What is the Sleeping Beauty Paradox?
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- čas přidán 25. 06. 2022
- An explanation of the Sleeping Beauty paradox for probability. Specifically it is a paradox for Bayesian Epistemology. The video looks a the views of David Lewis and Adam Elga, as well as the Rip Van Winkle paradox and the snake eyes sleeping beauty paradox.
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You should assign a 100 percent probability that the "prince" is a creep.
By saving her life?
@@Scrimparmy by feeding her sleeping pills and asking her weird questions.
In the original story, the prince doesn't actually wake her up. He has his way and leaves. 9 months later, she births twins, one of which suckles the finger with the spindle's splinter in it, waking her up.
This prince is not an improvement, imo, because he has a 50% chance to put her back to sleep
Aurora just can't catch a break
@@fenixmeaney6170 at least he isnt raping her, yikes.
@@fenixmeaney6170 ah hell nawww
Something seems very confused about the notion of this being paradoxical. When the scenario was first described, my initial thought was that you should of course guess that the coin landed tails, because 2/3 of the times you are awoken will be times when the coin landed tails. That is not a guess about *the fairness of the coin* though. It's a guess about the product of the probability of the coin landing one way and the probability of being awoken given the coin landed some way.
Consider for comparison another scenario. Your friend is going to go into another room and flip a coin. If it lands on heads, he will come out and ask you to guess how the coin landed. If it lands on tails, he will flip it again, and repeat that procedure until the coin lands on heads. You know this ahead of time. Now your friend goes into the other room, and comes back out and asks you to guess how the coin landed (on its last flip, natch). You're obviously going to guess "heads", because you know that you will only ever be asked the question if the coin lands on heads. But that doesn't mean you think it's a coin that always lands on heads. "What are the odds of this coin landing on heads per each flip?" and "What are the odds that the last flip before I asked you this question landed on heads?" are different questions, with different correct answers.
Assuming a fair coin, it's a 50/50 chance of it landing on either heads or tails. But that's not what you're being asked about; you're being asked how the last flip of the coin before you were awakened landed. Since you will be awakened twice as often for tails as for heads, the answer to that is 2/3 tails. But the coin still has 1/2 odds of landing tails per flip.
Interesting example. The issue is that you are changing the underlying probabilities by flipping the coin more than once. The coin is only ever flipped once in the original example. Think of it like this. If your friend asked you that question before they went in the other room, you would still say 100% heads. However if the prince asked you what the chances of the coin landing heads right now before he puts you to sleep and wakes you up, the right answer is clearly 50/50. The challenge is that nothing about the coin flip changes (he does not flip it more times) between your guess before the procedure and after it, but you think the chances should change.
That said, your broader point is an interesting one, that you could believe both that the coin would land 50/50 heads and tails, and at the same time that you should believe that it is more likely to land tails when you are awoken. That seems like a contradiction, or at best a breakdown in some of the principles of Bayesian epistemology.
@@CarneadesOfCyrene Another commenter on this video made a better example case than mine that sticks closer to the original scenario and doesn't involve multiple coin flips: '…if I know "The Prince never wakes a sleeper on Tails", I ought name Heads.' Like in my scenario, you know that 100% of the time when you are awoken, the coin landed heads - P(H|A)=1. But of course you still believe that there's only 50% odds of the coin landing heads - P(H)=1/2. The original scenario is just a less extreme version of that: you know that P(H|A)=2/3, and also that still P(H)=1/2.
And you know this ahead of time, you don't learn anything new upon awakening that you need to update your beliefs about. Before you go under, knowing how The Prince will behave, you should believe that the safest guess (by twice) to the question he will ask you is "tails", and that the actual odds of the coin landing tails on each slip is still only 1/2. After you wake up, your beliefs should rationally remain the same, because you haven't learned any new information. If you *did* learn any new information, for example if he tells you what day it is, *then* you should update your beliefs - if it's Monday the odds of the coin being tails drops to 0.5, if it's Tuesday they rise to 1 - but in the original scenario no new information is provided upon awakening.
I think this isn't really the same thing though, because the underlying probability that the coin eventually lands head, which is 100% given enough time, hasn't changed.
The thing is that you know more information than is being allowed for. I mean you know what the coin flipper is going to do for different outcomes of the first flip. This changes the odds.
It's similar to the three door problem, where you're asked if you want to change after the first is opened - you build in an understanding of the choices your adversary, or the coin flipper, will make.
So no contradiction with Bayesian logic as long as you take into account everything you know at the outset.
@@CarneadesOfCyrene
I tend to see other things. Suppose, on tails, the prince doesn't wake you up on Monday. He only says you did and that you bet the coin came up heads. If Monday is to be forgotten completely on tails, it may as well not exist, or just be a simulation. That leaves two possibilities: Monday heads, Tuesday tails.
The fact is that there are many ways of looking at this. And some of those vary on what should be considered "real."
Any claim that knowledge of the consequences of a coin flip ought not adjust my assessment of the odds of the coin position, is incorrect on its face. For instance, if I know "The Prince never wakes a sleeper on Tails", I ought name Heads.
This. Being awoken once or twice is literally dependent on the coin toss, so why would those ever be considered independent?
Thus, why I find it strange people even call this a paradox.
This has always been my favourite thought experiment. I learned about it from a video game called Zero Time Dilemma, and it's fascinated me since then
Cool! I didn't realize it was in a video game. Very interesting. :)
Great, another Zero Escape fan! Now to wait 9 years to meet another one...
The way I see it:
- The coin landing on a heads or tails is still 1/2 each, there's no doubt about that. The probability that I wake up on a Monday is 2/3 and on a Tuesday is 1/3. The probability I wake up on a day where the coin landed on heads is 1/3 (and 2/3 for tails).
- The big issue for me is what is this paradox asking for? The coin's probability for either outcome is clearly 1/2. The probability of me waking up on a day where the coin landed heads/tails is also clear (1/3 and 2/3 respectively). These aren't mutually exclusive, I just don't get what the question is asking for. If it's just asking for the probability of the coin landing on tails, then 1/2, but if it's asking for the probability that I wake up on a day where it landed on tails, then 2/3.
- Lets view this as a game of sorts where I get 1 point for guessing correctly. The question then becomes "if it lands on tails and I wake up twice, do I get 2 points for guessing tails each time I woke up?" If so, then the whole "what probability should you assign it" makes more sense and has a clear answer. If tails lands me 1 point if and only if I guess tails on both days, then the answer is also clear (as in it's basically the same 50/50 if I always choose one answer). The issue is that the value of an answer in the events hasn't been clarified yet.
Isn't there a 100% chance that you will be woken on Monday and then also a 100% chance that you would be woken on Tuesday?
@@hammockcamping2500 You won't wake up on Tuesday if the coin landed on heads (heads side up (although the coin landed on the tails side(down)))
@Katana 123 at 1:25 he says that she is awakened again on Tuesday.
@@hammockcamping2500 By again, he means for the 2nd time that coin flip not 2nd time waking up on Tuesday
@@Bugy64 I think you're talking about a different riddle. The coin is only flipped once.
This kinda reminds me of the "Judge says the prisoner will be executed and will be surprised about what day he dies" scenario where the premise boggles the actual point it's supposed to illustrate.
Prince wakes up Sleeping Beauty. He asks her "How did the coin land when I flipped it?"
She responds: "Who cares. I'm awake now, and well rested. If it was true love's kiss, you wouldn't be asking me stupid questions."
...well this was more mathematics than I was prepared to handle today.
I have always been skeptical of the use of statistical analysis applied to a single event (There are lies and damn lies, and then there are statistics). It often depends on some sort of conflation of objective facts and psychological tendencies, but more important, it proceeds from a casual mapping of a manifold, a multiplicity of events (the experimental sample, so to speak) onto a unit - the one event in question. It assumes that the state of affairs regarding the previous events - which state of affairs is a static entity - is, as you might say, convertible into a dynamic entity, namely the one actual event about to unfold. Further, it ignores that inconvenient fact that a single event has a probability of either one or zero - it either occurs or not, but never (for macroscopic events, at least) lands in that imaginary area where the probability p is such that 0
It does seem that probabilities are useful given that we fail to have sufficient information to fully predict future events. Someone may believe that it either is or is not the case that it will rain today, but might find probabilities useful when deciding to plan a picnic. And someone that says a coin is fair seems to be saying something real, something true even if a perfect physics engine might be able to predict exactly how that coin will land each time. Or, at least there seems to be an important difference between a fair coin and a weighted coin.
The fact that you are awake and being asked the question means that it cannot be Tuesday after heads - which is new information compared to before the experiment started.
The problem with this set up is that of the three events, one event produces another. Let me explain:
P(Awoken on Monday after heads):50%
P(Awoken on Monday after tails):50%
P(Awoken on Tuesday after tails):50%
This is fine because being awoken on Monday and Tuesday for tails are 100% dependent on each other. P(A or B) = P(A)+P(B)-P(A and B)
So, in any run, you will be awakened on a monday for 100% of the runs and tuesday for 50% of runs.
Now these are the probabilities of the general outcomes, not the specific instances, for which being woken up on a monday and tuesday on tails become mutually exclusive. So the probabilities become:
P(Heads and it is Monday)=50%*100%=50%
P(Tails and it is Monday)=50%*50%=25%
P(Tails and it is Tuesday)=50%*50%=25%
Adding up the Probabilities still show that there is a 50/50 probability unchanged!
However, going by pure probability may show the chance of being right, it does not show the full outcome of being right. Remember that if it is tails, you are awoken twice! This means that you can be right twice in a single go if you chose tails!
What this boils down to is that you have a greater expected value from tails as:
E(H) = 1 time correct * 50%= 0.5
E(T) = 2 times correct * 50% = 1
TLDR: The probability for heads or tails is still 50/50, but guessing tails gives a larger payoff (2x)
So, another way of looking at it is that the assumption of equal chances (1/3) for all three outcomes; Monday Heads, Monday Tails, and Tuesday Tails, is incorrect. For any given waking, there is a 50/25/25 chance for each outcome. Hence, regarding the "bets", 25% of the time, a bet of Tails will win 100% of the time (on Tuesday). Conversely, 75% of the time it is either Monday+Heads or Monday+Tails. 2/3 of that is Heads, and 1/3 is Tails. Between Heads and Tails, the expected "yield" for a given week would be as follows:
Heads: 2/3 (Monday) + (0 * 0.5) (Tuesday) = 66.7% yield
Tails: 1/3 (Monday) + (1 * 0.5) (Tuesday) = 133.33% yield
Ergo, for the total week's wins that fall on Monday, 66.7% are Heads, full stop. Whereas, if she picks Tails, there is a 33.3% chance it is Monday, plus a guaranteed 100% for a followup on Tuesday for 133.3% total yield. To put it simply, if she always picks Tails, then Tails pays 2:1 over Heads because, even at 50/50 chance to get Tails, you get paid out twice compared to Heads paying out once. The chances of each outcome cannot be divorced from one another because waking on Tuesday is contingent on getting a Tails on Monday, thus they must be considered together.
No, it doesn't. Neither in theory, nor in practice. If you run a simulation the 1/2 method approaches 50% of correct guesses. The 1/3 method approaches 55% correctness (with growing number of iterations). And the more awakenings you add after tails compared to heads the higher the chances that the 1/(n+1) approach will guess right.
Doesn't the 2/3rds response assume that the three options are independent? Given that waking up on Tuesday is very much a function of whether or not you flipped heads or tails, this doesn't seem to be a good assumption to make.
Yeah this seems like a pretty trivial probability problem, no? The odds of waking on each day should be 1/2 Monday, 1/4 Tuesday, 1/8 Wednesday, and 1/16 for Thurs & Friday.
@@Pianoblook I don't know about that. The way I see it you have to consider the events that are independent. If they flip heads then there's 100% chance that today is Monday. If they flip tails, then you've got a 50/50 shot of it being Monday or Tuesday. Assuming the heads and tails are independent of each other, this gives an overall 75% chance of it being Monday, and 25% chance of it being Tuesday.
I realized that if what I said is correct then, then if you play the game in reverse, with 75% chance for waking up on a Monday, and a 25% chance for Tuesday you actually get an overall probability of waking up having flipped tail of 62.5% which doesn't agree with either answer. I've got a feeling I went wrong somewhere.
It seems that the odds of coin aren't changing but rather the value of the wager. The problem could be rewritten to say that two people are betting on a coin flip if both people agree that if the coin lands on tails the value of the wager is doubled, would it be smarter to bet heads or tails. It would obviously be smarter to bet on tails but not because tails is more likely but rather because the consequences are doubled if the coin lands on tails.
Yeeaah, I still don't get the probability theorem... but got a very striking picture of just how REALLY, REALLY, *REALLY* CREEPY are the fairytales we tell our daughters since before they can walk. Good education, thanks.
Something, something, Monty Hall Problem.
Check this video out for a Bayesian's take on the Monty Hall Problem: czcams.com/video/Nn0OOYx4bNE/video.html
I was confused at first but i like the way you explained it
Apologies if it was confusing. I'm glad you liked the explanation. :)
What happens if you refuse to answer? Will he only put you to sleep upon a wrong answer? Also, what about cop out answers like the side of the coin that was showing was the one your eyes noted it to be?
I’ll admit I don’t really understand the idea this example is trying to explain, but it seems to me that if the coin was flipped only once you could decide to answer tails on Monday and heads on Tuesday. I also agree with a comment below: philosophy and statistics can be strange bedfellows.
I am asking too many questions as well. My mind isn't on the end of skitzo for this to be interesting.
The situation and sequence of events seems a tiny bit constructed. That reminds me of our good old high school math tests, I love it.
The prince is Anton Chigurh.
The actual probability depends on the important event. If every question counts, it's better to bet on tails because the reward would be higher even if the probability is the same. But if the first and second awakening count as one, it's 50/50. From another point of view it's 50 for tail/25 for heads/25 for being asleep on monday. It all depends on how the information is used.
They count as two, obviously, because you are asked in both cases and since your memory is erased you don't know when and where you are.
Suuuper interesting!
Thanks!
The greater question for me lies here in the estimation of the "real" probabilities. Lets say that the chances of this coin are not known (it could be 40/60 or 30/70; lets say you can't be sure its a normal coin or maybe it is a normal one). Would you be able to tell something about the probabilities of the coin itself? Would that change your estimation ? If i were to use the ratio the the coin flip is connected with me waking up i could estimate the 2/3 Heads and 1/3 tails, independently of what the "real" chances are. In my world (in this experiment) the coin is 2/3 / 1/3. And since i do not keep any new information the process is finished. I would need to have some sort of memory, so that i can learn from experience.
One thing that string of thoughts tell me is that we are only able to discover our own world. By exchanging information i can expand my world by a fraction of your world. But what is beyond that world might be never fully understand. All i know is that my world is a fraction of reality.
I don't know, I feel like we can't use define probabilities exclusively based on how much you would bet on them, because bets are made on expected value, which is not just a function of probability. Maybe I'm getting this terribly wrong, but it seems like a big oversight to me. The expected value of choosing tails is obviously twice as much as heads, no arguing here, but stating that this alone proves tails to be twice as likely does not follow since it implicitly assumes that the reward is the same in both cases. Rather than tails being twice as likely I'd say it's twice as rewarding.
The question here is: if you are wrong when choosing tails are you wrong twice, or are you simply twice as wrong?
Like, if I changed the paradox to something like "you win 1$ if you guess heads correctly and you win 2$ if you guess tails correctly" the probability of the coin having landed on tails when I wake up is clearly 1/2. If I changed it to "you win 1$ every time you guess correctly, but if it's tails the prince will cast a spell on you that will force you to repeat your answer" the probability is still 1/2. I'm just doubling the reward in one case, thereby doubling the expected value. Now, if erasing someone's memory means resetting them to the exact state before they gave their answer so that they will make the same choice (which is the whole point of the paradox), what's the difference between this scenario and the one where the prince forces you to repeat yourself? He might as well take a recording of your answer and then play it back to you once you wake up.
I don't think you are actually wrong twice (higher probability), I think you are twice as wrong (higher reward), but it's been made to seem as if you were wrong twice with the memory erasing shenanigans. The proof is that, unlike with heads, you'll never be able to win 1$ with tails: you either win 2$ or you win nothing, so there's no reason to arbitrarily set the reward for guessing tails as 1$ and double the probability to make things work.
There are three scenarios, true, but why are we assuming that they are equally likely? If the coin lands on head (P(heads)=1/2) you are waking up on Monday with probability 1, since it's the only option. If the coin lands on tails (P(tails)=1/2) when you wake up you might be on Monday or Tuesday, with a probability of 1/2 each. In short, the three options are:
P(heads, monday) = 1/2
P(tails, monday) = 1/4
P(tails, tuesday) = 1/4
Therefore, from your point of view, both heads and tails are equally likely, but if you had to bet you should bet on tails, because you know that if you are correct your answer will be counted twice regardless of what has happened or will happen. If you remove the double winning aspect by only rewarding you for guessing correctly on the last day you wake up (which is the exactly same thing as guessing correctly both days if the coin lands on tails) the paradox evaporates.
This paradox comes from the fact that, since your memory is erased, to you this situation looks essentially identical to one where you are equally likely to wake up in any of the three situations. But it's not, because the events are absolutely not independent. You either win 100% of the times you are asked or you lose 100% of the times, increasing the number of times I wake you up is only an artificial way to increase the expected reward in a way that looks like it's creating separate, independent events that have to be considered equally to calculate the probability. The point is that both interpretations will give the same expected value, so it's not immediate to see the mistake because "it works". It looks like the Monty Hall problem, but the three scenarios are not independent, so you can't just assign a 1/3 probability to each without thinking.
Funny how they're always telling the truth in these scenarios. Surely one of the variables should be whether or not the creepy prince will lie about the coin toss so he can keep you in bed indefinitely.
I feel like this would be a good scenario to test in an actual test to show what the statistics *actually* are.
It'd be pretty trivial, just loop through:
coin = random("h", "t")
if coin == "h":
head_correct = head_correct + 1 //monday
else:
tails_correct = tails_correct + 1 //monday
tails_correct = tails_correct + 1 //tuesday
The # of waking-times "heads" is the correct guess winds up half the # of waking-times "tails" is the correct guess.
Feels kinda akin to the Monty Hall problem
If we are talking about bets, we also need to stablish the rewards for being right on the guess. If coin lands on tails you will have 2 chances to guess "tails" right. If coin lands on heads, you will have 1 chance to guess "heads" right.
If payment is $1 guessing heads and $1 guessing tails, then it's obvious you should answer tails.
$H = 1/3 * $1 + 2/3 * $0 = $1/3
$T = 2/3 * $1 + 1/3 * $0 = $2/3
But if payment is $3/2 guessing heads and $3/4 guessing tails, then the payment balances out.
$H = 1/3 * $3/2 + 2/3 * $0 = $1/2
$T = 2/3 * $3/4 + 1/3 * $0 = $1/2
0:32 The paradox relies on Sleeping Beauty forgetting about Monday's interaction if the coin came down tails. I do not accept that this is necessarily true.
It is true by the definition of the problem. :) If you don't accept it as true then you are not solving the problem.
I mean, real talk, just say Tails. Under those circumstances, just bite the L of being right exactly 50% probable of being right. There is no consequence for failure for saying Tails and being wrong.
I honestly only wish the audio was a little louder.
Hold up its a probability of 3/4 heads. On Monday you have a probability of 1/2 (as normal to awake) otherwise another 1/2 on Tuesday times the already existing 1/2 for having tails the day before this is (1/2 + 1/4 ) = 3/4
Furthermore then the Rip Van should be 1/2^(1000) tails witch is a 1, 302 zeros behind the coma. A little more than 1/1000
So either
1. Philosophers are bad at math
2. I am bad at maths
3. Or I just didn't understand the question/scenario correctly
If 2 or 3 please explain
In my opinion, you should judge the probability that the coin landed on tails to be 2/3. If something has a probability of X%, then it should be true, on average, X% of the time. This is true regardless of whether you believe in frequentism or not - the only difference is that frequentists say this is the fundamental meaning of probability, while other interpretations say this is a result of whatever probability really means. Based on the experiment given, it will be true 2/3 of the time that the coin landed on tails.
The objection that you haven't been given any additional information on what side the coin landed on doesn't hold any water for me. You have been given more information: Specifically, we you are being woken up and asked the question by the prince, you are given the information that the prince is, at this very moment, waking you up and asking what side the coin landed on. Since this will occur twice as often when the coin lands on tails as it does when the coin lands on heads, the posterior probabilities should be adjusted accordingly. We can do the calculation explicitly:
If we let T be the event of the coin landing tails, H represent the coin landing heads, and A represent our being awake at this very moment, then we know that the prior probability is P(T)=1/2 (assuming a fair coin), and P(A|T)=2*P(A|H). Bayes's Theorem tells us that the posterior probability is P(T|A) = P(A|T)P(T)/P(A).
Since the coin lands either heads or tails, and not both, P(A) = P(T)P(A|T)+P(H)P(A|H) = 1/2*(P(A|T)+P(A|H))= 3/4*P(A|T). (In the last step, we used the fact that P(A|H)=P(A|T)/2). Putting this result back into Bayes's Theorem, we get P(T|A) = P(A|T)*(1/2) / (3/4*P(A|T)) = 2/3.
So what's the result you get when you actually do something like this in real life?
Didn't even know Van Winkle died
So the 50/50 camp thinks, that if the coin flip results in heads, then it's Monday and if the coin flip results in tails, then it's Monday or Tuesday:
(Coin flip is heads → it's Monday) ∧
(Coin flip is tails→it's Monday or Tuesday)
This amounts to no "significant information" about the coin flip's result. Hence, both coin flip results are equally likely. Therefore,
P(Monday) = P(Monday|Heads)•P(Heads)+P(Monday|Tails)•P(Tails)
= 1•0.5+0.5•0.5 = 0.75 = 75% and
P(Tuesday) = P(Tuesday|Heads)•P(Heads)+P(Tuesday|Tails)•P(Tails)
= 0•0.5+0.5•0.5 = 0.25 = 25%
So the _prior_ probability for the coin flip resulting in heads or tails should be 50/50 - equally likely.
The 3/4ths for tails camp thinks, that if it's Monday, then the coin flip resulted either in heads or tails and if it's Tuesday, then the coin flip resulted in tails:
(It's Monday → the coin flip resulted in heads or tails) ∧
(It's Tuesday→the coin flip resulted in tails)
This amounts to no "significant information" about which day it is currently. Hence, this day (at which the question is currently asked and addressed of "What the odds of the result of that coin flip is or should be?") is as likely to be Monday as to be Tuesday or to say, this current day to be equally likely to be either Monday or Tuesday. Therefore,
P(Heads) = P(Heads|Monday)•P(Monday)+P(Heads|Tuesday)•P(Tuesday)
= 0.5•0.5+0•0.5 = 0.25 = 25% and
P(Tails) = P(Tails|Monday)•P(Monday)+P(Tails|Tuesday)•P(Tuesday)
= 0.5•0.5+1•0.5 = 0.75%
So the _posterior_ probability for the coin flip resulting in heads or tails should be 25/75=1/4 - or 3/4ths in favor for tails.
The 2/3rds for tails camp thinks, that it's an equally likely trichotomy of the case being either it's Monday and the coin flip resulted in heads OR it's Monday and the coin flip resulted in tails OR it's Tuesday and the coin flip resulted in tails:
Ω= (It's Monday ∩ the coin flip resulted in heads) ∪ (it's Monday ∩ the coin flip resulted in tails) ∪ (it's Tuesday ∩ the coin flip resulted in tails)
This amounts to no "significant information" about current "situation"/outcome. Hence, all those three outcomes are equally likely. Therefore,
P(Heads∩Monday) = P(Tails∩Monday) = P(Tails∩Tuesday) = 1/3
P(Heads)= P(Heads∩Monday) = 1/3 and
P(Tails) = P(Tails∩Monday)+P(Tails∩Tuesday)
= 1/3+1/3 = 2/3
So the probability for the coin flip resulting in tails should be 2/3rds.
As a frequentist I think, that it's not, that we have no "significant information" about the coin flip result.
The most common coins are "fair coins" meaning, that the odds of a flip of such a fair coin resulting in either heads or tails are about the same or to say 50/50 for heads or tails. *So given most common coins to be such a "fair" coin it is very reasonable to assume that considered prince have been making or doing a coin flip with such a common "fair" coin and therefore the probability resulting that considered coin flip to be heads or tails **_probably_** being 50/50.*
Yes, as a frequentist I think, that for _a posteriory_ knowledge/reasons the _prior_ probability is _probably_ (inductively) 50/50 regardless of any "slippery slope"/games/thought-experiments with coins.
So yes, I am in the 50/50 camp - not for the reasons or the lack there off in this video, but for other positive reasons.
I'm aware of the Problem of Induction.
But as a frequentist I guess, that I just have to live with it as any body else must _probably._
This is typical of so many "paradoxes" in that it is created by poorly defined statement of the problem. It seems confusing because in fact either answer can be correct depending on what you mean. The answer is "1/2 of the time" if it refers to the number of times that the coin is tossed but "1/3 of the time" if it refers to the number of times that Sleeping Beauty is asked. Define your question and there is no controversy.
In philosophy people can talk endlessly and never reach a conclusion.
In math you can do the work to get to the one and only correct conclusion.
In math the definition of probability is "the number of times a particular event occurs divided by the total number of possible outcomes." (You can have a different definition of probability but this is the math definition.)
If you conduct this experiment 100 times with 100 different sleeping beauties, you will get 50 heads flips and 50 tails flips.
The 50 heads beauties will be asked 50 questions.
The 50 tails beauties will be asked 100 questions.
There will be A total of 150 questions asked.
Since there are 100 flips of a fair coin, there are no other numbers possible then the ones listed above.
Each of the sleeping beauties knows the above.
When each of the beauties are woken up they are asked "What do you believe is the probability the coin landed heads?"
Each of the beauties knows the coin landed heads 50 times.
Each of the beauties knows there will be 150 total questions asked.
So from the perspective of each of the beauties the probability that the coin landed heads is
50 ÷ 150 = 1/3.
You can look at this any way you want but the above (and the two other math proofs that reach the same conclusion) is the one and only way math looks at it.
It turns out the correct answer is 1/2, if you wish to see my explanation here is a link to my video. czcams.com/video/qn96gTECuBU/video.html
The betting arguments don't make any sense to me. It only makes sense to guess tails because the reward is higher if you do (you get paid on Monday and Tuesday instead of just Monday). That doesn't change the odds of the coin flip. If I told you I'm going to flip a coin and if you correctly guess Heads I'll give you a £1 but if you correctly guess tails I'll give you a £1 today and a £1 tomorrow then you'll obviously choose Tails because it's a greater reward. The odds remain 50:50. How does that differ with this puzzle?
The debate around this problem honestly keeps surprising me, and most of the arguments (sometimes for both positions) just don't make any sense - at least to me.
The best argument for the thirder position is based on the idea of the bet, and the fact that you would earn more money if you bet 1/3 for head every time. This is well explained in this video.
And this is indeed correct, there is nothing to say about the math here. But come on:
1. this clearly doesn't answer the question, since Sleeping Beauty is asked what she thinks the position of the coin is now - not how she would bet if she was to earn 1 dollar for every correct answer in a repeated experiment.
2. this is based on an hidden assumption that is not stated in the problem: that SB would earn a dollar PER answer, and not just 1 dollar at the end of the experiment (which is normally how you compute probabilities)!
So the thirders are giving a correct answer to the wrong question - they are changing the problem by introducing unstated assumptions, this simply cannot be considered a correct answer to the problem.
Change the assumption to make it 1 dollar per experiment rather than per answer, and the thirders will give the same answer than the halfers.
You cannot just change the problem like this, by introducing unstated assumptions, otherwise the problem can have any number of answers: just invent whatever you like.
If I, as your teacher, ask you what is 3 + 4, and you answer 5 because this is the hypotenuse of the rectangle triangle with sides of length 3 and 4 respectively, I don't care if you give me the best proof in the world: you are not getting any points, period.
Another argument that just don't hold water is this idea that halfers should update their belief to 2/3 when told that it is Monday. Excuse me?
Probabilities are not my area of expertise, but when presented with this kind of bullshit, either the theory has to be trashed, or (more likely) it is simply misused.
So, where is the mistake? I am not sure, but I would argue that this is trying to apply a thirder's view to the halfer position: the idea that each day has a specific probability (based on SB view) and that it can influence SB's belief.
While in fact, as far as I can tell, the halfer's position is exactly that the number of days and their respective 'probabilities' have no role to play when answering the question. This just cannot be used to update SB's belief.
We are considering that there are 2 paths, not 3 days, and that each path is equally likely.
Being told that it is Monday doesn't change anything, Monday has a 100% chance of occurring in both path. If you want to do math, you can divide 0.5 by 1, and you will still get 0.5 - but I don't think that this is the correct way to use Bayes' theorem anyway (or that it should be applied here).
In the end, I think that the best way to properly answer the “paradox” is to consider the “snake eyes SB” version of the problem. Except that instead of a 6 faced dice, imagine doing this with a random number between 1 and 1 billion.
Since thirders like bets, consider the following payouts:
- on a 1, you get 1 billion x 1 million dollars, making you the richest person in the history of the world!
- on any other number, you get a ‘mere’ 100 000 dollars.
You are asked: what is your credence that a 1 was rolled. And by the way, you are on a game show, this is a one in a life time opportunity.
What’s your answer?
If you are an indefectible thirder, you’ll leave the game with nothing. Halfters will be 100 000 dollar richer.
That’s life, and you only get points for correct answers. Unless, of course, you are extremely lucky!
You are not alone. I myself consider some of the arguments made by some people from both camps unsound.
While I find the 1/2er's position more reasonable than the 1/3er's position, its implications are highly *ABSURD* .
Specifically _"this idea that halfers should update their belief to 2/3 when told that it is Monday"_
I replied to you because of your first sentence, and because you have succeeded to point out what's wrong with the wrong arguments. So I think it is a good idea to show you this conversation. Check the highlighted comment and its replies. (It should appear as the first one)
czcams.com/video/lbTo7uxcKvo/video.html&lc=Ugwqjga8lAva2qYC4xp4AaABAg.9tXNLYwHc7Z9vGChn0X7pU
I hope you will find it interesting and broadening to your perspective.😄
Edit: Alright, looks like the link isn't working in the way I expected. Sort comments by "new first" and look for the one made by @ronald3836.
"dollar PER answer" is obviously the representation of the chances. I.e. and answer to what she is being asked for.
The 1/3 argument is based on the new information you get from the fact that you are awake and being asked. I.e. that because of that it cannot be Tuesday after heads. Which is the new information you didn't have before the experiment started - back when the chance would have been 2 out of 4.
I'm still not getting those Dutch Book arguments.
Sure. IF I bet my credence of the coin flip result being heads c=0.5 for 1$ - gaining the 1$ for 0.5$, when the coin flip result is actually heads, then I will losing money on the long run given me betting more times and more frequently, when the coin flip actually resulted in tails.
But then why reacces my credence for break even, when I could just not make that bet and instead make the bet for the coin flip being tails with my credence of the coin flip being tails 1-c=0.5 for 1$ - gaining the 1$ for 0.5$, when the coin flip result is actually tails?
By making such a bet then I will make money in the long run, since I will bet and win that bet more often, when the coin flip actually resulted in tails.
Why break just even, when you can simply break the system?
*If we can make both bets simultaneously - betting on heads with 0.5$ and simultaneously betting on tails with 0.5$ for 1$, whenever we are awakened, then we will not win but also not lose money on the long run, since the gain and loss of money will even themselves out in that case.*
I simply don't get these Dutch Books arguments.
Am I missing something here?!? What is that, what I'm supposed to be missing here?
The events for tails aren't independent. So it's 50% that it's tails. That also means that it's 50% tails when woken up on monday or on tuesdays. But it's also 50% heads, so it's also 50% heads when woken up on monday.
That doesn't mean there are suddenly 150% of outcomes, it just means that the 50% tails means to be waken up multiple times.
I don't understand how this is a paradox.
You say "the events for tails aren't independent." That is true for an observer who is outside the experiment. The issue behind the paradox you don't want to understand, is whether they are, or are not, independent for SB, the observer inside the experiment. And your claim that they are not leads to another paradox.
Let's say it is 50% Heads for now. And so 50% Tails. But it isn't both Monday and Tuesday, so it is 25% Tails&Monday, and 25% Tails&Tuesday. With me so far? Or do you think it is both Monday and Tuesday at the same time?
Now let's say, right after SB states her guess, we tell her what day it is. If we tell a halfer SB that it is Monday, she can now update her guess to 2/3 Heads, 1/3 Tails. Why? because in the 75% chance that it would be Monday, 50% belongs to Heads and 25% belongs to Tails.
BUT WAIT! We don't have to actually flip the coin until Tuesday Morning. That is the only time the coin matters to the experiment. So, when SB said there was a 2/3 chance of a Heads result, that is a prediction for a coin that has not yet been flipped.
Oops.
@@jeffjo8732 I simplified the experiement in my comment quite a bit, I see that now. For now, between the mainstream solutions my thought would probably be between a double halfer and that the question isn't defined clearly. But I came up with an individual novel solution, which will be a bit harder to explain...
Here a taste of it:
Find the highest point in the function
(a1 * (1 - coin(t)) * (1-p(t)) - a2 * (1 - coin(t)) * p(t) + b * coin(t) * p(t)^2 + c * 2 * coin(t) * (p(t) * (1-p(t))) - d * coin(t) * (1-p(t))) / (a1 * (1 - coin(t)) + b * coin(t))
with p(t) in [0,1]
The result will be E[X]
For this solution, the thirder position would essentially look like putting in the values
a1 = 1, a2 = 0, b = 2, c = 1 and d = 0 and E[X] = 2/3, when p(t) is 1
The lewisian halfer position would look like putting in the values
a1 = 1, a2 = 0, b = 1, c = 1/2 and d = 0 and E[X] =1/2, when p(t) is in [0,1]
The double halfer position would look like putting in the values
a1 = 1, a2 = 0, b = 1, c = 0 and d = 0 and E[X] =1/2 , when p(t) is exactly 0 or 1
And many more solutions could be calculated, each with more or less sensible assumptions about SB and the experiment. For example:
With a1 = 1, a2 = 0, b = 6/5, c = 4/5 and d = 0 then E[X] is 3/4 , when p(t) is 49/80
My solution is essentially a calculation of SB's expected happiness after she completed the experiment and looks back at her decisions.
So, How happy is SB, when she got it right and it was heads, how happy is SB, when she got it right once and it was tails, how happy is she when she got it right twice and it was tails minus how sad is she, when it was heads and she got it wrong, minus how sad she is when she got it wrong once and it was tails, minus how sad she is when she got it wrong twice and it was tails.
I do not believe this is a paradox at all. Yes, there might be more situations when she gets woken up with Tails, but those situations are not equally likely in outcome.
To illustrate that the chance HAS to be 50/50, look at it from this angle:
Imagine, we add a third party to the room, who takes over the role of "guesser". Instead of the princess, this person gets to guess which side the coin landed on, the rest stays the same.
If you add the guesser during any random day without telling them what day it is, they would rationally guess 50/50 for the probability. Because that would be the correct answer.
We can also let both the guesser AND the princess guess, to illustrate that they are betting on literally the same event. The princess waking up doesn't add any information to this puzzle and is just smoke and mirrors.
Correct me if I am wrong. But I think the guesser would not rationally guess 50/50.
Because the guesser knows the condition in which he would be called in. He could make a guess on what date it is based on the fact the he was called at all.
He knows he would not be called on a Tuesday if the coin was Tails but would be called on Monday if the coin was Heads or Tails. So if he is called at all, he would be able to guess that that day was 2/3 Monday and 1/3 Tuesday.
For an extreme example proving that information could be known by the guesser: He would know the day that he was called was not on a Wednesday or any other day of the week besides Monday and Tuesday.
What’s the reward for getting it right though? The question is irrelevant. Does he keep you awake if you guess right? No, if it’s tails he puts you back to sleep regardless.
What are you gaining from guessing right?
You may as well just say it landed on its edge and get on with your day.
That’s conditional probabilities for you. Not that difficult once you get the idea.
I also am skeptical about Bayesian Epistemology. To me, it looks like a discipline in search of a problem to justify itself. And this problem is a good example of it, since it is does not need epistemology.
If that sounds wrong to you, you need to re-read Elga's paper that introduced the problem. The problem he says he is solving does not mention Monday, or Tuesday, or a guaranteed waking in one situation but not the other. Those elements were introduced, by Elga, to describe his thirder solution. The actual problem statement says "During [the experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice)." He goes on to structure the problem, with the order you describe here. With the implication that it does not affect what the answer should be.
This structure is what introduces epistemology, but it is not needed. To see this, first consider what I will call the basic experiment: Two coins, C1 and C2, are manipulated in such a way that the four possible combinations {HH,HT,TH,TT} each have probability of 1/4. What is the conditional probability of HT, given the event {HT, TH, TT}? This should be answerable on the first day a student taught conditional probability. It is Pr({HT})/Pr({HT,TH,TT} = (1/4)/(3/4) = 1/3.
To see that this answers Elga's *_original_* problem, make a compound experiment out of my basic experiment. After SB is put to sleep, flip coins C1 and C2. If either is showing Tails, wake her and ask her for the probability that C1 is currently showing Heads. Then put her to sleep with amnesia, turn coin C2 over. Again, if either coin is now showing Tails, wake her and ask her for the probability that C1 is currently showing Heads.
The epistemology approach confuses SB's ability to observe the event {HH} with the event itself. What would, or would not, happen under event {HH} does not change the fact that anytime she is awake, SB knows that she is in an exact instance of my basic experiment. And that she know that the event {HT,TH,TT} has occurred. Her answer must be 1/3.
It seems like a computer simulation could answer this pretty easily.
1st
Congrats! The question is are you just now posting first, or did you post first, fall asleep, forget about it, wake up and post first again?
@@CarneadesOfCyrene I just got it, I am bursting laughs right now, hahahaha!