Well, perhaps because relative frequencies aren’t always probabilities?
neq1
The Sleeping Beauty Challenge
Maybe I’m naive, but I actually think that we can come close to consensus on the solution to this problem. This is a community of high IQ, aspiring rationalists.
I think it would be a good exercise to use what we know about rationality, evidence, biases, etc. and work this out.
I propose the following:
I will write up my best arguments in favor of the 1⁄2 solution. I’ll keep it shorter than my original post.
Someone representing the thirders will write up their best arguments in favor of the 1⁄3 solution
Before reading the others’ arguments, we will assume that they are right, and that reading it will only confirm our beliefs (this is hard to do, but I find that this approach can be helpful)
We cannot respond for at least 24 hours. (this will give us time to digest the arguments, without just reacting immediately)
We will then check to see if there is agreement
If we still disagree, we can have some discussion (say, via email) to see if progress can be made
We will post our original two arguments and conclusion here (maybe in a new post)?
What do you think?
I tried to set this up in such a way to reduce some of the known biases that prevent agreement. Am I missing something?
Possible pitfall: if we come to an agreement, people who disagree with our conclusion might say it’s because one of us was a poor representative of their viewpoint. However, I think we’d still move a step towards consensus.
What say you?
You make a good point. However, I’d argue that those in Thirder Land had nothing to update on. In fact, it’s clear they didn’t since they all give the same answer. If 50% of the population has cancer, but they all think they do with 0.9 probably, that’s not necessarily less accurate than if everyone thinks they have cancer with 0.5 probability (depends on your loss function or whatever). But the question here is really about whether you had evidence to shift from .5 to .9.
Thanks for your detailed response. I’ll make a few comments now, and address more of it later (short on time).
Your argument is, I take it, that these counts of observations are irrelevant, or at best biased.
No, I was just saying that this, lim N-> infinity n1/(n1+n2+n3), is not actually a probability in the sleeping beauty case.
The disagreement seems to center on the denominator; it should count not awakenings, but coin-tosses.
No, I wouldn’t say that. My argument is that you should use probability laws to get the answer. If you take ratios of expected counts, well, you have to show that what you get as actually a probability.
I definitely disagree with your bullet points about what halfers think
I said: “Just like Beauty waking up on Monday is the same as Beauty waking up on Tuesday. There is no justification for treating them as separate variables.”
You disagreed, and said:
What you say is true for any outside observers, and for Sleeping Beauty after the experiment is over and the logbooks analyzed. But while Sleeping Beauty is in the experiment, this option is simply not available to her. The scenario has been carefully constructed to make this so, that’s what makes it an interesting problem. The whole point of the amnesia drug in the SB setup (or downloadable avatars, or forking universes, random passersby, whatever) is that she has NO justification nor even a method for NOT treating any of her awakenings as separate variables, because the information that could allow her to do this is unavailable to her. By construction—and this is the defining feature of Sleeping Beauty—all Sleeping Beauty’s awakenings are epistemically indistinguishable. She has no choice but to treat them all identically.
Hm, I think that is what I’m saying. She does have to treat them all identically. They are the same variable. That’s why she has to say the same thing on Monday and Tuesday. That’s why an awakening contains no new info. If she had new evidence at an awakening, she’d give different answers under heads and tails.
The thing is, the argument in favor of the 1⁄3 solution on the Wikipedia page is flawed. I tried to explain the flaw, but perhaps I failed. It makes me cringe when I think that people are going to that page for the solution.
Also, not only did I critique the wikipedia page, but I critiqued parts of papers by Radford Neal and Nick Bostrom.
That’s not to say my post deserves more up votes. Others can judge the quality of my work. But I’m pretty sure I covered some new ground here.
(1) You’re right, I got the numbers wrong. Thanks.
(2) If she knows she is somewhere along a two week path, the probabilities are different than if she knows she is somewhere along a one week path. She’s conditioning on different information in the two cases.
Once you go more than 1 week it’s not the sleeping beauty problem anymore. Half the time she’s woken up once at night, 1⁄4 of the time she’s woken up 6 times in morn and once at night, 1⁄4 of the time she’s woken up 12 times in morn. This doesn’t have to do with when the coins are tossed. It’s just that, if you do it for 1 week you have the sleeping beauty problem; if you do it multiple weeks you don’t
Your original question was about one week, not two (I thought).
At the beginning of every week, a coin is flipped. If heads then rather than having 6 days diurnal and 1 day nocturnal, you just have 1 day nocturnal and six days in hibernation. If tails then you have 6 days diurnal and 1 day in hibernation.
Are we just doing this twice? What happens between the weeks? Do they know the experiment has started over?
Sorry, I’m not following. What are you doing with these two weeks’ worth of coins?
Not all awakenings are equally likely. 50% chance it’s one of the 6 morning awakening. 50% chance it’s the one night awakening.
I think the coin flip does change things. In fact, I don’t see why it wouldn’t.
In case 1, you know you are somewhere along a path where you will wake up on one night and wake up on 6 mornings. You can’t determine where along that path you are, so you guess morning has probability 6⁄7
In case 2, there is a 50% chance you are somewhere on a path where you wake up once at night (and never in the morning), and a 50% chance you are somewhere on a path where you wake up on 6 mornings and 0 nights. So, probability it is morning is 1⁄2.
Intuition Pump
Suppose 50% of people in a population have an asymptomatic form of cancer. None of them know if they have it. One of them is randomly selected and a diagnostic test is carried out (the result is not disclosed to them). If they don’t have cancer, they are woken up once. If they do have it, they are woken up 9 times (with amnesia-inducing drug administered each time, blah blah blah). Each time they are woken up, they are asked their credence (subjective probability) for cancer.
Imagine we do this repeatedly, randomly selecting people from a population that has 50% cancer prevalence.
World A: Everyone uses thirder logic
Someone without cancer will say: “I’m 90% sure I have cancer”
Someone with cancer will say: “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.”
Notice, everyone says they are 90% sure they have cancer, even though only 50% of them actually do.
Sure, the people who have cancer say it more often, but does that matter? At an awakening (you can pick one), people with cancer and people without are saying the same thing.
World B: Everyone uses halfer logic
Someone without cancer will say: “I’m 50% sure I have cancer”
Someone with cancer will say: “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.”
Here, half of the people have cancer, and all of them say they are 50% sure they have cancer.
My question: which world contains the more rational people?
Credence isn’t constrained to be in [0,1]???
It seems to me that you are working very hard to justify your solution. It’s a solution by argument/intuition. Why don’t you just do the math?
The experimenters fix 2 unique constants, k1,k2, each in {1,2,..,20}, sedate you, roll a D20 and flip a coin. If the coin comes up tails, they will wake you on days k1 and k2. If the coin comes up heads and the D20 that comes up is in {k1,k2}, they will wake you on day 1.
I just used Bayes rule. W is an awakening. We want to know P(H|W), because the question is about her subjective probability when (if) she is woken up.
To get P(H|W), we need the following:
P(W|H)=2/20 (if heads, wake up if D20 landed on k1 or k2)
P(H)=1/2 (fair coin)
P(W|T)=1 (if tails, woken up regardless of result of coin flip)
P(T)=1/2 (fair coin)
Using Bayes rule, we get:
P(H|W)=(2/20)(1/2) / [(2/20)(1/2)+(1)*(1/2)] = 1⁄11
With your approach, you avoid directly applying Bayes’ theorem, and you argue that it’s ok for credence to be outside of [0,1]. This suggests to me that you are trying to derive a solution that matches your intuition. My suggestion is to let the math speak, and then to figure out why your intuition is wrong.
If you are the person that was selected beforehand to be revived in the event of heads, then I agree with 1⁄11. Unfortunately, in variation beta we lose the ability to label someone ahead of time. This changes things.
I need more clarification. Sorry. I do think we’re getting somewhere...
The experimenters fix 2 unique constants, k1,k2, each in {1,2,..,20}, sedate you, roll a D20 and flip a coin. If the coin comes up tails, they will wake you on days k1 and k2. If the coin comes up heads and the D20 that comes up is in {k1,k2}, they will wake you on day 1.
Do you agree that P(H|W)=2/22 in this case?
- May 13, 2010, 9:01 AM; 0 points) 's comment on Conditioning on Observers by (
Based on my interaction with people here, I think we all are talking about the same thing when it comes to subjective probability.
I agree that you can use betting to describe subjective probability, but there are a lot of possible ways to bet.
Continuity problem is that the 1⁄2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment
Why is this a problem? I’m perfectly comfortable with that property. Since you really just have one random variable in each arm. You can call them different days of the week, but with no new information they are all just the same thing
By D do you mean W?
What’s happened is closer to E(H|D) = E(D|H) E(H) / E(D), over one run of the experiment, and this yields 1⁄3 immediately.
Is this how you came up with the 1⁄3 solution? If so, I think it requires more explanation. Such as what D is precisely.
Variation Alpha is unclear, as worded. Let’s say one of the 10 people is Sleeping Beauty, and the other people have different names. Sleeping Beauty was identified ahead of time, and she knows it. If she is not selected, then no one is interviewed. Then, if she is revived, she should think it was heads with probability 10⁄11.
But… if we will interview everyone who is revived, and no one was labeled as special ahead of time, then all each person that was interviewed knows is that at least one person was revived, which was a probability 1 event under heads and tails.
This is just the self-indication assumption situation.
Consider an example. Suppose we want to know if it’s common for people to get struck by lightening. We could choose one person ahead of time to monitor. If they get struck by lightening in the next, say, year, then it’s likely that getting struck by lightening is common. But… if instead everyone is monitored, but we are only told about one person who was struck by lightening (there could be others, we don’t know), then we have no information about whether getting struck by lightening is common or not.
It seems even weirder in the Xtreme Sleeping Beauty, where she’s awakened a thousand times : “my subjective probability for heads is 1⁄2, but I’m only willing to bet up to 6 cents”.
I guess I don’t see why it’s weird. The number of times she will bet is dependent on the outcome. So, even though at each awakening she thinks probability of heads is 1⁄2, she knows if it’s tails she’ll have to bet many more times than if heads. We’re essentially just making her bet more money on a loss than on a win.
Even in the limit not all relative frequencies are probabilities. In fact, I’m quite sure that in the limit ntails/wakings is not a probability. That’s because you don’t have independent samples of wakings.