleteThis kind of confusion is similar to the kinds that arise with anthropic reasoning about multiverse "populations" and "democratically selected" typical-observers etc, though I'll grant that it's harder to keep a clear head when infinite measures are involved.

In finite cases like with sleeping beauty, I can't see that there's much excuse for a "professional" getting it wrong though!

Here's an example of an infinite case of a probability measure "paradox" (countably infinite prisoners with black and white hats) which is fun:

https://cornellmath.wordpress.com/2007/09/13/the-axiom-of-choice-is-wrong/

*Really* stupid blog-post title, sorry, (well it's not my blog!). But there are a couple of great posts from Terry Tao in the comments section about the breakdown of probability additivity assumptions when infinite (and especially non-measurable) sets are in play.

Don't show it to DeLong and the 1/3-ers though, it's liable to blow their minds/cause smoke to come out their ears... ;-)

You only update the probabilities when you get new information. Imagine that *you* are lying in the bed, remember nothing about if you already have been woken up or not, and you know that half of the beds are red and half white. You will answer 50 percent chance for the color of your bed without additional information.

ReplyDeleteBut if you are told the day, like Monday, its different. First calculate the probabilities for this case to happen in the two cases. For one color of the bed you get 50 percent for the color multiplied by the probability that it is this day in the subspace you are in. What do you get? You get (as Lubos said) 0,5(the color red)*(the day, monday for this case)1 , and the probability for the other event is 0,5(the color white)*(the day, monday)0,5. So 50 percent and 25 percent. Now you know that it is Monday so you must "renormalize" to find the probability for the color. So 50 percent out of 50+25 percent is 2/3, which is your probability for the red bed when told Monday.

Luboš, I posted a comment just under you giving him a basic course in Bayesian inference. He deleted my comment then sent me an email saying "I don't think you read the problem correctly..." I sure did.

ReplyDeleteLOL, he also has an ego. I posted a comment mentioning that this new blog entry exists and it's still there. Some avoidance of anything controversial - if I avoid the term sycophancy - may help. ;-)

ReplyDeleteThat was exactly what my reply to him was about. He said it's unsupported so I used a matrix and then just the plain old Bayes Rule. He merely said I don't understand the problem.

ReplyDeleteTurning a Bayesian question into a game theoretic one might be an interesting exercise in AI, but it's not what it is now.

Oops I claimed this to be so simple and now I have succesfully confused myself. The multiplication of the probabilities and the 50 percent for Monday or Tuesday in the case where she is woken up both days, that are multiplied on, are treated as exclusive "events" in the above comment, either one happens or the other. But they don't exclude one another, both Monday and Tuesday will happen as time flows. So now I think telling her Monday won't change anything, because Tuesday wasn't an option yet, it doesn't "live" with Monday. Therefore Monday and Tuesday cannot be clumped together, so the probability remains 50-50 after being told Monday. Being told Tuesday will of course still complete reveal the color of the bed. Sorry, that I either am confused now or was confused a moment ago. How embarrassing.

ReplyDeleteThe sleeping beauty knows she is being asked. The information that she is awake and asked tells her she is not in the 1/4 of not awake beauties who are not asked. Among the 75 beauties who may be asked these are evenly the other three outcomes of day and flips. The outcome resulting in the question being asked eliminates the fourth outcome.

ReplyDeleteNice :-)!

ReplyDeleteA lot of naysayers will regret their naysaying if SSM is measured at the upgraded LHC.

ReplyDeleteAs I tried to write a reply to Rehbock, and repeating the calculation that the probability change when she is told its Monday, made me think that there is a mistake in the conditional probabilities concerning learning about it being Monday. The reason is that Tuesday and Monday don't "live together as possibilities at the same moment". If she was thrown into a Monday box always in the case of tails and a Tuesday box half the time and a Monday box the other half if it was heads, the calculation of the 2/3 for tails upon learning about it being Monday would be true. But because Monday and Tuesday are separated in time and Monday always happens and she is always woken up on Mondays I don't think telling her its Monday changes anything, so even this additional information keeps the probability for heads or tails as 50-50. Do you agree that the separation in time makes this difference?

ReplyDeleteSpeaking of three-sigmas (I usually use a max. of two for AMS dating, hopefully aided by Bayesian if they come-out in order in stratigraphic sequence), how about this for a three-Sigma outlier event, a recent political suicide (there was also a 'defenestration 10 days ago), by an ex-Yank ally:

ReplyDeletehttp://www.bbc.com/news/world-europe-31855700

I'm sure Jeffrey Epstein or Rajendra Pachauri would be willing to do the experiments if someone can supply the babes.

ReplyDeleteDeLong is not a big admirer of Marx's. He makes only three positive comments on Marx's economic work. First, he approves Marx's recognition there is a business cycle, not just inexplicable crises. Big but, though, he believes the business cycle has been corrected (yes, really, events of the last decade are just a mistake.) so this is purely of historical interest. Second, he approves Marx's advocacy of industrial civilization, which strikes me as approving someone for liking Mom and loving apple pie, whereas disapproving seems at best churlish. And the third he approves is some economic history. For economists, economic history is not even "of historical interest," so this is not high praise. The rest is negative.

ReplyDeleteAs for Sleeping Beauty? Again, the probabilities can be viewed as measures of the subjective certainty of outcome. As such, she can meaningfully calculate probabilities, that is, how certain her knowledge of possible outcomes is, even if she doesn't know the day. If she wants to know the magnitude of her uncertainty, she has to calculate. What she can't do, which I think is what you're saying, is calculate the real frequency from the information she has. The thing is, from a subjective probability standpoint, frequency is not probability, and "real" is moot.

I am sure that the whole thing lies in the way one understands the question. This is how I understand it: she is woken and asked to bet her money on whether it was tail or heads at the beginning of the particular week (or the previous Friday or whatever). She knows she is equally likely to have been woken heads/mo, tails/mo and tails/tu. If she always bets against heads, she will be winning 2/3 of the time. In other words, in tails weeks, she is given the chance to win twice, while in head weeks she only loses once... Am I wrong in the interpretation of the question?

ReplyDeleteI think that all the confusion is caused by the fact that the question for the sleeping beauty is a bit ambiguous. Probability of exactly what are we asking about?

ReplyDelete1) The whole experiment is the coin toss and nothing else. All the sleeping and waking stuff after that is irrelevant. The probability is 1/2.

2) The experiment consists of the coin toss PLUS some follow-up actions that depend on the toss outcome. It all ends with a "measurement" of the result. The scheme is such that only 1/3 of measurements will come out as heads, the remaining 2/3 will be tails. If you ask "what is the probability that any specific measurement will give a heads result", then the answer is 1/3. You're asking not just about the "coin+measure" system, but about the "coin+sleep/wake+measure" system.

It's telling that both authors formulate the question differently:

L: calculate the probability that the coin landed "heads" on Sunday.

B: what is your confidence that the coin flip winds up heads?

The first formulation is much more precise than the sloppy second one.

Sorry I was reading this "ℓ+ℓ−, a lepton pair – " and try to figure out what the hell you meant when I was distracted by an ad for womens underwear. Then I came to this page where I was distracted by an ad for womens underwear. For some reason I find woman in underwear distracting. If you want me to focus on physics you might want to have fewer ads.

ReplyDeleteDeLong is a ginormous asshole. His primary intention to discredit Lewis is obvious. He used you as a pawn expecting that quoting your usual non-PC style and global warming opinions will win him an argument with little effort on his side, since his entire post is otherwise an incoherent babble and confused self-gloating.

ReplyDeleteThat this elementary sophist strategy worked is clear from some responses.

Dear Jaroslave, accuracy is great and important but I don't see in what interpretation your "L" sentence and your "B" sentence could possibly differ, and I don't even see why you claim that one of them is formulated more accurately than the other. And I don't see what is the exercise that gave you the result of 1/3.

ReplyDeleteAs I see it, all your formulations of the problem are equivalent and accurate enough to be unambiguous. It is some of your reasoning that is totally sloppy and wrong!

Well, his is just another hard left blog so it attracts lots of people who are living with the left-wing ideology such as global warming.

ReplyDeleteBut at the end, I think that these blog posts have nothing to do with the climate or politics and I don't really care what those chronic politically defective commenters are saying - I had to care when I was living in an environment polluted by them from the floor to the roof but now I can freely despise and ignore these scumbags (and there are many over there, indeed!).

At the end, it's primarily blog posts about the statistical reasoning and it just seems that my explanations are totally getting through and dominate even among DeLong's readers despite his vigorous attempt to suggest that the correct solution is so stupid that it must only be an April Fools Day prank.

The whole thing lies in accepting the strict definition of probability. That is:

ReplyDelete- the number of events that realize a particular outcome (you winning, for example) divided by the number of all possible, *exclusive* events.

However, you waking up on Tuesday and winning (by betting on tails) is not an exclusive event w.r.t. the other waking events. It is clearly dependent on the event of your waking up also on Monday.

The tail event is what caused these two *different* but obviously dependent waking events to occur.

That is, the only thing that was really exclusive was that it was a tail since it obviously can't be both tail and head at the same time.

You won but you didn't win twice in the sense of two independent events. It was 100% sure that you would win on Tuesday once you won on Monday. That is really one event in the sense of the definition, even if physically it was two events. The definition requires to measure the events in the same units.

Otherwise, you could toss a coin once, then take several pictures of the result and claim that the number of your wins is equal to the number of pictures you have taken of the coin in its final state. Well, once the first picture shows the win, clearly all the subsequent pictures will show a win too. It still is a single win.

Yeah, but I still somehow think that the creators of the puzzle had in mind the subjective outlook of the Beauty: "given you were woken up, what coin toss would you bet your money on?" The Beauty needs not know more about the world than "given I was woken up, it is twice more likely for the toss to have been tails"... and why should she care that she might be double counting one particular toss? She is not asked about the chance of heads in one unbiased toss, she is I believe asked a different question. She is asked about her personal perspective of the situation.

ReplyDeleteI think the whole confusion is in language and meaning of the question. This is quote from BDL page: "When awakened and asked "what is your confidence that the coin flip winds up heads?", what answer should he give? If he is then told "it is Monday; what is your confidence that the coin flip winds up heads?", what answer should he give?" I believe that the question means "what answer should he give to be right as often as possible". This of course leads to the tails answer.

ReplyDeleteNote that the word "probability" is nowhere to be seen in the question.

ReplyDeleteDear Mažňáku, "credence", "confidence", and "[subjective] probability" are synonymous, so your assertion that the word "probability" may be true in some literal sense, but it is false morally.

ReplyDeleteI don't see what the change in terminology really changes in essence. If I am confident about something that means I attribute it a high probability. One can also say I consider that event more likely.

ReplyDeleteHowever, by redefining what it means to win, or what it means for a certain event to occur in relation to a bunch of other, somehow related events that may also occur, one can surely arrive at many different ways to 'calculate the probability'. I just wouldn't really call it a calculus of probability.

LOL, Tony, did you write your comment independently of my reply to the same maznak's comment below? Because we wrote almost the same thing. ;-)

ReplyDeleteYes, completely independently. My mobile browser was showing note that there is one new comment as I was about to press send.

ReplyDeleteGood - your comment had some extra content. But good to see this synergy. ;-)

ReplyDeleteBeing careful and distinguish similar words is great and often important - and people often err.

On the other hand, this objection is also often used in a wrong way when people insist on distinguishing things, concepts, ideas, and words that are actually the same. In those cases, their focus on accuracy and exact terminology is spurious and only increases chaos.

Many of them will probably regret silently, pretending that they have never naysayed anything. I sincerely hope that it won't be the case of Adam Falkowski who will owe me $10,000 at that point! ;-) My $100 for him in the case that SUSY isn't seen soon (when 30/fb of luminosity is surpassed and all data analyzed) is ready.

ReplyDeleteDear Tom, she was provided by Google to make the point sharper. The opposites attract and the two leptons have the opposite charge, like you and her.

ReplyDeleteOk, so you say that if the question was interpreted as "what should she say to be right as often as possible", it would be a wrong interpretation? Somehow it still seems to me to be the most natural one...

ReplyDeleteTony, I am not jealous about the question you got. What to do with it?

ReplyDeleteThe question is "what is your confidence that the coin landed heads?" Why should this totally clear sentence be "interpreted" as the incomprehensible sentence "what should she say to be right as often as possible"?

This "interpretation", Maznak, is much more confusing than anything else that was said before. You may think that you are making things better or more accurate but you are doing exactly the opposite.

The word "often" in your version of the question is also enough to see that you completely misunderstand the probability calculus. We are talking about the *confidence*, i.e. the subjective probability, and with this interpretation, the situation is really occurring only *once* in a lifetime. So the word "often" is nonsense. It is *you* who is completely distorting the discussion and making it chaotic and misleading.

So are you saying that it makes a difference whether it is a one-off situation or a repeated experiment? Not trying to be difficult, just trying to learn something, honestly.

ReplyDeleteTom, the ads here are tailored to each viewer differently based on their Google profile. Thanks for letting us know you better :)

ReplyDeleteI got one for home insurance and one for car insurance. Shows what an interesting life I lead.

ReplyDeleteYou mean thirders are wrong that she is different in the information she has? She is different because she knows she is awake and is therefore to choose among the three tosses that result in her being awake.

ReplyDeleteThe exercise giving me a 1/3 result is for example this one:

ReplyDeleteEvery day I knock on your door and give you a sealed envelope. Inside is a piece of paper that says either HEADS or TAILS. What is the probability that the content is HEADS, assuming the following rules?

1) On Day 0: I toss a coin and write the result into the envelope.

2) On Day n + 1: If I tossed a coin on Day n and the result was TAILS, then I perform no toss today and write TAILS. Otherwise, I do a new toss, like on Day 0, and write the result.

3) You never remember any of the previous results.

The ratio of HEADS:TAILS in this sequence is 1:2, so the probability of HEADS on any day n > 0 is 1/3, isn't it? Can this 1/3 be interpreted as a probability, even though the values in the sequence are not independent? I don't see clearly why not.

Of course, this exercise is not exactly the original sleeping beauty problem, but rather an "iterated sleeping beauty", where I repeat the experiment over and over and wake her up once or twice every week.

Right, 1/3 of the words in a long sequence of yours will be "heads" because your rule may be simplified: toss contains essentially once a day but if you get "tails", write the result twice.

ReplyDeleteIf one watches the paper on a "random" day, then the fraction 1/3 may be interpreted as the probability of "heads".

In the case of the sleeping beauty, which is a different problem, the whole tension about the right result is about the counterpart of the concept of the "random day".

Among "random days", the halfers say that 1/2 are the days when the coin shows "heads" and 1/2 are the days when it shows "tails", so that "tails days" have to be included with 1/2 weights of the heads' weights, while the "thirders" naively insert all the awakening days from different histories with the same weight, and that's why they end with the wrong result 1/3 for the probability.

This mistake is equivalent to saying that the possibility that one gets "tails" occurs with a doubled probability just because a woman is awaken twice during the possibility. But it doesn't. The probability doesn't get modified as a function of awakenings of any woman - or any man, either.

Exactly, Maznak. One may learn quite something. People with similar complaints like Tom should realize that it is not the blog owner or the readers who are choosing the ads for him. It's a rather sophisticated automatic program.

ReplyDeleteDear Maznaku, you may repeat the experiment and if everything will be the same, the subjective probability will always be the same.

ReplyDeleteBut the subjective probability will *not* be equal to some frequentist ratio of cases that go one way or another if that ratio is calculated sloppily.

To calculate the ratio properly needs to assign the correct weights to different scenarios when the experiment is repeated.

So to get the right probabilities from the frequentist ratios, the weeks following the "heads" toss must be repeated twice, so that the number of awakenings remains the same in each week. In other words, the "tails" weeks have to be given the weight 1/2 times the weight of the "heads" - exactly to cancel the fact that she is awaken twice on that week.

When you do repeat these experiments with these right weights, you will get the correct probability of heads after awakening, P=1/2, even from the ratio of awakenings that show heads after many weeks.

If you get P=1/3, it's because you used wrong weights.

Not sure whether I understand your comment, Rehbock - almost certainly No.

ReplyDeleteI am saying something very simple. They treat the three combinations - heads-Monday, tails-Monday, tails-Tuesday - as if they were three equally likely places in space and if the woman were choosing at which of these places she is located.

But this Z_3-like symmetry doesn't exist. In reality, the tails-Monday and tails-Tuesday awakenings describe the same woman in the same history - in the same "parallel world", if someone loves to think about the many worlds interpretation of probability both in classical and quantum mechanics - so they must always be viewed as following from the same probability associated with the past events (that decided that this history takes place), and if one also wants to decide "when" she woke up, the different times when it could have happened (Monday or Tuesday) must divide this fixed probability (the number of beds) - they can't "steal" it from completely different histories.

This has been said so many times in so many creatively different ways (like the beds) that I don't believe that it makes any sense to continue this discussion. Who hasn't gotten it by now will probably never get it.

All I am trying to understand flaw in this: - sleeping beauty is awake in only three of four coin outcomes. She knows and we know that she is awake because she answers question. So the possible outcomes drops to three when she is asked omitting the coin day combination in which she is not awake.

ReplyDeleteRight, there are 3 coin_state-day arrangements, but they are *not* equally likely. WTF?

ReplyDeleteI thought maybe generalizing the problem might help explain it. Consider the 8 mutually exclusive events:

ReplyDeleteE000: tails, sleep monday, sleep tuesday,

E001: tails, sleep monday, awake tuesday,

E010: tails, awake monday, sleep tuesday,

E011: tails, awake monday, awake tuesday,

E100: heads, sleep monday, sleep tuesday,

E101: heads, sleep monday, awake tuesday,

E110: heads, awake monday, sleep tuesday,

E111: heads, awake monday, awake tuesday,

with the sum of the corresponding probabilities equal to 1. Given that an event includes at least one awakening, what is the probability that it is a heads event? Clearly, it's

P=(pE101+pE110+pE111)/(1-pE000-pE100)

where the p-prefix means probability. Now suppose we know a priori that pE000=pE100=0. If we also know the coin is fair, it follows that pE101+pE110+pE111=1/2=P.

The conclusion is reached even without knowing exactly what the probabilities of six of the events are.

Here is my last try. I am sorry that I am unable to explain my point. Perhaps I am misunderstanding what we can know. Here was why and what I thought they are equally likely:

ReplyDeleteThe coin has two choices and the question(s) may be on either day. That means only four combinations of day and coin HM TM and HT TT are equally possible. Equally possible it is M or T and equally H or T. Until she is awake she knows only that it is one of those four equal possibilities two for each side. When she is awake she knows it is not the one of four combinations in which she is left asleep. That new information allows her to choose among the remaining three equally likely outcomes of the combination of coin and day when answering.

I am not trying to be obstinate or otherwise perturb you. I still miss why we don't have three equal possibilities based on her learning she is awake.

Sorry Lubos, I can't resist either ;)

ReplyDeleteWhat should she say to be right as often as possible?

Well, the best thing then is to say what would make her 100% correct each and every time! For example, she can become a lover with one of the guys who are waking her up. He would secretly signal her the actual result of the coin toss and she would say the right thing.

After 10 coin tosses which happen to be, say exactly 5 heads and 5 tails, she would be right 15 times, a whooping 150%!

I'm joking of course, just to demonstrate how ridiculous conclusions can be reached if one mixes and/or shifts the interpretations, definitions and conditions.

If one has to determine the probabilities of the heads-Monday, tails-Monday, and tails-Tuesday combinations (possible arrangements of the state of the coin and the day when she is woken up), the two tails possibilities have to share the 50%, so if Monday and Tuesday are equally likely for tails, the most sensible arrangement is 25% for tails-Monday and 25% for tails-Tuesday.

ReplyDeleteThe probabilities are 50-25-25 for the three arrangements which also allows you to say that the probability is 50/(50+25) = 2/3 that it's "heads" if she's told that it's Monday, 25/(50+25) = 1/3 for "tails" if she's told it's Monday

This is wrong. If she's told it's Monday, the probabilities are (still) 50/50. And the 50-25-25 "arrangement" doesn't correspond to anything meaningful.

It may be counterintuitive or subtle but if she is told it's Monday, the probability of heads jumps to 2/3 because in that case, she *does* learn new information she didn't know on Sunday.

ReplyDeleteIt was possible that the wakeup was on Tuesday, which would mean that the coin landed tails, but because she learned that it's not Tuesday, tails became less likely, and their probability dropped to 1/3.

At any rate, be aware that both Elga and Lewis agree with me that the probability jumps by 1/6 if she learns it's Monday, search e.g. for E6 and L7 in Lewis' paper

http://www.joelvelasco.net/teaching/3865/lewis%20-%20sleeping%20beauty%20reply%20to%20elga.pdf

Did you see my comment where I updated my stance on this? I am now also claiming that the probability remains 50-50 upon learning Monday (and of course 0-100 on being told Tuesday). The reason being that Monday and Tuesday do not exist as potential options at the same moment, (and this separation in time is the crucial thing) so being told Monday eliminates Tuesday from the thinking. One is now just lying in a bed knowing that 50 are red and 50 white, and also knowing that it is Monday. What were to be done to one in the future doesn't matter because its known where one is in time. My other comment was this:

ReplyDeleteAs I tried to write a reply to Rehbock, and repeating the calculation that the probability change when she is told its Monday, made me think that there is a mistake in the conditional probabilities concerning learning about it being Monday. The reason is that Tuesday and Monday don't "live together as possibilities at the same moment". If she was thrown into a Monday box always in the case of tails and a Tuesday box half the time and a Monday box the other half if it was heads, the calculation of the 2/3 for tails upon learning about it being Monday would be true. But because Monday and Tuesday are separated in time and Monday always happens and she is always woken up on Mondays I don't think telling her its Monday changes anything, so even this additional information keeps the probability for heads or tails as 50-50. Do you agree that the separation in time makes this difference?

Drawn:

---------White bed------Monday-------Tuesday

/

---------Red bed----Monday

50-50 for learning monday.

vs:

-----------Monday

--------White bed

-----------Tuesday

------Red bed ------------Monday

2/3-1/3 for learning Monday

waking Tuesday is never possible on a Monday, and being told Monday makes the place in time sharp and destroys the clumping together of days in the probabilities.

My last drawing was deformed a bit by posting, sorry, it was better before

ReplyDeleteThere *must* be a way to divide 100% to these possibilities for her, right?

ReplyDeleteI don't think so. I mean you can do it the way that the thirders do, which is to sample across the space of all awakenings, but then you get their answer, which has nothing to do with the likelihood that the coin is heads. But otherwise the whole idea of trying to assign consistent probabilities to each scenario is a trap as far as I can tell.

Lubos - I think dreamfeed is correct. Bob Walters says much the same here: http://rfcwalters.blogspot.it/2014/08/the-sleeping-beauty-problem-how-some.html - 'There is a complication I didn't mention explicitly in the last post. In the case of heads the beauty knows that only one step has occurred. Hence she should calculate the probability of the heads case using the probabilities in one step, not one or two steps. Hence she should estimate the probability of Monday as 1/2 after heads.'

ReplyDeleteBy the way, what is going on with upvote/downvote by guests? Today it's not possible. A few days ago it was possible. Prior to that only upvoting was possible. I don't know what probabilities to assign. Am I the unwitting subject of some experiment?

Cheers!!!

I think this also makes leads to a 50/50 "double half-er" conclusion as well. Given that an event includes a Monday awakening, what is the probability that it is a heads event? It's P2=(pE110+pE111)/(pE010+pE011+pE110+pE111). Now, turning to the specific Beauty problem, we have pE100=pE101=pE111=0, pE110=1/2, pE000=pE001=pE010=0, pE011=1/2, giving P2=1/2.

ReplyDeleteI wonder if Brad Delong answered 50% to the Monty Hall problem.

ReplyDeleteThe problem said 'sometimes' on Monday.

ReplyDeletedreamfeed - I agree with the spirit of your remarks here, but I think that the trap lies elsewhere.

ReplyDeleteOf course one can assign consistent probabilities to the awakenings! The mistake lies in assuming, or requiring, them to be mutually exclusive. The weighted average used by Lubos and Walters is an unneccessary 'fix' to a non-existent problem and allows them to address the 'thirders' on the 'thirders' terms. It's only virtue is that it yields the correct answer to the *original* question.

The probabilty of (mon,heads) = probabilty of (mon,tails) = probabilty of (tue,tails) = 1/2. This is perfectly consistent because the probablity of ((mon,tails) or (tue,tails)) = 1/2 as well. No weighted average is needed, just an elementary calculation.

To illustrate, I refer you to the Markov chain diagram in this post: http://motls.blogspot.co.uk/2014/08/sleeping-beauty-thirders-rudimentary.html?m=1. The probabilities of the awakenings, or nodes, given the initial state, are the product of successive transition probabilities. (for more general diagrams it would be a sum over all possible ways of arriving at a particular node from the initial state - much like quantum mechanics) When a transition probability is equal to 1 the conditional probability of the second w.r.t. the first is 1; this is part of the elementary calculation mentioned above.

I view this as a pseudo-problem, with most of the arguments being about whose misconceptions should prevail.

I'm also leaning towards the opinion that dreamfeed/Walters are correct here, Perhaps Lewis made a minor error in his paper stemming from giving too much credence to the other side's "ensemble of potential consciousness" argument...

ReplyDeleteI think it can be seen from "double guantanamo-izing" the problem.

Instead of tossing a fair coin, the experimenter will roll a fair 1000-sided dice. If it comes up 1-999 beauty will be woken once on monday, if it comes up 1000 she will be woken a million times over the coming days with induced amnesia etc. She will be asked to guess whether or not the dice came up 1000.

Additionally the experimenter informs her that he has arbitrarily predetermined that he will inform her what day it is on either none or all of any subsequent awakenings in the experiment.

Clearly if she is woken and told that it is monday she should still guess that the dice came up in the range 1-999 and not 1000, regardless of the putative existence of a million other awakening events that are (most probably) counterfactual in any case.

I think the subtlety here is that although she has gained new information on being informed that it's monday, (she now knows herself to be at the first, and probably last, step of the process), the extra information is redundant when it comes to assigning probabilities to which of the two branches she might be on.

The thirder wants to answer a different question - namely, what if a large ensemble of runs of the experiment exists, and a "random consciousness injection genie" will inject his consciousness into one of the beauties at a random awakening event - in which case "the dice came up 1000" is overwhelmingly likely to be the correct guess.

But this is changing the definition of what are the "independent events" whose likelyhood we are considering, in way that may be logically consistent, but that has nothing to do with any physical events or correlations.

(Normally we would consider the tails-monday, tails-teusday awakenings of a given beauty to be maximally correlated and not independent events!)

The random consciousness injection genie may be fun to think about, but morally he corresponds to an (admittedly interesting) "unphysical solution", and I wish physicists would leave him alone!

I wonder... if a truly committed thirder was playing the double-guantanamo version and was informed that he would be killed in the event of making a wrong guess, would he still have the balls to guess that the dice came up 1000 - even if he was informed it was monday? :D

I agree and I used frequentist definition because it is more familiar to people in simple cases like coin toss. Weight is of course correct and better term than saying that events must be "in the same units". I think that Maznak got that part anyway, but isn't willing to accept, or is not knowledgeable enough to see the merit of well defined, consistent set of rules.

ReplyDeleteI've been giving this some thought, and I'm also leaning towards the opinion that dreamfeed/Walters are correct here, Perhaps Lewis made a minor error in his paper stemming from giving too much credence to the other side's "ensemble of potential consciousness events" argument...

ReplyDeleteI think it can be seen from "double Guantanamo-izing" the problem.

Instead of tossing a fair coin, the experimenter will roll a fair 1000-sided dice. If it comes up 1000 beauty will be woken once on monday, if it comes up 1-999 she will be woken a million times over the coming days with induced amnesia etc. She will be asked to guess whether or not the dice came up 1000.

Additionally the experimenter tells her that he has arbitrarily predetermined that he will inform her what day it is, on either none or all of any subsequent awakenings in the experiment.

Clearly if she is woken and told that it is monday, she should still guess that the dice came up in the range 1-999 and not 1000, regardless of the highly likely existence of a million (minus one) other awakening events that she can rule out being on, but that are as yet still unobserved/counterfactual.

Knowing that it's monday and that moday awakenings will be a small proportion of all awakenings in a typical run of the experiment, is not an indicator that the dice came up 1000 with probablity approaching 1.

To sharpen this let's add that beauty will be killed if at any point in the experiment she makes a wrong guess, (Her goal is to survive).

Clearly if she adopts a predetermined strategy that if awoken and informed that it is monday, she will guess that the dice came up 1000, she's going to be in big trouble (99.9% probability to die), regardless of the fact that she in that case knows it's not tuesday or any of the subsequent days when she will (most probably) be woken!

I think the subtlety here is that although she has gained new information on being informed that it's monday, (she now knows herself to be at the first, and probably not last, step of the process), the extra information is redundant when it comes to assigning probabilities to which of the two branches she might be on.

If we invert the setup so that 1-999 results in 1 awakening, and 1000 results in a million awakenings, it also sheds light on what's going on in the mind of the convinced "thirder".

The thirder wants to answer a different question to what was the likelyhood of the dice coming up 1000 in a given experiment - namely, what if a sufficiently large ensemble of runs of the experiment exists, and a "random consciousness injection genie" will inject the thirder's consciousness into one of the beauties' streams of consciousness at a random awakening event - in which case "the dice came up 1000" is of course overwhelmingly likely to be the correct guess,

But this is changing the definition of what are the "independent events" whose likelyhood we are considering, in a way that may be logically consistent, but that has nothing to do with any physical events or correlations.

(Normally we would consider the tails-monday, tails-teusday awakenings of a given beauty to be maximally correlated and not in any reasonable sense independent events!)

The random consciousness injection genie may be fun to think about, but morally he corresponds to an (admittedly interesting) "unphysical solution", and I wish physicists would leave him alone!

I wonder... if a truly committed thirder was playing the inverted double-guantanamo version and was informed that he would be killed in the event of making a wrong guess, would he still have the balls to guess that the dice came up 1000 - even if he was informed it was monday? :D

I have a problem with some statements, which may be due to a typo or my minterpreting (or both;)).

ReplyDeleteFirst, all awakenings are exclusive in the sense that you are either awaken on Tuesday or on Monday, when toss resulted in either head or tail. Pairs (day, toss result) are also exclusive. However, some pairs are dependent on each other. This is a typo that I had in one of my posts as well.

Second, the total of all probabilities of all exclusive events should be 1. We need to use weights, or renormalize, to get 1.

Well, initially so did Erdős Pál, one of the greatest mathematicians who ever lived :D :D

ReplyDelete(He changed his mind when it was explained to him of course)

Probability is hard...

Tony - Given the statement of the problem, and the Markov chain model, awakenings are not exclusive. If she is awakened on Monday after a throw of tails then she is certain to be awakened on Tuesday. Of course she doesn't know the day or the throw, so it doesn't matter to her.

ReplyDelete'Monday' and 'Tuesday' are labels, or indices, indicating the number of steps taken in the process, i.e. n=1 and n=2. Along with the result of the toss they can also label the nodes. Of course they must be distinct, but they do not serve any other purpose. One might say they are parameters and not observables. The state of the system at any given time, or day, or value of n. after the initial state is a pair of nodes (one from either trajectory), each with a probability of 1/2 and these nodes are mutually exclusive (as are the trajectories themselves, and any pair of nodes - one from each trajectory, or even a node from one trajectory and a pair of nodes from the other). However, all pairs are not exclusive - (mon, tails) and (tue, tails) are not exclusive, while (mon,heads) and ((mon, tails) or (tue, tails)) are exclusive. Nodes on the same trajectory are not exclusive and weighting them does not make them exclusive.

Practically the first thing one learns about probability is the addition rule, and this is all that is required here. The weighting simply panders to the 'thirders' misconceptions about the importance of awakenings.

Let's first agree on what we mean by exclusive, shall we?

ReplyDeleteYou seem to use it in the sense that one event (awakening on Monday after tail) doesn't *prevent* another (Tuesday with tail) from happening later on.

I use it in the sense that if it is Monday then it is not Tuesday. Unless, of course, we start considering other reference frames, etc. which is probably not a kind of complexity that we want to include in the analysis of this problem.

We are going in quite different directions. You are seeing paths and chains while I am looking at the probability space if total measure 1 and seeing two delta functions sitting on head and tail respectively, then tail splitting into two, one with label Monday another with label Tuesday.

If I look at awakenings label (which is a pair) I see three delta functions (M,h), (M,t) and (T,t), if I look at coin toss results label I see two, where I added (M,t) and (T,t) into one, the second being (M, h).

The set of three or the set of two, all exclusive elements of the corresponding set, given the criteria which makes them distinct and non-overlapping (they are delta functions after all)

Yes, I am going back to the axioms:

ReplyDeletehttp://en.wikipedia.org/wiki/Probability_axioms

- First axiom

The probability of an event is a non-negative real number: P(E) where E is event in event space Omega which is always finite.

- Second axiom

P(Omega) = 1

This is the assumption of unit measure: that the probability that some elementary event in the entire sample space will occur is 1. More specifically, there are no elementary events outside the sample space.

This is often overlooked in some mistaken probability calculations; if you cannot precisely define the whole sample space, then the probability of any subset cannot be defined either.

- Third axiom

This is the assumption of additivity:

Any countable sequence of disjoint (synonymous with mutually exclusive) events P(E1), P(E2) ...satisfies the formula : probability of the union is the sum of probabilities of each element of the union.

The light just came on. Sorry. Of course the three waking are no longer equal for same reason - i was just locked into that four box

ReplyDeleteThe assumption that N choices are equally likely i.e. the probability of each is 1/N is rather tempting.

ReplyDeleteOne funny example was this Walter Wagner on Comedy Central who claimed that the LHC will destroy the world with 50% probability - because it will either happen, or not, so it's 50-50.

But somewhat less dramatically, I remember how my classmate at the basic school, a great historian - and today a professional one - vigorously fought with our mathematics teacher and he claimed that if you toss 2 identically looking dice, there are 15 choices what you can get, so the probability of 5-6 is the same as 6-6 and it is 1/15.

If the believer in such a thing is stubborn, and he was, it is not easy to convince him that 5-6 is actually 2 times more likely than 6-6 because 5-6 includes 5-6 and 6-5 with the dice which are in principle distinguishable but in practice not, and 5-6 and 6-5 and 6-6 in this sense are equally likely, with P = 1/36 each.

More generally, when people get used to the fact that the probabilities are not this uniform and always equal to 1/N, they want some "easy universal physical visualization" of probabilities.

So they imagine that probabilities are always directly proportional to some volume, or duration, or some mass, etc. But in general, probabilities are given by numbers and formulae that are much more complex and general than any of these "extensive" things. The laws of physics e.g. quantum physics are hard and they produce the probability or probability amplitude from complicated unitary operators integrated in complicated ways etc. None of the simple rules for probabilities is right universally, for all problems in the whole Universe.

A priori, we don't know what the probability of something is, and sometimes we can calculate it if we are lucky. The "naive" believers think that probabilities are something trivial, given by counting or weights or timing etc. But probabilities are in no way trivial: everything that is complicated about the world may be interpreted as complexity of some probability distributions - everything that quantum mechanics predicts, in particular, are probabilities that may be very difficult and structured.

Hi Tony/RAF III/dreamfeed.

ReplyDeleteI'm now sure that this is indeed correct.

The upshot of my "wordy" post above is that if clarifies the situation to consider deformations of the probability of Heads versus Tails away from 50-50.

If I set P(Heads) = 1/n with n arbitrarily large, then choose a number of tail-awakenings N such that N >> n, then by the Elga/Lewis counting this gives me P(Heads|Monday) -> 1 as N -> inf.

So i can makes heads arbitrarily unlikely, but according to the Elgin/Lewis

on waking and being told it's monday I can be certain the the coin came up heads!

It's useful to consider deforming probabilities away from 50-50 because you can no longer get away with sloppy arguments based on counting branches or nodes, (like the QM MWI crowd who can't cleanly interpret wavefunction amplitudes).

I agree that the error here with 50-25-25 is that Monday and Tuesday are not independent events... P(Tuesday|Monday) = 1 not zero! The Markov chain interpretation of this as the transition probability along the branch makes the situation perfectly clear, and the canonical labeling of the nodes is 50-50-50 not 50-25-25. This doesn't violate normalisation as the second two are perfectly correlated events not independent events!

50-25-25 can be useful as post-hoc weighting to restore even betting or give nice weights for the thirders' "consciousness injection genie" to respect, but thats adding to the model.

Walters says it best:

"In a Markov process one cannot in general talk about the probability of being in a state. One must talk about the probability of passing from state x to state y in n steps. (If the process is ergodic then in a limiting distribution there

is some sense of talking about the probability of being in a state, but

that is not the case here.)"

I guess physicists can get confused because they're used to considering thermal ensembles in an ergodic phase space where all states are "equally likely".

Thirders can of course ask the question "whats the probability of heads given a random node on the Markov chain" and it's well defined, but not physically meaningful/interesting.

To show a thirder the error of his ways, have him consider deforming so that P(Tails) is small and N awakenings is large and see what happens... ;)

I think the told it is Monday probability is 1/2.

ReplyDeleteTo see this, consider the other question, but with the coin tossed after she was awakened on Monday to determine if she will be awakened on Tuesday. I think the probabilities are the same in both versions, and makes it clear that the awakening probability should be 1/2.

However, for being told it is Monday, you are saying she should argue that this means the probability of an upcoming coin toss will be 2/3.

Tony - Please consider this a reply to both of yours above.

ReplyDeleteI would first like to say that I envy the clarity with which you express your thoughts and reasoning. Even though (I think) I somehow knew all along how you and others were thinking about this, the clarity of your remarks makes it much easier for me to respond.

Now the bad news -

Your construction of the probability space is illegitimate. The fact that the conditional probabilty of (tue,tails) given (mon,tails), P((tue,tails) | (mon,tails)) = 1, equals one should have tipped you off. The two cannot be disjoint in the probability space and must, in fact, overlap.

The problem (and the Markov chain, which accurately represents the problem) demands the following representation as a probability space; a random variable labelled Monday from the space to the set {heads, tails} which partitions the space into two regions of equal measure and another random variable labelled Tuesday which does the same; furthermore, these two random variables must be identical to account for the conditional probability mentioned above.

Sleeping Beauty knows all this when she awakens so her subjective estimate of the probabilities must be the same as given in the problem. The awakenings are irrelevant.

Finally, by the addition rule I meant P(A or B) = P(A) + P(B) - P(A)P(B|A) which for A=(mon, tails) and B= (tue, tails) gives 1/2 = 1/2 + 1/2 - 1/2.

I hope you find this reply as clear as I found yours.

If you are allowed to sample a certain event twice, and if we take that sampling happens in time, with our usual concepts of time, then sure : the second sample will follow the first, they are dependent events, etc. We can now talk also about the sampling events, instead of only the coin toss events.

ReplyDeleteIn my mind this just introduces unnecessary complexity and confusion when the simple fact is that you are oversampling.