Of course, there's a Wikipedia page about it, which I'll borrow to provide the background:
"Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening.
A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.
Any time Sleeping Beauty is awakened and interviewed, she is asked, 'What is your belief now for the proposition that the coin landed heads?'"The problem is discussed at length on this CrossValidated thread. As the person who posted the question explains, there are two common reactions to this problem:
The Halfer position. Simple! The coin is fair--and SB knows it--so she should believe there's a one-half chance of heads.
The Thirder position. Were this experiment to be repeated many times, then the coin will be heads only one third of the time SB is awakened. Her probability for heads will be one third.The thirder position is correct, and I think the argument based on long-run averages is the most persuasive. From Wikipedia:
Suppose this experiment were repeated 1,000 times. It is expected that there would be 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in one-third of the cases would heads precede her awakening. This long-run expectation should give the same expectations for the one trial, so P(Heads) = 1/3.But here's the difficulty (from CrossValidated):
Most, but not all, people who have written about this are thirders. But:
- On Sunday evening, just before SB falls asleep, she must believe the chance of heads is one-half: that’s what it means to be a fair coin.
- Whenever SB awakens, she has learned absolutely nothing she did not know Sunday night. What rational argument can she give, then, for stating that her belief in heads is now one-third and not one-half?
I wonder where exactly in Bayes' rule does the formula "fail". It seems like P(wake|H) = P(wake|T) = 1, and P(H) = P(T) = 1/2, leading to the P(H|wake) = 1/2 conclusion.I have come to a resolution of this problem that works, I think, but it made me realize the following subtle point: even if two things are inevitable, that doesn't make them equally likely.
Is it possible to get 1/3 using Bayes' rule?
In the previous calculation, the priors are correct: P(H) = P(T) = 1/2
It's the likelihoods that are wrong. The datum is "SB wakes up". This event happens once if the coin is heads and twice if it is tails, so the likelihood ratio P(wake|H) / P(wake|T) = 1/2
If you plug that into Bayes's theorem, you get the correct answer, 1/3.
This is an example where the odds form of Bayes's theorem is less error prone: the prior odds are 1:1. The likelihood ratio is 1:2, so the posterior odds are 1:2. By thinking in terms of likelihood ratio, rather than conditional probability, we avoid the pitfall.
If this example is still making your head hurt, here's an analogy that might help: suppose you live near a train station, and every morning you hear one express train and two local trains go past. The probability of hearing an express train is 1, and the probability of hearing a local train is 1. Nevertheless, the likelihood ratio is 1:2, and if you hear a train, the probability is only 1/3 that it is the express.
[UPDATE 11 November 2015] Peter Norvig writes about the Sleeping Beauty problem in this IPython notebook. He agrees that the correct answer is 1/3:
The "halfers" argue that before Sleeping Beauty goes to sleep, her unconditional probability for heads should be 1/2. When she is interviewed, she doesn't know anything more than before she went to sleep, so nothing has changed, so the probability of heads should still be 1/2. I find two flaws with this argument. First, if you want to convince me, show me a sample space; don't just make philosophical arguments. (Although a philosophical argument can be employed to help you define the right sample space.) Second, while I agree that before she goes to sleep, Beauty's unconditional probability for heads should be 1/2, I would say that both before she goes to sleep and when she is awakened, her conditional probability of heads given that she is being interviewed should be 1/3, as shown by the sample space.
[UPDATE June 15, 2015] In the comments below, you’ll see an exchange between me and a reader named James. It took me a few tries to understand his question, so I’ll take the liberty of editing the conversation to make it clearer (and to make me seem a little quicker on the uptake):
James: I'd be interested in your reaction to the following extension. Before going to sleep on Sunday, Sleeping Beauty makes a bet at odds of 3:2 that the coin will come down heads. (This is favourable for her when the probability of heads is 1/2, and unfavourable when the probability of heads is 1/3). She is told that whenever she is woken up, she will be offered the opportunity to cancel any outstanding bets. Later she finds herself woken up, and asked whether she wants to cancel any outstanding bets. Should she say yes or no? (Let's say she doesn't have access to any external randomness to help her choose). Is her best answer compatible with a "belief of 1/3 that the coin is showing heads"?
Allen: If the bet is only resolved once (on Wednesday), then SB should accept the bet (and not cancel it) because she is effectively betting on a coin toss with favorable odds, and the whole sleeping-waking scenario is irrelevant.
James: Right, the bet is only resolved once. So, we agree that she should not cancel. But isn't there something odd? Put yourself in SB's position when you are woken up. You say that you have a "belief of 1/3 in the proposition that the coin is heads". The bet is unfavourable to you if the probability of heads is 1/3. And yet you don't cancel it. That suggests one sense in which you do NOT have a belief of 1/3 after all.
Allen: Ah, now I see why this is such an interesting problem. You are right that I seem to have SB keeping a bet that is inconsistent with her beliefs. But SB is not obligated to bet based on her current beliefs. If she knows that more information is coming in the future, she can compute a posterior based on that future information and bet accordingly.
Each time she wakes up, she should believe that she is more likely to be in the Tails scenario -- that is, that P(H) = 1/3 -- but she also knows that more information is coming her way.
Specifically, she knows that when she wakes up on Wednesday, and is told that it is Wednesday and the experiment is over, she will update her beliefs and conclude that the probability of Heads is 50% and the bet is favorable.
So when she wakes up on Monday or Tuesday and has the option to cancel the bet, she could think: "Based on my current beliefs, this bet is unfavorable, but I know that before the bet is resolved I will get more information that makes the bet favorable. So I will take that future information into account now and keep the bet (decline to cancel)."
I think the weirdness here is not in her beliefs but in the unusual scenario where she knows that she will get more information in the future. The Bayesian formulation of the problem tells you what she should believe after performing each update, but [the rest of the sentence deleted because I don’t think it’s quite right any more].
----
Upon further reflection, I think there is a general rule here:
When you evaluate a bet, you should evaluate it relative to what you will believe when the bet is resolved, which is not necessarily what you believe now. I’m going to call this the Fundamental Theorem of Betting, because it reminds me of Sklansky’s Fundamental Theorem of Poker, which says that the correct decision in a poker game is the decision you would make if all players’ cards were visible.
Under normal circumstances, we don’t know what we will believe in the future, so we almost always use our current beliefs as a heuristic for, or maybe estimate of, our future beliefs. Sleeping Beauty’s situation is unusual because she knows that more information is coming in the future, and she knows what the information will be!
To see how this theorem holds up, let me run the SB scenario and see if we can make sense of Sleeping Beauty’s beliefs and betting strategy:
Experimenter: Ok, SB, it’s Sunday night. After you go to sleep, we’re going to flip this fair coin. What do you believe is the probability that it will come up heads, P(H)?
Sleeping Beauty: I think P(H) is ½.
Ex: Ok. In that case, I wonder if you would be interested in a wager. If you bet on heads and win, I’ll pay 3:2, so if you bet $100, you will either win $150 or lose $100. Since you think P(H) is ½, this bet is in your favor. Do you want to accept it?
SB: Sure, why not?
Ex: Ok, on Wednesday I’ll tell you the outcome of the flip and we’ll settle the bet. Good night.
SB: Zzzzz.
Ex: Good morning!
SB: Hello. Is it Wednesday yet?
Ex: No, it’s not Wednesday, but that’s all I can tell you. At this point, what do you think is the probability that I flipped heads?
SB: Well, my prior was P(H) = ½. I’ve just observed an event (D = waking up before Wednesday) that is twice as likely under the tails scenario, so I’ll update my beliefs and conclude that P(H|D) = ⅓.
Ex: Interesting. Well, if the probability of heads is only ⅓, the bet we made Sunday night is no longer in your favor. Would you like to call it off?
SB: No, thanks.
Ex: But wait, doesn’t that mean that you are being inconsistent? You believe that the probability of heads is ⅓, but you are betting as if it were ½.
SB: On the contrary, my betting is consistent with my beliefs. The bet won’t be settled until Wednesday, so my current beliefs are not important. What matters is what I will believe when the bet is settled.
Ex: I suppose that makes sense. But do you mean to say that you know what you will believe on Wednesday?
SB: Normally I wouldn’t, but this scenario seems to be an unusual case. Not only do I know that I will get more information tomorrow; I even know what it will be.
Ex: How’s that?
SB: When you give me the amnesia drug, I will forget about the update I just made and revert to my prior. Then when I wake up on Wednesday, I will observe an event (E = waking up on Wednesday) that is equally likely under the heads and tails scenarios, so my posterior will equal my prior, I will believe that P(H|E) is ½, and I will conclude that the bet is in my favor.
Ex: So just before I tell you the outcome of the bet, you will believe that the probability of heads is ½?
SB: Right.
Ex: Well, if you know what information is coming in the future, why don’t you do the update now, and start believing that the probability of heads is ½?
SB: Well, I can compute P(H|E) now if you want. It’s ½ -- always has been and always will be. But that’s not what I should believe now, because I have only seen D, and not E yet.
Ex: So right now, do you think you are going to win the bet?
SB: Probably not. If I’m losing, you’ll ask me that question twice. But if I’m winning, you’ll only ask once. So ⅔ of the time you ask that question, I’m losing.
Ex: So you think you are probably losing, but you still want to keep the bet? That seems crazy.
SB: Maybe, but even so, my beliefs are based on the correct analysis of my situation, and my decision is consistent with my beliefs.
Ex: I’ll need to think about that. Well, good night.
SB: Zzzzz.
I think it's interesting to note that contrary to the frequentist's argument, sleeping beauty has actually learned a piece of information after she's being woken: she is being interviewed. That allows a Bayesian update indicating that it's not Tuesday Heads (unless she thinks that the experimenter is a trickster and might be interviewing her anyway, which means we also should be taking into account her prior degrees of belief toward that possibility and updating all of that information).
ReplyDeleteThe problem states that she will be interviewed on Monday and Tuesday if the coin is tails, and on Monday only if it's heads. So being interviewed doesn't tell her anything about which day it is.
DeleteBut you are partly right: the posterior differs from the prior, so the datum (waking up) does provide information, as strange as that seems.
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DeleteOops. I misread Mark's comment. Sorry, Mark, and thanks, Paul. You are right that SB learns that it is not Tuesday Heads, which refutes the halfer argument that SB learns nothing.
DeleteShe already knew on Sunday that she would not be interviewed on Tuesday if the coin toss was heads. She didn't learn anything new on waking.
DeleteI'd be interested in your reaction to the following extension. Before going to sleep on Sunday, Sleeping Beauty makes a bet at odds of 1.5:1 that the coin will come down heads. (This is favourable for her when the probability of heads is 1/2, and unfavourable when the probability of heads is 1/3). She is told that whenever she is woken up, she will be offered the opportunity to cancel any outstanding bets. Later she finds herself woken up, and asked whether she wants to cancel any outstanding bets. Should she say yes or no? (Let's say she doesn't have access to any external randomness to help her choose). Is her best answer compatible with a "belief of 1/3 that the coin is showing heads"?
ReplyDeletePersonally I think that to write "the thirder position is correct" may be overstating things. As a mathematical problem, it's not well-posed; there are various ways to formulate it rigorously as a conditional probability question, and different formulations have different answers. As a philosophical question, I find the 1/3 interpretation more appealing, but I don't know if that's a strong enough basis for saying one answer is "correct" and the other "incorrect"....
Yes, she should cancel the bet. Based on her prior, the bet is favorable, so she was right to accept it. But then she gets more information, because the event she observed (waking up) was more likely under one hypothesis than the other. So she updates her beliefs and, based on her posterior, the bet is no longer favorable, so she would be right to cancel it.
DeleteBut the policy "never cancel" leads to her winning money on average. (Specifically, if she stakes $100 for a return of $250 on heads, then her average gain is $25 - let's say the bet is resolved on Wednesday when she is woken up for good at the end of the experiment). This is better than "always cancel" in which case she never wins any money. So there is something odd about the logic which is leading to "always cancel".
DeleteThere would also be something strange about logic which would say she is right to make a bet that she knows in advance she is going to cancel. We could even suppose there is a non-refundable transaction fee of $5 to make the bet in the first place. Is she still right to make the bet (since her expected gain is more than $5), even though she knows she is going to cancel it? This seems like throwing $5 away.
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DeleteIf the bet is only resolved once (on Wednesday), then SB should accept the bet (and not cancel it) because she is effectively betting on a coin toss with favorable odds, and the whole sleeping-waking scenario is irrelevant.
DeleteBut if (as I assumed) the bet is resolved each time she wakes up, then I stand by my answer: if she bets based on her priors, she would accept the bet; then, after she gets more information, she would cancel it. There's nothing irrational about that, is there?
You are right that is it strange to accept a bet she knows she will cancel, but that's because it's unusual to know in advance that you will get information in the future. If you like, SB could think, "Based on what I know now, this bet is favorable, but I know that I will get information in the future that will change my beliefs and make this bet unfavorable, so I might as well decline it now."
Right, the bet should only be resolved once, sorry for not making that clear. (Otherwise, it's actually not right for her to accept it originally, if she loses twice whenever she loses, but only wins once whenever she wins).
DeleteSo, we agree that she should not cancel. But isn't there something odd? Put yourself in SB's position when you are woken up. You say that you have a "belief of 1/3 in the proposition that the coin is heads". The bet is unfavourable to you if the probability of heads is 1/3. And yet you don't cancel it. That suggests one sense in which you do NOT have a belief of 1/3 after all.
SB is not obligated to bet based on her current beliefs. If she knows that more information is coming in the future, she can compute a posterior based on that future information and bet accordingly.
DeleteEach time she wakes up, she should believe that she is more likely to be in the Tails scenario, but she also knows that more information is coming her way.
Specifically, she knows that when she wakes up on Wednesday, and is told that it is Wednesday and the experiment is over, she will update her beliefs again and conclude that the probability of Heads is 50% and the bet is favorable.
So when she wakes up on Monday or Tuesday and has the option to cancel the bet, she could think: "Based on my current beliefs, this bet is unfavorable, but I know that before the bet is resolved I will get more information that makes the bet favorable. So I will take that future information into account now and keep the bet (decline to cancel)."
Again, the weirdness is not in her beliefs but in the unusual scenario where she knows that she will get more information in the future. The Bayesian formulation of the problem tells you what you should believe after performing each update, but in this case you have some flexibility about when to perform the update.
In this betting example, Sleeping Beauty can take advantage of the extra information provided by each waking event to do better by canceling with a certain probability. I am still a novice at Bayesian stats, so before I tried calculating the optimum value using stats I ran a proof of concept test in Python. Sure enough, by canceling about 25% of the time Sleeping Beauty can increase her return from 50% to about 56%! If I get a chance in the next week or so I may sit down and generalize a formula for the optimum cancellation rate for an arbitrary number of waking events and bet payoff ratio.
DeleteThe betting example can't show anything unless you reconcile having a variable number of collection opportunities. But there is a way to make the problem easier to handle. Wake SB both days, but only ask her for her belief in Heads if it is Monday, or the coin landed Tails. And give her an hour to ponder what her answer will be, if she is asked (she knows both details on Sunday).
ReplyDeleteIn that hour, she knows that there is a 1/2 probability for either Heads or Tails, a 1/2 probability that it is Monday or Tuesday, and that these random variables are independent. So each combination has a 1/4 probability. That is, the prior probabilities are P(H&Mon)=P(H&Tue)=P(T&Mon)=P(T&Tue)=1/4.
She also knows some conditional probabilities: P(Askl|H&Mon)=P(Ask|T&Mon)=P(Ask|T&Tue)=1 and P(Ask|H&Tue)=0. From these, she can use Bayes Rules to say P(H|Ask)=1/3. (I'm not going to write it out; it's simple, but long in this format).
The thirder solution to the original problem is essentially the same. Different people take different steps trying to avoid saying that Tuesday, after Heads, exists as a potential observation opportunity for SB. They avoid it simply because she will be unconscious when, or if, it occurs. But it does exist, and her information content is the same in my variation when she is asked for her belief, as it is in the original when she is awake. How, or if, she would observe it is irrelevant when she observes it isn't the case.
The halfer solution is saying that Tuesday, after Heads, does not happen. The other three possibilities represent the entire sample space for her observation possibilities, and a prior-probability space must be constructed to reflect that sample space.
The sleeping beauty problem is ambiguous because it does not say what sample space she is using. Probabilities are defined on a per sample space basis. The sample space of the coin toss is {H,T} and the sample space for the questions about the coin state is {MH,MT,UT} where H=heads, T=tails, M=Mondays and T=Tuesdays. The probability of heads for the first sample space is 1/2 and the probability of heads for the second sample space is 1/3, since they are both equiprobable sample spaces. To see equiprobability, just notice that out of every 1000 coin tosses about 500 will be heads, 500 will be tails, and about 1500 questions will be asked about 500 which will occur when it is Monday and heads, another 500 which will occur when it is Monday and tails, and the remaining 500 which will occur when it is Tuesday and tails.
ReplyDeleteShe should use the probability for the sample space she assumed and the problem doesn't tell what sample space that is. The problem is bad because it introduces two different sample spaces without clarifying which one is operative. For example, if the problem also stated that for betting purposes on repeated trials of the experiment she should bet as much money as possible then it would be clear that she should use the sample space for the questions about the coin state to get the probability. But if instead of that we added to the original problem that she give the probability for repeated tosses of the coin then money won or lost is irrelevant and she should use the sample space for the coin toss to get the probability. The sleeping beauty question is ambiguous because it is asking about belief in the frequency of the truth value of occurrences of the PROPOSITOIN "the coin landed heads" not the proposition that the coin's probability of landing heads is 1/2. That is, the question doesn't make clear if it is asking about the probability of the proposition being true during repeated coin tosses or if it is asking about the probability of the proposition being true during repeated questioning in many repetitions of the experiment. These are not the same thing because when the coin is tails she is questioned twice but when the coin is heads she is questioned only once.
Now she knows the proposition is true one out of every three times she is asked and she is not going to mistake that for the fact that the coin comes up heads one out of every two times during coin tossing. So, for the proposition "the coin landed heads" the frequency of this proposition being true during repeated questioning in many repetitions of the experiment is different than the frequency of it being true during repeated coin tosses. If the coin toss actually came up heads then the proposition "the coin landed heads" is true but if the coin toss actually came up tails then the proposition "the coin landed heads" is false. How often the proposition is true or not depends on the circumstances. So adding two different prepositional phrases onto the original question highlights the ambiguity of that question:
(case 1)
What is your belief now for the proposition that "the coin landed heads" in the case of repeated questioning in repetitions of the experiment?
(case 2)
What is your belief now for the proposition that "the coin landed heads" in the case of repeated tosses of the coin?
The conclusion: the sleeping beauty problem is ambiguous because case 1 and case 2 use different sample spaces and if one removes the phrase "questioning in repetitions of the experiment" from case 1 and removes the phrase "tosses of the coin" from case 2 then the ambiguity of the original question is exposed.
"The sleeping beauty problem is ambiguous because it does not say what sample space she is using." I disagree. Whether or not she will be asked more than once, she only knows about one "questioning" whenever she is asked. And there is only one coin flip. Your alleged ambiguities are red herrings.
DeleteThe problem people encounter is indeed one of sample space, but it is because the sample space OF THE EXPERIMENT has four possible outcomes. They are {MH,MT,UH,UT} where H=heads, T=tails, M=Mondays and U=Tuesdays (I think that is what you wanted to U to be). Tuesday still happens, even if Heads is flipped; SB just does not observe it.
To see this better, use four Sleeping Beauties.SB1 will be left asleep under MH, SB2 will be left asleep under MT, SB3 will be left asleep under UH, and will be left asleep under UT. (The original SB is SB3 here.)
So three will be wakened on Monday, with the Monday Sleeper swapped for one of them on Tuesday. They will be put into a room together to discuss the question "What is the probability that I will be wakened exactly once?" Each SB knows her own combination, but is prohibited from telling the others.
When wakened - as each of them will be - each knows that two of their group will be wakened twice, and one just once, so their answers should sum to 1. Each knows that the information she possess is equivalent to the information possessed by the other two, so their answers have to be equal.
Their answers must be 1/3.
The answer is 1/2. Sleeping Beauty knows she is going to be awakened, so the fact that she is awakened does not change the probability that the coin is heads. (When she awakes, the probability that it is Monday is 3/4, and if it is Monday, then the probability that the coin landed heads is 2/3. 3/4 x 2/3 = 1/2.)
ReplyDeleteYou are looking at only one fact among the set of facts that are needed to represent her sample space. There are others. And even if she can't know them, she has to represent them when she creates her sample space.
DeleteOne way to demonstrate this is to wake her on Monday after Heads, and on Tuesday or Wednesday (but not Monday) after Tails. Do you agree that this can't change the probability of Heads for her when she is awake?
When she is wakened, she does not know whether it is Monday, Tuesday, or Wednesday. Since she was equally likely to be wakened on each of those days, and can't tell them apart, doesn't she have to say each has a probability of 1/3?
Then there is the irreconcilable paradox in your last statement. The coin does not have to be flipped until Monday Night. If you tell her that it is Monday, by your logic she must say that the probability that the fair coin you WILL flip tonight is 2/3.
How does the coin know to change, and what is the change?
"If you tell her that it is Monday, by your logic she must say that the probability that the fair coin you WILL flip tonight is 2/3."
DeleteIf you tell her that the coin hasn't been tossed, then she should rightly conclude that the probability of the (future) toss coming up heads is 1/2. But in providing her with that information, you have completely eliminated consideration of the possibility that the coin toss could have resulted in Beauty waking on Tuesday. In other words, you have changed the conditions of the "experiment" to where the coin toss cannot have any affect on the day.
If you don't inform her of status of the coin toss (performed or not), then she still has to take into account in her estimate that the toss could have come out tails. Since she knows that it's not Tuesday, that eliminates a possibility that could have occurred during a result of tails. The odds now favor heads by 2:1.
"One way to demonstrate this is to wake her on Monday after Heads, and on Tuesday or Wednesday (but not Monday) after Tails. ... When she is wakened, she does not know whether it is Monday, Tuesday, or Wednesday. Since she was equally likely to be wakened on each of those days, and can't tell them apart, doesn't she have to say each has a probability of 1/3?"
But she is NOT equally likely to be wakened on each of those days. One way to demonstrate the flaw in your reasoning is to ditch the coin and have her buy a lottery ticket with a one-in-a-million chance of winning instead. If her ticket is a loser, wake her on Monday; if her ticket is lucky, wake her on Tuesday and Wednesday (but not Monday). Since she could have been woken on any of those days and can't tell them apart, do you now believe that she has a 66% chance of winning the one-in-a-million lottery? Not unless you're a fool, you don't.
“But in providing her with that information, you have completely eliminated consideration of the possibility that the coin toss could have resulted in Beauty waking on Tuesday.”
ReplyDeleteIn what way? If it comes up Heads, she will not be wakened on Tuesday. If it comes up tails, she will. No conditions of the experiment have been changed. I can't fathom what you mean.
“But she is NOT equally likely to be wakened on each of those days.”
Yes, she is. There is a 50% prior probability to be wakened each day. That's what I meant. That's different than the probability that today is Monday, given that she is awake.
If there is a chance P that she will be wakened on Monday, and so 1-P that she will be wakened Tuesday and Wednesday, then the probability that today is Monday, given that she is awake, is P/3. For Tuesday or Wednesday, it is (1-P)/3.
“One way to demonstrate the flaw in your reasoning is to ditch the coin and have her buy a lottery ticket with a one-in-a-million chance of winning instead.”
The betting scenario is indeterminate because it does not address the significance of winning once, or winning twice. I showed how to get rid of that issue on August 28. Please, address that one.
“Do you now believe that she has a 66% chance of winning the one-in-a-million lottery?”
How did you get that number? You aren’t using conditional probability correctly.
On Sunday, she has a 1/1,000,000 chance that she will be wakened Monday, and a 999,999 /1,000,000 chance of being wakened Tuesday or Wednesday. So when she is awake there is a 999,999/3,000,000 chance that it is Monday and she is a loser, and a 1/3,000,000 chance for either Tuesday or Wednesday and she is a winner. So she perceives a 2/1,000,001 chance that she is a winner. In general, the formula is 2/(N+1) for a two-awakening event that has a 1/N chance.
BUT THIS PERCEPTION EXISTS ONLY INSIDE THE EXPERIMENT. It does not indicate the chances she will walk away with $1,000,000 after the experiment is over.
"If there is a chance P that she will be wakened on Monday, and so 1-P that she will be wakened Tuesday and Wednesday, then the probability that today is Monday, given that she is awake, is P/3. For Tuesday or Wednesday, it is (1-P)/3."
DeleteBut the sum of the probabilities of the only three mutually exclusive possibilities should be 1, right?
P(Monday) + P(Tuesday) + P(Wednesday) = 1
P/3 + (1-P)/3 + (1-P)/3 = 1
Solving for P gives P = -1. Congratulations, you've just asserted that the chance that she will be wakened on Monday is minus one. Can't you see that your numbers don't work?
"The betting scenario ..."
Who said anything about money or betting? I certainly didn't. This is just the probability of a lottery ticket being selected or not.
In any case, you messed this one up too. Let's say n="1 million" to shorten and generalize the numbers. There are three possibilities: Monday, Tuesday, or Wednesday. You have said that when she is awake P(Monday) = (n-1)/(3n) and P(Tuesday) = P(Wednesday) = 1/(3n). But
P(Monday) + P(Tuesday) + P(Wednesday) = (n+1)/(3n)
As before, the sum of the probabilities of the possible outcomes should be 1, which doesn't happen unless n=1/2. It's clear that your numbers are off.
The correct answer is P(Monday) = (n-1)/n and P(Tuesday) = P(Wednesday) = 1/(2n). Here you can see that
P(Monday) + P(Tuesday) + P(Wednesday) = 1
regardless of what n is. That's the general answer. Now, take it back to being a coin toss by letting n=2.
P(Monday) = 1/2 and P(Tuesday) = P(Wednesday) = 1/4. Since Monday is synonymous with "losing" (either the lottery or the coin coming up tails when heads is called) in your three-day problem, the probability of tails is 1/2 and so the probability of heads (the only other possibility) also is 1/2.
Thank you all for these comments. I will generally publish any comment that is relevant to the topic, but I encourage everyone to maintain a civil tone. I think you are more likely to find and resolve the point of contention if you keep the responses focused and polite.
Delete"But the sum of the probabilities of the only three mutually exclusive possibilities should be 1, right?" I’m going to use the more formal terms “exhaustive” and “disjoint.”
ReplyDeleteIf you have an exhaustive set of disjoint events, yes. If there is a condition you know about, which alters these properties of the event space, no. Then you need to create a new exhaustive set of disjoint events, and divide each probability in the new (posterior) set by the sum of their probabilities in the old (prior) set.
Usually, such a condition eliminates events from the prior set, and what you divide by is less than 1. However, in the Sleeping Beauty Problem, the condition takes two events that are the same event in the prior, and makes them mutually exclusive in the posterior. It can't be Monday and Tuesday at the same time *inside* the experiment, but the lab tech responsible for waking SB knows to be present on two days if the coin lands on Tails.
To SB, on the inside, the sum of the prior probabilities of the posterior sample space is 3/2. Each event in this posterior sample space has a prior probability of 1/2, so dividing then by 3/2 makes each 1/3 as a conditional probability.
Now, I know this is an unorthodox approach. But if you examine it, you will find it completely valid and self-consistent. It was when I tried to re-cast it as a more orthodox problem that I came up with the four-volunteers version. I would really like to see you address that one.
It would be nice if you would use some sort of mathematical notation. Your words merely call for more questions.
Delete"If there is a condition you know about, which alters these properties of the event space, no."
In the case of the SB problem, what is this additional information that you know? You never explain this.
"However, in the Sleeping Beauty Problem, the condition takes two events that are the same event in the prior, and makes them mutually exclusive in the posterior. It can't be Monday and Tuesday at the same time inside the experiment, ..."
Wait a minute. Are you saying that it CAN be Monday and Tuesday at the same time before the experiment begins?! Once again, a little notation would help, because you've just taken a statement that is obvious (a day cannot be both Monday and Tuesday) and then implied that under certain conditions it is not true. Once again, this is where a little mathematical notation would go a long way.
"I would really like to see you address that one."
You can't even explain effectively what you want to say here. How can I address any other of your "recastings" of the problem?
If I can read a little into what you've written, I think that you're trying to say that one can take a naive view of the problem for the priors -- such as that all three disjoint events in the exhaustive set are equally likely -- and then adjust for new information. The problem is that you don't have any new information in this problem.
The best you can do is to reexamine the probabilities after the day is known. In that case, using the naive priors above, you get P(tails) = 1 and P(heads) = 0 if it's Tuesday, and P(tails) = 1/2 and P(heads) = 1/2 if it's Monday.
But can you have confidence in those results? To do Baysean analysis responsibly, you need to check the sensitivity of the results (posteriors) to the choice of inputs (priors). In this case, the results for Tuesday are very robust. The posterior probabilities are unchanged except in the most extreme priors that posit that the probability of the event that beauty wakes on Tuesday after a result of tails is zero.
The probabilities calculated for Monday, however, are extremely dependent on the guesses supplied for the priors. The probabilities are P(tails) = P(heads) = 1/2 if all of the priors are 1/3, but if the (correct) priors -- i.e., P(heads & Monday) = 1/2 and P(tails & Monday) = 1/4 -- are used instead then the probabilities are P(tails) = 1/3 and P(heads) = 2/3. It's a case of garbage-in/garbage-out.
So the standard Baysean technique of guessing and trying to apply what data is available doesn't work in this case. One must start from the beginning and work the problem through in a more "orthodox" way to get a result that one can have confidence in, but it's not too difficult. This is a fairly simple problem.
“In the case of the SB problem, what is this additional information that you know?
ReplyDeleteThat she is inside the experiment, on an unknown day, and on a single day. This changes the conditions of her sample space from one where being awakened on Monday and Tuesday can be considered to be the same state of the experiment (OUTSIDE), to one where Monday and Tuesday must be considered to be different states (INSIDE).
“Are you saying that it CAN be Monday and Tuesday at the same time before the experiment begins?!”
You are not trying to see my meaning.
A random variable is a variable whose value is determined by chance or randomness (whatever you think that means; I take it to mean that you can’t know the value). There are two random variables when viewed from outside the experiment, COIN and WAKE. They can take on the values Coin={H,T} and WAKE={Mon,Tue}. Obviously, they are not independent. Calling the “outside” probabilities PO, PO(COIN=H)=PO(COIN=T)=PO(WAKE=Tue)=1/2, and PO(WAKE=Mon)=1.
Note that DAY, or TODAY, or NOW, or whatever you want to call the random variable representing what day “it can be“ is not a random variable of the experiment from this viewpoint. "It is" Sunday. Or Wednesday. So yes, you can, and do, know that a WAKE result will occur on both Monday and Tuesday if Tails gets flipped. Your confusion is thinking that means I’m saying the days can occur “at the same time.” In fact, this confusion is the root of the controversy over this problem.
I’ll stop using the more formal “Variable=Value” event notation if the variable is obvious. It’s tedious, and makes the text too long to read, imo.
You can also express unions of these events, PO(H&Mon)=PO(T&Mon)=PO(T&Tue)=1/2. These don’t sum to 1, because T&Mon and T&Tue are the same event OUTSIDE THE EXPERIMENT, where the random variable representing time is WAKE. But note that by using these events, I have isolated the individual days as separate events.
There are also two random variables inside the experiment, but one has a different meaning. They are COIN and TODAY, with the same possible values. TODAY has a different meaning inside, than WAKE did outside, since today can only be one day. Specifically, while (COIN=T)&(WAKE=Mon) and (COIN=T)&(WAKE=Tue) represent the same event outside the experiment, (COIN=T)&(TODAY=Mon) and (COIN=T)&(TODAY=Tue) are different events inside the experiment.
Now note that these “inside” events, with “TODAY,” as the variable, represent the same events as the “outside” events with WAKE but that have been isolated to individual days. So their prior probabilities – the probabilities where the exhaustive, disjoint set does not have to sum to 1 – are the “outside” probabilities for equivalent outside events. PI(H&Mon)=PI(T&Mon)=PI(T&Tue)=1/2. These normalize to 1/3.
Of course, I could have said the same thing much simpler by saying that SB cannot detect a difference between the three possible situations on the inside, and that each situation had the same chance to arise on the outside. But then, I already did, and Elga said as much in his original argument.
And with all due respect, I think what I have said has been clearly expressed. The four-volunteer variation, in particular, cannot be misinterpreted. Its answer can only be 1/3. Please tell me if you agree, or why you don’t.
"One must start from the beginning and work the problem through in a more 'orthodox' way to get a result that one can have confidence in, but it's not too difficult." Actually, it is impossible, because of the change or variables. Lewis' halfer argument ignored it. Elga's thirder argument dodged around it, probably because of the potential for the confusion you exhibited above. But the four-volunteer version removes that concern.
Here's a summary of that last, overly-wordy response: Every halfer argument confuses the result "SB will be wakened on Day X" with the result "I, SB, am awake and it is Day X." Whether or not you agree with my treatment of this difference, and whether or not the correct answer is 1/2 or 1/3, these halfer arguments are invalid because they ignore this difference.
ReplyDeleteBut my variation of the problem, where four volunteers participate and a different one is left asleep under each possible combination of the coin flip and the day, removes the need to address the difference. Three will be awake at any time, they are functionally indistinguishable to each other when they are awake, and for exactly one of those three the actual coin result will be her assigned coin result. The confidence that any of them should have, that this describes her, can only be 1/3.
"This changes the conditions of her sample space from one where being awakened on Monday and Tuesday can be considered to be the same state of the experiment (OUTSIDE), to one where Monday and Tuesday must be considered to be different states (INSIDE)."
DeleteYou call "outside" really means "will (eventually) be woken on" and what you call "inside" really means "has woken on." But these are completely different things. One cannot be used as the prior for the other. That's not how Baysean analysis works! The former is just a stating of whether the coin toss is heads or tails. The latter includes information about what day the subject (Beauty) wakes on.
"Of course, I could have said the same thing much simpler by saying that SB cannot detect a difference between the three possible situations on the inside, and that each situation had the same chance to arise on the outside."
This is a typical (flawed) line of reasoning used by the 1/3 crowd. It almost seems reasonable when the odds of the random event are even, such as a coin toss. The flaws in the reasoning become obvious when the chance of the coin coming up tails is replaced with something much more extreme, like picking this week's lucky lottery numbers. If your reasoning is that "SB cannot detect a difference between the three possible situations on the inside" then you must conclude that her chance of picking the lucky lottery numbers is 2/3. She believes that she has two-to-one odds of being a lottery winner. I predict that she will be happy for a while, and then she will be very sad when she learns that she is not actually the winner.
"But then, I already did, and Elga said as much in his original argument."
No. Elga's argument focused on what happens on Monday. He reasoned (incorrectly) that, since the coin could be tossed either before or after the subject was woken without affecting the experiment, the toss has nothing to do with what happens on Monday. Therefore, the odds on Monday are always 50/50% heads/tails. This is fallacious reasoning, however.
If Beauty is told that the coin has not yet been tossed, then she should conclude that P(Heads) = P(Tails) = 1/2, due to the fair coin assumption, but in this case, she has gained additional information. If she doesn't know what the status of the coin toss is, then she must consider the possibility that it has already occurred, and if it has already occurred then it could have come up tails and if it came up tails Monday could already be gone and it could actually be Tuesday.
If she is told that today is Monday, then she can eliminate the last possibility, which means that the probability of the coin coming up tails decreases from 1/2 to 1/3. The probability of heads correspondingly increases to 2/3.
Elga's mistake was not to take this into account. Then he reasons backwards through the probabilities to reach the conclusion that Beauty should assume that the probability of heads is 1/3 -- even though this is inconsistent with the own assumptions that he has established in his own restricted form of the problem (i.e., that we are considering what happens on Monday).
"And with all due respect, I think what I have said has been clearly expressed. The four-volunteer variation, in particular, cannot be misinterpreted. Its answer can only be 1/3. Please tell me if you agree, or why you don't."
DeleteRelabeling the problem does nothing to bolster your arguments. The alternate case put forward (independently) by Arntzenius and Dorr a dozen years ago is better constructed and somewhat more convincing, but it is still flawed. In their scenario, all four possibilities are originally possible -- i.e., heads or tails, Beauty wakes up on both Monday and Tuesday. But the kicker is that if she wakes on Tuesday when the coin toss was heads, she soon gains information about the coin toss and the day. (How this magic occurs differs between Arntzenius and Dorr, but for our purposes the result is the same.)
So they reason that she originally had to assign a 1/4 probability to each possibility (i.e., even odds). Then when one possibility (the combination of heads and Tuesday) is removed, the remaining alternatives also share the same probability. They ignore, however, that the outcome of Event A (the coin toss) completely determines the probabilities of the outcome of Event B (which day SB wakes). Once the proper order of influence is established, it's obvious that eliminating the possibility of heads and Tuesday means that whatever chance of that now-impossible event occurring has to be folded into the probability of heads and Monday.
Your twist on this flawed argument is not as convincing as theirs. Sorry.
I have posted two arguments here: One unambiguously and clearly shows that the answer must be 1/3. It is different than Arntzenius and Dorr's in that it doesn't have the flaw you think you identified there. But it is not a mathematical derivation.
ReplyDeleteThere have been many attempts at such derivations, and since they get different answers, there must be something wrong in at least half of them. It could be both, so finding what one thinks is a flaw does not prove one is correct. You think you found one in the thirder arguments. My unorthodox relabeling was an attempt to show one in the halfer argument. Neither is conclusive.
But I still have unambiguously and clearly shown that the answer can only be 1/3. You conveniently avoiding saying why you thought it was flawed - you just asserted that it is. So whether or not I identified the halfer flaw to your satisfaction isn't as relevant as the fact that the I have shown the answer can only be 1/3.
The closest you came to identifying a flaw is this instance of affirmignteh consequent: "It almost seems reasonable when the odds of the random event are even, such as a coin toss. The flaws in the reasoning become obvious when the chance of the coin coming up tails is replaced with something much more extreme, like picking this week's lucky lottery numbers."
Just because the Principle of Indifference can be applied incorrectly, it doesn't mean I did. This is the type of argument that makes me feel you are fitting your arguments to your conclusion. The lottery example is a flaw because the Principle of Indifference is not that, when have N disjoint cases, the chances of each is 1/N regardless of the circumstances under which they arise. It applies only when those circumstances can be demonstrated to be identical. AS I DID.
Besides, and like you pointed out before, if the answer to the original SB problem is 1/2? Then unless you can point to a way that any of three awake volunteers is NOT answering the original SB question, you have three disjoint cases where the sum of the probabilities is greater than 1.
So I repeat: In my four-volunteer variation, an awake volunteer is in a circumstancially-identical situation with two others. There is an experiment state that applies to exactly one of them. And its probability is the same probability as the probability in the original SB problem.
That probability can only be 1/3.
"So I repeat: In my four-volunteer variation, an awake volunteer is in a circumstancially-identical situation with two others. There is an experiment state that applies to exactly one of them. And its probability is the same probability as the probability in the original SB problem."
DeleteSo what happens when you have four volunteers and the coin toss comes up heads all four times? (It will one out of 16 times you try this.)
Your "unorthodox" solution has no relevancy because you are not considering the original problem. You're not even in the same neighborhood. (Hint: You're not allowed to pick your "volunteers" for this exercise.) If you've removed all of the randomness from the problem then you don't get to talk about probability. That's just the way it works.
Sorry, but you're not even wrong. You've decided to go bark up a completely different tree.
"So what happens when you have four volunteers and the coin toss comes up heads all four times?"
ReplyDeleteThere is only one coin toss, not one for each volunteer. Each volunteer is wakened once, or twice, based on the same coin toss. Each will be left to sleep under a different combination of the day, and that one coin toss. Since there are four combinations, and four volunteers, each combination is represented exactly once.
Three volunteers will be wakened each day, depending on the day and that one coin toss. Each is asked for the probability that the one coin toss result is the one in her "leave asleep" combination. So for each, the question is equivalent to the question in the original SB problem. She may always wake up on Tuesday, and after Heads, but those are equivalent versions.
The three volunteers cannot give different answers, because it is effectively the same question that is asked of each. The sum of the three answers must be one. The answer is 1/3.
"Your 'unorthodox' solution has no relevancy because you are not considering the original problem. You're not even in the same neighborhood. (Hint: You're not allowed to pick your "volunteers" for this exercise.) If you've removed all of the randomness from the problem then you don't get to talk about probability. That's just the way it works."
With all due respect, there isn't a single thing that makes sense to me in any of that. This isn't about the "unorthodox solution," which is a way of trying to derive a mathematical solution. It is a demonstration that the probability must be 1/3. I am not "picking" a volunteer; and even if I were, there is no reason I can't. I haven't removed the randomness from the problem in any way, I just used the same randomness in different, but equivalent, ways for each of the four.
I thought I had expressed the four-volunteer problem unambiguously. I guess I hadn’t, since it was misinterpreted. I still don’t see how it could be misinterpreted, but here is a fixed version:
ReplyDeleteYou, and three other women, have volunteered for the following two-day experiment. Each of you will be wakened once, or twice, based on the same flip of a fair coin. Two of you will be in a set assigned Day=1, and the other two will be in a set assigned Day=2. One in each of those sets will be assigned Coin=H, and the other in each set will be assigned Coin=T.
After being drugged into indefinite sleep on the evening of Day 0, the fair coin will be flipped. Represent the result with the variable C. On Day 1, all of the volunteers will be drugged awaken EXCEPT the one assigned Day=1 and Coin=C. The volunteers will not know what day it is. After they answer a question, they will be given an amnesia drug and be drugged back into indefinite sleep.
On Day 2, this process will be repeated, except that the volunteer left asleep will be the one assigned Day=2 and Coin=C.
The question that the awake volunteers are asked is “What is your confidence (i.e., the probability) that C is the coin value you were assigned?”
On each day, three volunteers will be wakened. Since they don’t know what day it is, each has the same information upon which to base her confidence. So each confidence must be the same. But C will be the value assigned to exactly one of these three, so their confidences must add up to 1.
That confidence can only be 1/3. And the original SB problem is exactly the same as the problem posed to one of these volunteers.
"But C will be the value assigned to exactly one of these three, so their confidences must add up to 1."
DeleteWhy?
You have four people (with one still sleeping) who are in four different situations. If you had ONE person who could be in either state A OR state B OR state C, with no other possibilities, then yes the probabilities of these three states should sum to 1, but that is not the case here. There is no such requirement.
I agree that all three should give the same answer, and that answer is that the probability of a fair coin toss is 1/2. Just because there might be someone napping in the other room doesn't change anything.
Here, I should also add that I am assuming that each volunteer does not know the assignments of the other volunteers. Otherwise the three who are awake could take note of who is missing and know BOTH the day and the result of the coin toss at once.
Delete“Here, I should also add that I am assuming that each volunteer does not know the assignments of the other volunteers.” Yes, you should. I said as much explicitly the first time I posted the problem. But it is getting increasingly difficult to find ways to prevent any possibility, however slight, of misinterpretation.
ReplyDelete“Why [will C be the value assigned to exactly one of these three, so their confidences must add up to 1]?” I refer you back to your comment of October 14, 2015 at 5:24 AM: “But the sum of the probabilities of the only three mutually exclusive possibilities should be 1, right?”
You are ignoring that you have evidence in this problem. That there is "someone napping in the other room." It changes something, despitge your assertion that it does not. It represents a state that does not apply to any of the three awake volunteers, but does depend on the random variable Coin. That's what "new information" about Coin means.
Say you wake up on a small tropical island, with no idea how or why you are there. You find two other people on the island, in the same state of confusion. Posted on a palm tree, you find a message saying “Two of you were chosen to be here by method X. One was chosen by method Y. To be rescued, correctly determine the probability, for each of you, that you were the one chosen by method Y.”
It does not matter what methods X and Y are. It does not matter what the prior probability, that either would have picked you, is. It does not matter what group of people - i.e., those "napping in the other room" - you were chosen from. It does not matter if you will be rescued only to find yourself on another tropical island in the same state. It does not matter if any of the others will be rescued only to find themselves on another tropical island in the same state. It does not matter if the random events used in these methods are the same, or different, or correlated. It does not matter if, as in the original SB problem, the definitions of what constitutes a random event are blurred by such a repetition.
There are three of you, each with identical information as it pertains to the question of method X or method Y. One of you is there because of method Y. Two are not. The chances it is you are 1/3. This is trivial result.
And it is exactly the situation the three awake volunteers find themselves in. One of which is exactly the original SB problem.
You keep trying to re-cast this problem into the external problem, when the entire point is to remove that consideration.
“If you had ONE person who could be in either state A OR state B OR state C, with no other possibilities, then yes the probabilities of these three states should sum to 1, but that is not the case here.” You didn't say why you think it isn't - you just asserted it, as you have asserted all of your points in this discussion.
It is the case here. The three states are “Assigned this day and Coin=~C, assigned the other day and Coin=~C, and assigned the other day and Coin=C.” One of you was assigned Coin=C. What was assigned to another person, is irrelevant (except as it helps to define what was assigned to each of you).
"You are ignoring that you have evidence in this problem. That there is 'someone napping in the other room.' It changes something, despitge your assertion that it does not."
DeleteNo, as Bertrand's box has shown us, not all additional information is useful. In that case, it was the type of coin drawn from a box, which provided no information about the probability of choosing box 3. In this case, the fact that someone is still sleeping tells one nothing about the result of the coin toss.
http://en.wikipedia.org/wiki/Bertrand's_box_paradox
It's very easy to fool yourself that it does, which is why Bertrand's analysis is still described as a "paradox."
You have a penchant for overly complicated examples that even you yourself admit that you explain poorly. Please let me simplify your thought experiment to something that should be more easily comprehensible.
We have four volunteers. We'll assign each of them one of four unique values: HH, HT, TH, and TT. The "H" (heads) and the "T" (tails) correspond to the results of two coin tosses, and the order is important. For example, "HT" corresponds to heads on the first toss and tails on the second. Each volunteer knows his own assigned value, but not any of the others. We have a researcher flip the two coins outside the room where nobody else can see. He then enters the room and points to the volunteer corresponding to the result of the double coin toss, who then leaves the room.
Now, we have three people left in the room. What should be their estimate that the FIRST coin toss came up heads? What should have been their estimate if nobody had been asked to leave the room?
In both cases, it should be 1/2. Whether someone leaves the room or not tells them nothing about the toss.
Upon further reflection, it seems pointless to me to even consider this "four volunteer" problem, because it is a completely different problem than the original SB problem. The difference can be summed up as follows: in the original problem Sleeping Beauty doesn't get to sleep in on Tuesday if the coin is heads.
ReplyDeleteReturning to my two-coin problem above, I forgot to include the essential part of the SB problem that the first coin toss affects the outcome of the second. That is, taking the first coin toss to be the coin toss in the SB problem and the second to be the day, if the first comes up heads, it will ALWAYS be Monday. No second toss is necessary, and any problem that assumes that a Tuesday is somehow possible (and had been eliminated) is not the same problem as the Sleeping Beauty problem.
So forget what I wrote above. It doesn't apply to the Sleeping Beauty problem. The "four volunteer" problem that has been posited here is completely irrelevant to the Sleeping Beauty problem and is pointless to discuss any further.
My understanding is that there is only one toss of the coin:
Delete"A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday."
So the coin is tossed, and 50% of the time she's woken once (Heads), while the other 50% of the time she's woken twice (Tails). Ergo she's more likely to find herself being woken up after the coin landed Tails, and her belief should correspond to that probability. What with her being perfectly rational or something like that.
"Ergo she's more likely to find herself being woken up after the coin landed Tails ..."
DeleteNo. You're assuming that all awakenings occurs with the same probability. They do not. If she wants to guess the results of the coin toss more often than not, then she should always guess tails. This is not because the coin is more likely to come up tails than heads, however. It's just that her "payoff" of guessing the toss correctly is paid with odds of 2 to 1.
Half of the times that the coin is tossed, she'll end up guessing wrong.
"Unknown" is assuming that the three possible states where SB is awake all occur with the same probability. Not "all awakenings." The thee states are {H,Mon}, {T,Mon}, and {T,Tue}.
DeleteTwo of those states correspond to what Brian seems to want to call "the Monday awakening," and one to "the Tuesday Awakening." So "Unknown" is assuming "the Monday awakening" is twice as likely to occur.
There is a fourth state that exists, {H,Tue}, but it does not correspond to an awakening. It has the same probability to occur as the other states, but since SB is not awake she does not observe it. It is the elimination of this state, for an awake SB, that constitutes "new information."
“As Bertrand's box has shown us, not all additional information is useful.” If that is your takeaway from the Bertrand Box Paradox, then you completely missed its point.
ReplyDeleteThat point can be equated to “you need to consider all of the information, whether or not you think it will affect the result.” In that problem, there was a 50/50 chance to draw a gold or silver coin, and Bertrand’s point was that you need to include that proportion in your calculation. Read the part of the Wikipedia article you cited where the actual solution uses Bayes Rule, not the part that starts “Alternatively, one can simply note…”. The alternate is a demonstration of how the 1/2 answer can be right, not a proof that it is, and quite similar to where I previously suggested an alternate and you disagreed with.
In the SB problem, it isn’t “the fact that someone is still sleeping” that we use. It is the fact that there are exactly three equally-likely paths that can lead to being awake at the moment, even though there are four paths in total.
“It's very easy to fool yourself that it does, which is why Bertrand's analysis is still described as a ‘paradox.’” What Bertrand called a “paradox” is how he used two seemingly-valid approaches to get different answers. Not the problem itself, as you imply here.
His point was that one solution didn’t use available information, so it was wrong. Not that the answer was wrong, even though it was. It was a cautionary tale that is largely ignored in modern times. For example, there is an equivalent problem (four boxes, not three, two having one of each kind of coin) that most modern treatments get wrong: “Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?” The same two paradoxical solutions can be applied here. The one Bertrand called incorrect gives the answer most modern treatments get, 1/3. The one he called correct gives 1/2 - and no, it is not the same as the “older child is a boy” variation.
The author of this particular problem, Martin Gardner, even admitted this paradox, and retracted his original answer of 1/3. But most people remember only that original answer.
“You have a penchant for overly complicated examples that even you yourself admit that you explain poorly. Please let me simplify your thought experiment to something that should be more easily comprehensible.” My examples are well explained. What I admit is that they get misinterpreted. I don't think it is deliberate, but a result of not wanting to accept the result and so finding any way imaginable to misinterpret it.
"[In my two-coin example] we have three people left in the room. What should be their estimate that the FIRST coin toss came up heads?” It depends on what each person’s combination is, and if they know it, which makes this different from my example. But you still get the answer wrong.
HH and HT (if they are in the room still, and know their combination) should say 1/3. TH and TT should say 2/3. If they don’t know their combinations, each would say 1/2. But if you make it closer to my variation of the SB problem, you would ask for the probability the first coin matched their own first coin. On that case, everybody answers 1/3.
Make a grid of the possibilities, if you don’t believe me. I did. Even though I knew the answers without such a grid, I made it anyway. I don’t trust anybody, let alone myself, enough to ignore any information that exists.
“Upon further reflection, it seems pointless to me to even consider this ‘four volunteer’ problem, because it is a completely different problem than the original SB problem.” For the volunteer assigned {H,Tue}, it is 100% identical. For the others, it is functionally equivalent.
ReplyDelete“In the original problem Sleeping Beauty doesn't get to sleep in on Tuesday if the coin is heads.” In my variation, the volunteer assigned {H,Tue} “gets to sleep in,” or “doesn’t get to sleep in” (whichever you meant), under the same circumstances, on either day, in the exact same way. There is nothing different about her experience, real or possible, except that she knows of the other three volunteers.
You can try to find a difference along this line of thought if want. But IMO it only proves you do not want to accept the result, and are grasping for a reason to disbelieve it. If you disagree, please explicitly identify the difference for the volunteer assigned {H,Tue}, and why it should affect her answer.
"I forgot to include the essential part of the SB problem that the first coin toss affects... the day]." And this is the classic error made by all halfers.
The (first) coin toss does not affect the second coin toss (or the day, as I changed it to in order to show your intended meaning). It affects your ability to observe the second toss (or day) under the conditions of either experiment. The event still has a possible existence that was made impossible by the observed circumstances. As such, it constitutes "new information," and it must be removed from the sample space to calculate the probability once that information is obtained.
What seems to confuse halfers is not knowing what the information is, only its form. It still exists, and the form is important because it rules out possible states.
+++++
AGAIN: In my example, you are one of three people who are awake due to what can only be considered to be equivalent circumstances based on the full set of information they each possesses. Exactly one of these three is in the random state that the probability question asks about. The probability that it is you can only be 1/3.
You can assert that it isn't equivalent all you want, but I have shown that it is.
You asked me to evaluate your "four volunteer" problem, and now I have. The answer to the question you posed is 1/3 (although from different reasons those you're stating), but it doesn't matter because it is an entirely different problem than the Sleeping Beauty problem. There. I'm done.
DeleteYou're making all of the classic mistakes that people make when they try to analyze this simple problem, almost as if you were checking them off some list. You claim that because there are three possibilities, their probabilities should be the same. You mistakenly conclude that the result of the coin toss should be 50/50% if it is Monday. You make up whole new scenarios and (wrongly) claim that they are the same as Sleeping Beauty. Now, you've gone off on a completely different tangent and dragged up incorrect stuff that you had been posting on another part of this blog.
Your solutions don't properly generalize to the situation when the coin toss is replaced with some other random event with different odds. The answer you gave when I brought up the general problem made absolutely no sense and was devoid of any consistency, as I demonstrated. But as you have done time and time here, you just keep changing the subject.
Frankly, there's no where else to go. Thanks.
Yes, I asked you to evaluate it. I had hoped for an actual evaluation; not mere statements that it is wrong, with no explanation for why you think so. Yet it clearly is an exact analogy to the SB problem. If you want to disagree, please show me what you think the difference is.
ReplyDelete"You claim that because there are three possibilities, their probabilities should be the same." No, I demonstrated that their situations are symmetric, which is sufficient to *prove* that the probabilities must be the same. You did not contest this demonstration, which means that you can't call it a mistake to make that claim.
I also showed how one of the volunteer's state is exactly the original SB problem, without even needing to resort to the symmetry that exists. You have not contested this, either.
"You mistakenly conclude that the result of the coin toss should be 50/50% if it is Monday." That has nothing to do with the four-volunteer problem, although it is easily provable the same way. You conclude it should be 2/3 based an assertion - not a proof - that the answer when it is not know it is Monday is 1/2.
"Your solutions don't properly generalize to the situation when the coin toss is replaced with some other random event with different odds." Yes, they do. In fact, treating the day as "some other random event" is what Elga did, and is what leads to his 1/3 answer. It is necessary to treat it like it isn't "some other random event" to get 1/2.
Your approach to that generalization was outright wrong, which is easy to see by simply listing the cases. There are four equally-likely ways to flip two coins: HH, HT, TH, and TT. If the volunteer you assigned HH remains in your room, see knows that the combination HH was not flipped. But the other three remain equally likely, so her conditional probability for the first coin having landed Heads is 1/3.
Any student in his or her first class in probability can get that result. Yet you say it is 1/2, again without justifying it. And then you say I am wrong? Without justifying a different answer?
"The answer you gave when I brought up the general problem made absolutely no sense and was devoid of any consistency, as I demonstrated." Since you made just one reply since I addressed the generalized problem, and you did not attempt to demonstrate anything in that reply, I am again at a loss to see what you could possibly mean. The only thing I was "inconsistent with" was your wrong answer to it.
Yes, I agree there seems to be nowhere to go. You will not seriously consider any argument that leads to an answer you don't like, to the point of refusing to understand simple explanations.
"'You claim that because there are three possibilities, their probabilities should be the same.' No, I demonstrated that their situations are symmetric, which is sufficient to *prove* that the probabilities must be the same. You did not contest this demonstration, which means that you can't call it a mistake to make that claim."
DeleteNo, not at all. After I got over your very poor explanations and finally understood the problem that you had constructed, I agreed with you. Yes, the answer to the your problem is 1/3, but having to resort to four volunteers and symmetry is overkill and a very clumsy way to reach this conclusion. You only have to consider one person -- see below.
"I also showed how one of the volunteer's state is exactly the original SB problem, without even needing to resort to the symmetry that exists. You have not contested this, either."
My original mistake in trying to figure out your problem was not to challenge this assertion. Once I finally understood what you were asking, I realized that THIS is where you made your mistake. Your toy problem is not the same as the Sleeping Beauty problem, but it is relevant to another toy problem that you have mentioned -- see below.
"Your approach to that generalization was outright wrong, which is easy to see by simply listing the cases. There are four equally-likely ways to flip two coins: HH, HT, TH, and TT. If the volunteer you assigned HH remains in your room, see knows that the combination HH was not flipped. But the other three remain equally likely, so her conditional probability for the first coin having landed Heads is 1/3."
Yes, that is correct. What is hilarious is that you're being completely inconsistent and don't even realize it. Change "H" to "B" (boy) and "T" to "G" (girl). Now you have a modified version of the two-girl problem (BB, BG, GB, and GG -- BB is out), which you've sworn over and over again has a probability of 1/2, not 1/3.
The Sleeping Beauty problem is not the same as the two-girl problem. To put it in terms of boys and girls, you'd have to postulate the following: a couple really wants a boy, but they don't want more than two kids. Thus, they plan to have children until either (1) they have a boy or (2) they have two children, regardless of their sex. The question is what is the probability that their first child was a boy?
Naturally, the answer is 1/2, which is also the probability that SB should assign to the coin coming up heads unless she gets more information than she has upon waking up.
Paraphrasing directly from Martin Gardner, what is the probability that the following men have two children of the same gender if:
ReplyDeleteYou know that Mr. Adams has two children. You ask him if at least one is a girl. He says “Yes.”
You know that Mr. Brown has two children. You ask him to tell you something about one of them. He says “I have a son who is staring in his school play.”
You know that Mr. Smith has two children, and that at least one is a girl.
The answers are:
#1: 1/3, as you derived.
#2: 1/2. This version is Bertrand’s Box Paradox, with four cases instead of three. Mr. Brown could have told you about a daughter, or a son. His choice cannot affect the probability of matching genders, which started out as 1/2 and so must remain 1/2. The technical solution is P(BB|TellB) = P(TellB|BB)*P(BB) / [P(TellB|BB)*P(BB) + P(TellB|BG)*P(BG) + P(TellB|GB)*P(GB)]. Now, P(TellB|BB)=1, but the fact that it was his choice says P(TellG|BG)=P(TellB|GB)=1/2, making the answer 1/2.
#3: It’s ambiguous, because it depends on how you know about a girl. It could be the first method, or the second, or even a different method with a different answer. HOWEVER, if you must give an answer, and so must make an assumption about the method, the method in #2 is the only one you can assume. (More accurately, you assume only the conditional probability P(TellG|BG)=P(TellB|GB)=1/2.) The answer is 1/2.
I have talked about this in other forums, but not here (except to mention that Solution #2 is a variation of the Box paradox) before now. And I have been completely consistent about these answers in all places. Your two-coin generalization is #1.
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You are right, there are some similarities between the Two Child Problem, and the Sleeping Beauty Problem. But they are not the same. However, it is the Version #1 of the Two Child Problem, as I described it above, that has this similarity. For it to be Version #2, she would be wakened on only one day, with a choice after Tails.
“Having to resort to four volunteers and symmetry is overkill and a very clumsy way to reach this conclusion.” I agree, it is clumsy. But halfers have made it necessary by confusing the definitions of the events that SB can use when she is awake, with the ones she can define before she is first put to sleep.
On Sunday Night, being awake on Monday after Tails is the same future that has her awake on Tuesday after Tails. They are the same event, which has probability 1/2. When she is awake, being awake on Monday after Tails is a different present than being awake on Tuesday after Tails, so these are disjoint events in her current frame of reference. They each have prior probability 1/2, but so does a present where she is awake on Monday after Heads. As does the present where she is asleep Tuesday after Heads. There is no contradiction in the fact that these probabilities add up to 2, since in this description I reference two samples – a present on Monday, and a present on Tuesday.
The awake SB knows that her present corresponds to only one present, is not Tuesday after Heads, and does not reference a possible future or past awakening. She has “new information” – it is not Tuesday after Heads - and so can calculate P(Heads|Awake)=1/3.
That was a technical solution, and it is correct. But it takes one side in the ambiguity of the time frames that is the source of friction between halfers and thirders. You say my descriptions have been confusing, but it is only because you keep using the hafler definitions that you find it so.
The four-volunteers version, clumsy as it is, is the only way to remove the need to address these differences. The possibilities that remain are all represented in the present, not an ill-defined past or future. But for the volunteer who was assigned {H,Tue}, her situation is identical to the original Sleeping Beauty Problem. If her answer in the four-volunteer version is 1/3, it has to be 1/3 in the original.
#3 is not ambiguous. It's exactly the same as #1.
Delete#2 ends up being 1/2 because you are using additional information that you don't have in the other two that distinguishes the two children and puts them into two exclusive categories. You are using the information that Mr. Brown spoke about one child -- AND (implicitly) that he DID NOT speak about the other. Now you can identify the two children as "spoken about" and "not spoken about" and you know that the one spoken about is a boy. The ambiguity of which child is a boy has been removed. Only the other child needs to be considered and it's 50/50 a boy or a girl.
You could have used the information about the school play if you had complete confidence that the other child is NOT and CANNOT be in the school play. Once again, if there is additional information, it can be used.
In #3, there is no additional information about which child is a girl, so it is reasoned exactly the same way as #1, which is the same problem. You can't just throw in information that you think you MIGHT have known. You get to work only with what you've got. For example, I can't assume in the problem that Mr. Brown might really have only one child because there's a finite chance that the other child was hit and killed by a bus and I just don't know it.
"She has 'new information' -- it is not Tuesday after Heads -- and so can calculate P(Heads|Awake)=1/3."
But she doesn't have any new information. She knew for certain on Sunday night before she went to sleep that she was not going to wake on Tuesday if the coin came up heads. She reasoned then that the probability of heads is 1/2. Nothing has changed. The probability stays the same.
In the October, 1959 issue of Scientific American, Martin Gardner wrote: "Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information 'at least one is a boy' is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says 'at least one is a boy.' If both are girls, he says 'at least one is a girl.' And if both sexes are represented, he picks a child at random and says 'at least one is a ...' naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics."
ReplyDeleteGardner's last sentiment has be reaffirmed many times, as the best mathematicians continue to make the same mistake over and over again. The mistake is the assumption that "I know X about case Y" means that "I specifically sought out an example of case Y where X is true." This is clearly an an example of the logical fallacy called "affirming the consequent," where you take the true statement "If A then B" and deduce, incorrectly, that "If B then A" is also true.
"In #3, there is no additional information about which child is a girl, so it is reasoned exactly the same way as #1." This, however, is correct. There is no new information whatsoever pertaining to the question "do both children have the same gender?"
ReplyDeleteSince the probability of that was 1/2 before the information "at least one girl" was obtained, and we have agreed that this isn't "new information," then the answer must still be 1/2.
To change it to 1/3 requires "new information," which would be obtained if, for example, we had asked Mr. Adams "Is at least one a girl?" So in my versions of the Two Child Problem above, #3 is indeed fundamentally different than #1.
Brian, I really do want to try to discuss these things openly and civilly. But it doesn't help when you keep misrepresenting everything I say.
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The volunteer assigned {H,Tue} is in an experiment that is, for all purposes pertaining to the information she has, identical to the original Sleeping Beauty Problem. Brain has agreed that she should say the probability of Heads is 1/3 when - not if - she is awakened in that version.
This does require her to identify "new information." That information is "On Sunday Night, I knew for a fact that I would exist in exactly one of four different states at any time during the experiment. They are {H,Mon}, {H,Tue}, {T,Mon}, and {T,Tue}. I also knew for a fact that, at any point in time, I was equally likely to be in any of these four states. But when I find myself awake, I have learned that I am not in the state {H,Tue}, but nothing new about the other states. In fact, the supposition that I have learned nothing about the set of four states is absurd, since those who suppose it, use it to change P({H,Mon})=P({T,Mon}) to P({H,Mon})=2*P({T,Mon}), and P({H,Mon})=P({T,Tue}) to P({H,Mon})=2*P({T,Tue})."
Brian, this might make more sense to you if you do what you suggested earlier: spread the days out. Make it an eleven-day experiment. If Heads rolls, let her sleep an entire week, and wake her on the next Monday (Day 8). If Tails flips, wake her on the first Monday and Tuesday (Days 1 and 2). When she is awake, she has no information that makes it more, or less, likely to be Day 1, Day 2, or Day 8.
And don't say that Day 8 has to be more likely than Day 1 or Day 2. You are starting with your answer, P(Heads|Awake)=1/2, every time you say something like that. Before the experiment, the was a 50% chance she would be awake on Day 1, a 50% chance of Day 2, and a 50% chance of Day 8. She has learned nothing that lets her change THE RELATIONSHIP BETWEEN any two of those possibilities.
"In the October, 1959 issue of Scientific American, Martin Gardner wrote: ... "
DeleteYeah ... there's nothing surprising there. It is exactly what I have been saying: To take the answer from 1/3 to 1/2, additional information is required. In this case, knowledge or assumptions about how someone is answering a question. Asked about the two children, the informant chooses not to talk about a child that doesn't exist (i.e., a boy when there are two girls or a girl when there are two boys). This is exactly scenario #2.
It's like the trick to the Monty Hall problem. The answer hinges on additional information that is almost never explicitly stated, but is usually inferred about the puzzle, that Monty is not going to open up a door with the prize. Since he knows where everything is, he always chooses a door with a goat. (The other assumption that is usually inferred is that, if there are two goats behind both remaining doors, he picks a door completely at random with equal probabilities.)
If you don't use this information -- that is, you assume that Monty is always just randomly picking doors just like you are -- then there is no advantage to changing your original choice. Your guess is just as good as his. The odds are even.
Your #3 is "You know that Mr. Smith has two children, and that at least one is a girl." Aside from substituting "girl" for "boy," how is this any different than the information given by Gardner when he says, "If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer [the probability that both were boys] is 1/3." I disagree with Gardner that if you posit "another procedure" then it "leads to exactly the same statement of the problem." Not at all. He left out the additional information necessary to analyze the problem in the second way. Words matter.
Having been sloppy once, he ends up being sloppy twice, but he mostly got it right.
There are reasons why someone would assume that Monty would want to avoid revealing where the prize is, so this omission of information is somewhat forgivable for this brain teaser. There is no reason at all to assume how the knowledge about the sex of the children was obtained, so the only rational analysis assumes that this information is simply not available. We're back to picking families and children at random, subject to the two-children/at-least-one-girl criteria.
"[Sleeping Beauty's new] information is 'On Sunday Night, I knew for a fact that I would exist in exactly one of four different states at any time during the experiment. They are {H,Mon}, {H,Tue}, {T,Mon}, and {T,Tue}. I also knew for a fact that, at any point in time, I was equally likely to be in any of these four states. ..."
No, unless she is mentally defective, she knows that "{H,Tue}" does not occur during the course of the experiment, because that is the specification of the problem. She'll NEVER be asked about the state of the coin toss in that situation, and she knows this Sunday night. She can eliminate that possibility before she goes to sleep and before the first random value is determined. She never has to take it into consideration.
For a guy who has just spent a lot of quotes, words, and hot air arguing that HOW someone is asked a question results in a probability of 1/3 reverting to 1/2 (even if your DON'T KNOW how that person was asked), you sure have spent a lot of trouble and a lot of ill-formed logic trying to convince me that the 1/2 should really be 1/3 in this particular case.
Is it too much to ask for a little consistency? My analysis is at least consistent. Cosmetic changes like what you switch to calling the days doesn't do anything to undermine that consistency or any other argument I have. It's usually the philosophers who get caught up in the details of how this could happen, what this should be called, and how this toy problem affects Metaphysics.
@JeffJo and @BrianMays: I encourage you to wrap up the debate. The Internet is running out of bits.
ReplyDeleteAgreed. I'm done.
DeleteSlightly edited: “To take the answer from X to Y, additional information is required. In this case, knowledge or assumptions about how someone is answering a question.” This is a true statement. It describes the relationship between the prior probability X and a posterior probability Y. You are using it inconsistently to describe the difference between two posterior probabilities.
ReplyDeleteIn a two child family, X is either 1/2 (if the probability is for two children of the same gender), or 1/4 (if the probability is for two girls). So to take it to 1/3, as you do, you need “knowledge or assumptions about how someone is answering a question” that is not implied in the statement of the problem. I do not “take the answer from 1/3 to 1/2,” because those two answers represent posterior probabilities derived from the same prior probability under different circumstances. I say that you have no information about this question except what the answer is. And just like Monty Hall, which I didn't bring up, there may, or may not, be two equivalent answers to the unknown question (Open D2 or D3, know there is a boy or a girl). It is allowing for the alternate answer, which you can't assume is impossible from the problem statement alone, that makes the answer to the MHP "switch." Ignoring the possibility that makes it "why switch?". Similarly, it is allowing for the alternate answer that makes the answer to the TCP 1/2. Not allowing for it that makes it 1/3.
They are both examples of Bertrand's Box Paradox, where the paradox is that looking at just the observed answer, when the question could have had a second equivalent answer, seems to change X to a different Y when it cannot be shown to provide relevant information.
But you are not going to admit that it is you who is being inconsistent, no matter how much I point out how. You are not going to admit that I created a perfect analogy of the Sleeping Beauty Problem that shows her answer can only be 1/3. You are not going to admit that I identified the “new information” you claimed is missing there. You are not going to admit that “if there are two goats behind both remaining doors, [Monty Hall] picks a door completely at random with equal probabilities” is the same assumption as “if there is a boy and a girl, the informant picks a gender completely at random with equal probabilities.” You are not going to admit it is you who is treating these two problems inconsistently. You are not going to admit that I never consider the possibility of Monty Hall opening the contestant’s door, or ignoring where the goat is. In fact, all you have ever considered here is that my results must be wrong, so anything you can misrepresent must be a mistake I made.
I agree with Allen – there is no point in depleting any more bits, because you won't consider them. I'll end it with this one:
“Unless she is mentally defective, she knows that ‘{H,Tue}’ does not occur during the course of the experiment, because that is the specification of the problem. “
So, if Tails flips, you say Tuesday doesn’t happen? I think it does. I made that clear in my discussion of it. SB sleeps through it, but in all my experience, each day of the week follows it traditional predecessor whether or not there is a person who sleeps through it. And it is the fact that Tuesday still exists, but she can rule it out as existing “now” because she is awake, that constitutes her new information.
Allen, I had hoped at some point you would weigh in with an opinion on the four-volunteer analogy.
Let's extend the experiment to provide a direct test of Beauty's credence for Heads. She'll learn these new rules on each wake-up, and will be made to forget them upon being put back to sleep.
ReplyDeleteUpon wake-up, she is presented with two enormous jars, "A" and "B", of indistinguishable jellybeans and deadly poison pills. She must pick once from one jar and set the choice aside until the end of the experiment. At the end of the experiment, she must eat her choice(s).
She is told that the jars were filled according to the coin flip. If it was Heads, the proportion of jellybeans in "A" is 1 and in "B" is 0; if it was Tails, the proportion of jellybeans in "A" is 1/7 and in "B" is 6/7.
Thirders: can you explain how, with a credence 1/3 for Heads, Beauty correctly calculates that "A" is her safest choice?
JeffJo (below) - I'll save Creosote the trouble. As I have said above, the answer that you're looking for is 1/3, but this is a different problem than the Sleeping Beauty problem and therefore is irrelevant.
DeleteAccording to your (different) problem, SB is 24% more likely to be poisoned by always choosing A instead of B.
The correct answer to Creosote's problem is that SB is 25% LESS likely to be poisoned by always choosing A instead of B.
Creosote - Interesting problem, but it lends itself to the criticisms that are launched against the various betting modifications that are often used to analyze the Sleeping Beauty problem, because the results are not tabulated until the end of the experiment. Why not always use the same jar of jellybeans/poison, but have her eat it right away? Then she can use the knowledge that she couldn't have been poisoned on Monday (if the day happens to be Tuesday) to adjust her credence to the result of the coin toss.
Another possibility is the modification proposed by Roger White: SB is not automatically woken every day, but instead there is a chance c<1 that she will be woken (and a corresponding chance, 1-c, that she just sleeps through the day). Then SB has additional information that she can use -- i.e., that she was woken at least once during the test.
Halfers will (correctly) realize that this makes a difference to her credence to the coin coming up heads. The less likely she is to be woken each day, the more the importance of having two chances to wake (on a toss of tails) will affect the estimate.
Thirders will conclude that the chance c makes no difference, because the chance is the same every time she could wake. For example, in JeffJo's alternate, four-person experiment, the probability of everyone being woken on each day changes nothing. They either all three wake or they don't. The situation is still symmetric and everyone concludes 1/3.
Responding to the on-topic need for clarification and ignoring all the rest ...
Delete1. The beans variant of the problem has deferred consequences in order not to have any effect on Beauty's credence during the experiment.
2. Beauty should calculate, when she has credence p for Heads, that "A" offers a survival probability of p + (1-p)/49, whilst "B" gives a survival probability of 36(1-p)/49.
My question remains: what calculation do Thirders use to correctly reach the conclusion that "A" is the safest choice?
Responding to your branch of the main topic, even though you didn't respond to my branch:
DeleteThe Halfer and Thirder approaches differ in the *assumption* that there is no "new information" upon which to "update" the probability. Halfers never define what "new information" is, they only assert that it doesn't exist, which is why it must have the status of "assumption." The 1/2 answer, and any extension such as yours, all depend on that assumption.
Thirders say "new information" does exist, but can still prove (Elga in particular) that the answer is 1/3 without relying on any similar assumption in any way. Every attempt at rebuttal still relies on it - in fact, the supposed contradictions are always based on a comparison that uses it. So any conclusion, even if it could be shown to be correct, is a non sequitur.
For example, your calculation uses the Thirder's posterior probability for heads, 1/3, in the place where a prior probability is required if "new information" does exist. So it is incorrect within the thirder approach.
Probability can be thought of as a rate over a time parameter. But the Sleeping Beauty problem modifies the scale of the time parameter in unorthodox ways. The point of my extension, is that it removes the change of scale, and the assumption about "new information," from any solution.
To see this, extend your poison-bean version by using four volunteers as I did, each with their own set of jars as you described them. Of the three awake at one time - that is, using posterior probabilities in the changed time reference - if each picks jar A the expected number of survivors is 12/7. If each picks jar B, it is 9/7. So viewed this way, A seems to be the better choice. That is what you found - that Sleeping Beauty has a better chance to not be poisoned by a bean she will pick from Jar A, than from jar B. But if you look at all four volunteers, after the experiment, you are back in the prior time reference. Jar B is better.
@Brian Mays: You keep asserting my version is different, yet you refuse to provide a reasoned explanation of what is different. Please tell me which of the following four statement you disagree with, since you must disagree with at least one, providing clear explanations:
Delete1) In the version I presented to Creosote, the experiment that volunteer H2 finds herself in identical to the one the original Sleeping Beauty was in. So her answer must be the same as the original's.
2) The experiments that H1 and T2 find themselves in differ from H2's only in the value of one random variable. Since the random variable has two equally-likely values, their answers must be the same as H2's. T1's has two such changes (or one, when compared to H1 or T2), and also must be the same.
3) Of the three awake volunteers on either day, the answer to "Does the coin flip result correspond to your label?" is "yes" for exactly one of them.
4) It does not matter if an awake volunteer sees the other two who are awake, or if they even exist in the experiment.
In this branch of the comments section, I'll consider only the "beans experiment" as first proposed.
DeleteI can't quite tell, JeffJo, what calculation you are proposing Beauty uses. By "a prior probability is required", I think you're saying that Beauty proceeds as follows:
(1) survival probability is p+(1-p)/49 for "A" and 36(1-p)/49 for "B"
(2) with prior probability p=1/2 for Heads, those survival probabilities are 25/49 for "A" and 18/49 for "B"
(3) so pick "A"
That calculation makes no use of Thirder Beauty's credence, 1/3, for Heads. So, in the Thirder world, "credence" appears to be inconsequential for decision making.
Sigh ... One more time and that's it. In these problems, there are two random outcomes: X1 - the result of the coin toss, and X2 - the day of the week. In the Sleeping Beauty Problem (SBP), the distribution of X2 completely depends on the outcome of X1. This is an essential part of the problem. If X1=Heads, X2 can have only one value (X2=Monday). If X1=Tails, X2 can have two. Thus, X2 cannot be determined until X1 is known.
DeleteIn your different problem, X1 and X2 are completely independent random quantities. Neither can affect the other. That, in itself, makes the two problems different, but you try to pull off a sleight of hand after the fact by introducing "new information" -- i.e., that the combination X2=Tuesday and X1=Heads did not occur, so that there are only three possibilities left corresponding to the possible outcomes in the genuine SBP. Then you adjust the different problem to accommodate this new information and arrive at your answer. This sleight of hand has fooled several people, but in the REAL SBP, that combination of X1 and X2 was NEVER a possibility because that is how the SBP is defined. X1 and X2 were never independent, and there is no "new information" to be discovered that was not already known on Sunday night.
This is also why the Thirders get a different answer to White's modified problem. Intuitively, if it is no longer guaranteed that SB will wake each day, then the coin flip takes on extra significance: if the coin is tails, she's less likely to sleep through the entire experiment, because she has two chances, instead of one, to wake up. Thus, the fact that she didn't sleep through the experiment should mean something, and that something should depend on the probability c of waking up each day, but because the Thirders are analyzing a different problem, they end up concluding that it does not. The value of c has no effect on their answer, and it doesn't matter which approach they are using to arrive at this answer.
I'm sorry that you are incapable of understanding this. Anyhow, that's the last time I try to explain it to you.
"In your different problem, X1 and X2 are completely independent random quantities." No, they are not.
DeleteIn my problem, there is a third random variable, X3, for the volunteer being considered. And the three are inter-dependent, along the lines you describe. Using that distribution, the answer to my question is clearly 1/3, and I think you have agreed.
When you consider the joint distribution for X1 and X2 with any specific value of X3, the distribution is either identical to the one you describe for SBP (for X3=H2), or symmetric with it. But the answer of 1/3 can't change.
I'm sorry you seem incapable of even attempting to understand this, but it is a common technique in probability.
I can tell you what is wrong with White's analysis as well, but I don't want to expand the issue further.
Q Creosote: I must admit I didn’t look at your formula, and I misread your intent. I took your stated proportions to be of poison jelly beans, to all jelly beans. That changed the problem in more ways than one.
DeleteBut the simple answer is that of course thirders should use the prior probability P(Coin=H)=1/2 to determine the makeup of the jar, not the posterior probability (PCoin=H|Awake)=1/3. Using 1/3 properly – and I am too rushed at the moment to do it – will only tell you the probability that *this* jelly bean will be poison, not the probability that you will get one over the course of the experiment. That is the only thing “credence” can apply to.
Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way, i.e. her credence for the coin flip being Heads or Tails. For Thirders, those credences are 1/3 and 2/3 respectively.
DeleteIf you insist that Beauty makes her choice based on credence 1/2 for the Heads way of filling the jars then you are by definition a Halfer.
"Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way,."
DeleteNo, she really shouldn't. If the coin landed Tails, she will pick two beans, not one. So there is a difference between her confidence that *this* bean is safe, and that the set of beans she will consume on Wednesday will be safe.
But Beauty is not asked for what her confidence was on Sunday, or will be on Wednesday (which is what you are solving for). She is asked for her confidence *NOW*. This is exactly the question my variation - the one you ignore because its answer is 1/3 - solves for.
AGAIN: In my variation, H2 finds herself awake and in the exact situation SB is in. The other two awake volunteers are in symmetric situations. Exactly one of their labels matches the coin. The confidence any one of them should have, that it is herself, must be 1/3.
Creosote, I'll explain why your calculation (which you didn't explain, but is easy to understand if one tries) was INCORRECT after you explain this apparent Halfer paradox (see note 1 below):
ReplyDeleteFour labels H1, T1, H2, and T2 are randomly dealt (note 2) to four volunteers (note 3). They are put to sleep, and then a single fair coin is flipped. Three of the four volunteers are wakened on the first day after the flip, and a different (but overlapping) set of three are wakened on the second day. The amnesia drug is used between days.
On each day, the volunteer who is left asleep is the one whose label is CD, where C is the result of the coin flip ("H" or "T"), and D is the day number ("1" or "2"). That is, H2 is left asleep on the second day, if Heads is flipped.
When wakened, each is asked for the probability that the coin result was the one that corresponds to her label.
I think that each of the following observations is incredibly obvious, but others disagree. But they won't clearly state what the issue is. If you disagree, please try to be specific and I will clarify.
Obs1: The volunteer labeled H2 is undergoing the exact same experiment, with the same question, as the volunteer in the original Sleeping Beauty experiment.
Obs2: The other three undergo an experiment that is equivalent to the original, with the only differences being symmetries in the labels.
Obs3: The question each is asked is the same question, but it gets modified by those same symmetries.
Obs4: On either of the days, the coin result corresponds to exactly one of three awake volunteers.
Obs5: Since each has the same information about whether the question describes herself, each must give the same answer.
Halfers: How can the sum of the three answers be greater than 1?
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Note 1: Really, I'm not avoiding your question. I intend to answer it. But that answer requires a resolution of my question.
Note 2: "Dealt" just like cards. They are random, but each of the four volunteer gets a different one of the four labels. I feel silly explaining this, but it is one of the points people have claimed is hard to understand.
Note 3: They can't know each others' labels. The way I set it up, it really doesn't matter if they know their own, as long as they understand the processes; but it does take an extra step to show each volunteer's state is equivalent to the original Sleeping Beauty's.
Maybe this will help clarify what I'm trying to say:
ReplyDelete1) Brain Mays solves the SB problem with a solution with an answer of 1/2. Call this solution B, for Brian.
2) I presented a variation of the SB problem that I say, and I think Brian agreed with, has the answer 1/3. Call its solution J.
3) Brian insists that solutions B and J address different problems. I agree.
4) What the "four volunteers" example is supposed to show, is that the original Sleeping Beauty finds herself in exactly the same set of circumstances as one of the four volunteers, and symmetrically equivalent circumstances to the other three. Brian has not contested this, he has only said it needs a different solution than solution B.
5) Those circumstances are addressed by solution J.
So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty? Because she says the probability that the coin matches her label - i.e., landed Heads - is 1/3. A agree that this changes the problem from the one you solved - I'm saying that you solved the wrong problem.
"So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty?"
DeleteWhat part of "that's the last time I try to explain it to you" do you not understand?!!
I've already explained it to you, again and again. Since you cannot accept my explanation, you counter with:
1) "'In your different problem, X1 and X2 are completely independent random quantities.' No, they are not."
Oh really? No reason given.
2) "When you consider the joint distribution for X1 and X2 with any specific value of X3, the distribution is either identical to the one you describe for SBP (for X3=H2), or symmetric with it."
At this point, I seriously doubt you even know what a joint distribution is, but ... (sigh) ... giving you the benefit of the doubt, if the joint distributions were identical then we wouldn't be having this discussion -- would we? We would agree on the result.
3) "I can tell you what is wrong with White's analysis as well, but I don't want to expand the issue further."
In other words, White's analysis is way over your head.
4) "Using 1/3 properly -- and I am too rushed at the moment to do it -- will only tell you the probability that *this* jelly bean will be poison ..."
At this point, you have become a parody of yourself. Creosote is correct: you are by definition a Halfer.
Please leave us alone now. You are the last person who should be demanding anything from anybody.
For what appears to be an excellent and far-reaching survey of the available literature, readers might enjoy perusing http://www.sleepingbeautyproblem.org/tiki-index.php?page=Project%20Portal . Seems like it's a work in progress, so should get even better over time.
ReplyDelete"What part of 'that's the last time I try to explain it to you' do you not understand?!!" The fact that your explanation does not address my issue. It dodges it. The issue isn't whether my solution is to a different problem than yours, we agree it is. The issues is which corresponds to the SB problem.
ReplyDeleteH2 is identical to SB. H2's answer to the same question is 1/3.
I'll avoid ad hominem attacks.
In the original problem, say a six-sided die is also rolled on Sunday Night after SB is asleep. If it comes up even, an alarm is programed to go off at noon on Monday. If odd, it will go off at noon on Tuesday, but she may not be awake to hear it. SB is instructed to enter her confidence in Heads into a computer at precisely 12:30PM.
ReplyDeleteOn Sunday Night, she recognizes that the coin and the die represent independent events, and the alarm is equally likely to go off under any of the four possible situations {H,Mon}, {T,Mon}, {H,Tue}, and {T,Tue}. Of course, she will not be awake to hear it under {H,Tue}. Since she is not assured of hearing it, if she does it constitutes “new information” and allows her to update her confidence in Heads to 1/3. But not hearing it is also “new information,” and also allows her to update her confidence in Heads to 1/3.
What if SB is, unknown to the experiment director, unable to hear certain frequencies and so can’t hear the alarm? She can still update her confidence to 1/3 because the alarm was in only one state of going off, or not going off, and both allow the same update. Essentially, this is now a form of Bertrand’s Box Paradox. She can update her confidence to 1/3 under every possible state of the alarm, so her confidence must be 1/3 without considering the alarm. But please, please, please note that this answer only applies when the status of one day is all that is known.
The point I didn’t want to get into when Brian mentioned independence, is that independence changes as a result of the amnesia drug. On Sunday Night, being awake under T1 and T2 represent the same future, so they are dependent events. Brian’s solution treats them as dependent and gets 1/2. As such, it is addressing the problem from Sunday’s (or Wednesday’s) viewpoint only; that is, a state where all possible awake days are considered. But when SB is awake, she has no information about the possible “other” day, T1 and T2 represent different presents, and they are independent contrary to how Brian treats them. The Thirder solution treats them as independent to get 1/3. It address the problem from the viewpoint of either Monday or Tuesday, but not both as Brian does.
Both the Alarm-Clock version, and my four-volunteer version, are designed to isolate state where T1 and T2 are independent events. Yes, it is a different problem than the one Brian solves when he treats them as dependent. The question Brian refuses to address is, which is more appropriate to the original question?
@JeffJo @Creosote @BrianMays My general policy is that I'll publish a comment as long as it is on point and not abusive. But I encourage you to keep the tone civil. And if it looks like you can't reach consensus, you might have to agree to disagree.
DeleteOkey-dokey, I'm done. I look forward to other Thirders attempting the "beans experiment" in future, and will be interested to see what conclusion (if any) is eventually reached at www.sleepingbeautyproblem.org .
Delete"The question Brian refuses to address is, which is more appropriate to the original question?"
ReplyDeleteNonsense! I have said over and over that the one that is more "appropriate" is the one that corresponds to the definition of the problem. X2 depends on the outcome of X1. That is explicitly stated without any ambiguities.
By the way, you still haven't addressed White's "generalized Sleeping Beauty Problem." Google it if you want to find the original article published in Analysis in 2006. Or you can click on the link that Creosote provided above. It probably has a link that will take you there.
There are two "definitions of the problem." One is what the experiment director sees (and SB sees on Sunday), and a different one is what SB sees when she is awake. That one exactly matches the situation that H2 finds herself in, where the answer is 1/3. Which should she base her confidence on? I've shown you several ways why it must be 1/3, and all you do is repeat the other solution.
ReplyDeleteSo again, I am less interested in seeing your solution that gets 1/2, then knowing why you think she can treat Monday and Tuesday as the same day. Which is what White did.
Digging out my notes from years ago (and I'm not re-deriving it), I see that White stated that P(H) = (c/2)/[(c/2+(1-(1-c)^2)/2] = 1/(3-c). But his event "H" really is "H&Mon." You can use his exact same logic to get P(T&Mon)=P(T&Tue)=1/(3-c). These should add up to 1, and they don't.
No. His result is for P(H | SB didn't sleep through the experiment).
DeleteMonday and Tuesday have nothing to do with it. SB doesn't know whether it is Monday or Tuesday, because it could be either day. All she knows is that she didn't sleep through the entire experiment.
The credence for heads actually becomes P(H)=1/3 (the Thirder's answer), but only in the limit that it is extremely unlikely for her to have ever wake at all -- i.e., c->0.
There are three events of interest to a "thirder" SB when she is awake: H1=Today is Monday and Heads flipped, T1=Today is Monday and Tails flipped, and T2=Today is Tuesday and Tails flipped.
DeleteWhite called his event H, but he really was treating it as H1. His expression for P_(H|W) is more accurately called P_(H1|W) even though they represent the same thing. That's why the term (c/2) is used: the prior probability of being awake on Monday after Heads is c, and the prior probability of Heads is 1/2. He multiplied them to get his (c/2).
Note that the denominator of his equation [P_(H) P_(W|H) + P_(~H) P_(W|~H)] has nothing to do with the fact that he is calculating P(H), it is merely an expression of the Law of Total Probability using the partition {H,~H}. So the same denominator can he used in a similar calculation for P_(T1|W) and P_(T2|W). Since the numerators for these calculations are P(H1)*P_(H1|W) = P_(T1)*P_(W|T1) = P_(T2)*P_(W|T2) = (c/2), you get the same result for each calculation. Like I said.
I know you disagree with this approach. But it is what a thirder would use. What White accused a thirder of using is not.
+++++
What you, and White, ignore is that H2=Today is Tuesday and Heads flipped *is* *also* *an* *event* *in* *the* *experiment*. SB can't observe it, but it is still an event. On Sunday night, it has the same status, and prior probability, as H1, T1, and T2.
Her "new information" when she is awake is that H2 is not her current state.
The partition of SB's posterior event space contains four independent events, not two. Since she samples twice, with amnesia in between, the sum pf the prior probabilities for these four events is 200%, not 100%. The laws of probability work just fine on these events.
The late great philosopher David Lewis was a halfer. I'd be interested in any reactions to his paper on it. http://fitelson.org/probability/lewis_sb.pdf
ReplyDeleteThanks for asking, and for letting me know about this paper. I posted my response in a followup article: http://allendowney.blogspot.com/2016/09/sleeping-beauty-and-red-dice.html
DeleteAm I wrong saying this...
ReplyDeleteProbabilistically, the question is quite simple. Every time the coin is flipped, it IS 1/2 odds (it's a coin). But every time she is WOKEN UP it's 2/3 tails and 1/3 heads. Is it really that different to just saying something like, "I'm going to slap you twice on a tails, and once on a heads. How likely is it, at each point I've slapped you, that you got a heads or a tails?". Does the "amnesia" element really change it that much? It's the same here... if you are betting on the coin flip, it's still 1/2, but if you are being slapped, then it's more likely it's because you got a tails since 2/3rds of the slaps will end up being attributable to tails and only 1/3rd of them to heads.
So to me... it seems like although the "wager" problem is interesting, you are indeed "overthinking" it. Surely the thing that makes the difference is that she is possibly being asked twice about what the coin landed on when she wakes up, but when the bet is resolved, she is only being asked once.
Simply, if you do this, say, 6 times, and get 3 heads and 3 tails, and she bets on heads, she will earn money by being correct about the wager, but will also be MOSTLY correct about saying tails is more likely - because she has been asked about the coin NINE times (3 on heads, 6 on tails) and been correct SIX times.
TheRingshifter> Does the "amnesia" element really change it that much?
DeleteThe amnesia is necessary in order to argue about Beauty's credence within the experiment itself, i.e. to prevent Tuesday Beauty saying "oh, hang on, you woke me yesterday ... this must be a Tails time-line".
But should Beauty feel any differently about the coin flip within the experiment than she did prior to it? That's a key sticking-point of the Sleeping Beauty problem.
Alright, let's assume Beauty believes the thirder perspective and decides that she will answer "I think it landed on Tails, the odds are 2 in 3" every time she is questioned because she thinks she'll be correct more often. She gets asked twice if it comes up Tails after all, and all these mathematicians can't be wrong.
ReplyDeleteAs soon as she enters sleep/stasis and the coin is flipped there emerge two timelines, T1 where the fair coin came up Heads and T2 where the fair coin came up Tails. Since it's a fair coin T1 happens 50% of the time and T2 happens 50% of the time. This means that in T1 Beauty says Tails and is incorrect, whereas in T2 Beauty says Tails and is correct.
Doesn't this mean that she's correct in 50% of timelines and incorrect in 50% of timelines? There are only two possible timelines and they happen 50% of the time each since it's a fair coin. In fact she could be asked a thousand times in T2 and since her memories are wiped and reset she'll always answer Tails. Furthermore, doesn't this mean that if you repeat the experiment 1,000 times, you get 500 Heads and 500 Tails, and she answers Tails every time... that she's correct in 500 experiments and incorrect in 500 experiments?
The fact that she answers incorrectly once in T1 and correctly twice in T2 doesn't make her answer twice as correct. You're only tracking whether she answered correctly, not how many times she gets it right. I mean, you're not giving her 100$ every time she gets the answer right. If you did then she'd always say Tails because she has a 50% of getting 200$ and a 50% of getting 0$, whereas if she said Heads she'd have a 50% chance of getting 100$ and a 50% chance of getting 0$. What you're looking at is her certainty that the coin came up Heads or Tails. And if my previous logic is correct, that means that guessing Tails makes her correct in 50% of timelines... so she'd break even instead of being ultimately correct approximately 666-667 times out of 1,000 experiments.
I can't believe that the process of being interviewed would lead her to believe that the odds of it being Heads or Tails had changed. This isn't a Monty Haul problem, she's not getting any new information. She knows that she's going to be woken up at least once and has no capacity to distinguish between Heads(Monday), Tails(Monday), and Tails(Tuesday)... or the 997th Tails day for that matter.
Am I missing something here?
Christopher> Am I missing something here?
DeleteNo-one's disputing the maths:
* [A] One half of experiments in which Beauty wakes are those in which the coin is Heads.
* [B] One third of Beauty's experimental wake-ups are those in which the coin is Heads.
Beauty knows the facts, but she is obliged to pick just one of them as the basis of her "credence for Heads". Halfers insist she pick [A], Thirders insist on [B]. She's not allowed to state the reference class (w.r.t. experiments/wakeups) in her answer: her "credence" must be unqualified and absolute.
So there's an interesting philosophical question lurking at the heart of this: when Beauty can see both perspectives, [A] and [B], what then do we mean, fundamentally, by her "credence"? If there is a correct answer to the Sleeping Beauty problem then it's a philosophical one, not bunch of probability calculations.
Have you seen my four-volunteer version? Each will be wakened at least once, and maybe twice, based on the same coin flip. Each will be left asleep in only one set of conditions, different for each, as defined by the cross product {Monday,Tuesday}x{Heads,Tails}. Each will be asked for her credence that the coin landed on the side that would let her sleep through one day.
DeleteOne of these volunteers is undergoing the identical problem as in the original problem. Three are undergoing a functionally equivalent one, that must have the same answer.
On any day, exactly three volunteers will be wakened. On any day, exactly one of the three awake volunteers is in the set of conditions she is asked about. On any day, each of the three awake volunteers has the same information upon which to base her credence.
Yet that credence is found by a "bunch of probability calculations." It is 1/3.