For the Chloe problem here's a way to motivate the alternate answer. Allow a betting, insurance or future's market and a sufficiently large number of repetitions.
Let's also say instead of creating a billion billion clones in the case of rolling 1 million only a million million are created, but each of them receive a million if they correctly guess (so a million million, million (10^18) dollars are paid out by the happy parents).
The bet will be as follows:
1) to accept the bet Chloe voluntarily shakes the bookie's hand and then does a speech act, the speech act for Chloe accepting the bet is guessing that 1 million was rolled but counter-intutively she will thus be betting that 1 million was not in fact rolled
2) if 1 million is not rolled Chloe wins the bet and receives 1 million dollars from the booky, but nothing from her parents as she did not guess correctly.
3) If the million is rolled Chloe loses the bet (but receives the million dollars from her parents) and pays out a single dollar to the booky.
Thus on the average 1 000 000 iterations (where 1 million is rolled once) the booky pays out $999 999 million but receives $ million million dollars and so the booky will be ahead $ 1 million dollars.
Neither Chloe nor the booky (I'm imaging the booky as a R0G3R type battle droid, or a million million in the case where there are a million-million identical bookies to let Choe take the bet) know what the die roll was. Yet given what they know both do something fairly reasonable (arguably Chloe is slightly shortchanged on the deal, but only by a millionth of a dollar and so we can chock it up to risk aversion and necessary friction to compensate the bookie).
This reflects the fact that from the point of view of the galaxy the expected utility to Chloe guessing the 1-in-a-million is much greater than the utility of her guessing not that. This also reflects the fact that for every Chloe-like point of view where a number other than a million was rolled there are a million Chloe-like points of view where it was (assuming enough iterations to average it out). Although this assumes the million-million Chloes would not exceed the carrying capacity of the Galaxy and throw the Galaxy into a hard scrabble struggle for survival. Or that the parents are not just basically printing money to give Chloe and so creating a massive inflation.
On Linch’s blog I advocated “utility side SIA”; basically use halfer probability but multiply the utility by the # of copies. More generally, I suggested you could take the view that “only PU matters”, and freely shuffle factors between P and U depending on what makes things easier to think about.
Whenever there’s a “canonical” final outcome (as in your examples), then it’s usually obvious what you should do and as you show sometimes this looks like the halfer move and sometimes line the thirder.
In the anthropic dilemmas I find interesting there is no canonical outcome. Eg for the presumptuous philosopher you just have to say if you care more about being responsive to the experimental evidence or about how many of the philosopher’s copies are right. I’m sympathetic to the latter because the SIA move is what you’d choose given a “view from nowhere”, but there are problems with this (eg as I raised with Lich-sometimes universes with many copies seem too “cheap” wrt standard priors). Feels like there should be a solution, but idk what it is.
> In the previous version, we determined that the chance is 1 in 1 million. This time, though, the long tail of wakings finally comes into play and the chance is 1 in a billion billion.
p(1 million and day 1) is 1 in a million billion billion (on halfer logic), p(day 1) ~ 1, so conditionalise and you get 1 in a million billion billion (to a close approximation).
To condition, you divide the joint probability by the marginal: P(1 million and day 1)/P(day 1). P(day 1) is approximately (1 million -1) / 1 million (because in only 1 case out of a million do you wake on any other day, and in that case you are almost almost guaranteed to wake up on another day). So you're dividing 1 in a million billion billion by a tiny bit less than 1.
Woops, I thought you were referring to my second version if this, but I just checked the quote, and for the first version P(Day 1) = 1 / 1 million, isn't it?
Also, not sure about P(1 million and day 1)/P(day 1), because that assumes independent variables? Or does it not? My stats 101 is admittedly rusty.
Oh no, I thought you were talking about the original problem with day labels....
The conditioning formula is valid regardless of dependence, however.
Ok I've read the day labelling procedure more carefully and I still think P(day 1) \approx 1. If the dice does not land on 1 million, then P(day 1) = 1. But that happens with probability (1 million - 1)/1 million, so P(day 1) has probability of at least that. Why do you think it is 1/1million?
"The original question is ambiguous2 about which goal you should be maximizing for, but for my variant, the goal isn't “How do I prevent myself from being wrong the most number of times?”. The goal is, “What should I guess to maximize the likelihood that I’m not wrong right now?”"
The original SB asks what your credence upon awakening that the coin landed on heads *ought to be*. Does this depend on what goal you have? Perhaps, but if your goal is to be rational in your beliefs then perhaps that disambiguates.
There'a a famous LessWrong blog post by Yudkowsky about how the rational thing to do is to "make your beliefs pay rent in anticipated experiences."
Perhaps you ought to believe there is an invisible dragon in your garage if you value believing in dragons and don't care about truth. But if you want to be rational and don't actually anticipate that the bucket of flour you are holding will reveal the outline of a dragon when you dump the flour out into the garage, and actually anticipate that it will fall to the floor as if the garage is empty, then the rational thing to believe is that there is no invisible dragon in your garage.
So upon waking in Sleeping Beauty, is the rational thing to believe that the coin landed on heads with p=1/2 or p=1/3? Well that depends on what you anticipate experiencing. Do you anticipate that you are just as likely to learn that the coin landed on heads as you are to learne that the coin landed on tails? Or do you anticipate being twice as likely to learn that the coin landed on tails?
Extremizing it to p=0.001 (halfer view) and p=~0.999 (thirder view) by using a 1,000-sided die instead of a fair coin and having one million awakenings if and only if the die lands on 1,000 (and one awakening otherwise), the question is thus, upon awakening, do you anticipate that when you look over at the die to see what it landed on that you will almost definitely see that it landed on 1,000 (the SIA thirder view) or do you anticipate almost definitely seeing that it landed on some value from 1-999 (the SSA halfer view)?
I take the halfer view because I would anticipate seeing the die on some random value like 812. That is, I would almost definitely expect to see that it landed on 1-999.
The arguable ambiguity of the original phrasing of the question is something I skimmed over because (1) I didn't want to lose any readers who might disagree over it and (2) I cared more about arguing my case for the variant that makes most sense to me.
But... I think if you were to imagine the experiment being re-run many times, the "thirder" position sounds true to me, because you'd expect the unlikely result to happen at least once, and then it becomes overwhelming. Surely your "credence" must be impacted by the observation selection effect at that point.
What throws me, though, is the case where the experiment only gets run once. Does it make more sense to base your credence off the coin flip / dice roll, or to base it off the expected value of your observations? Maybe the "halver" position makes more sense here, but if that's true, and it's true that enough repeats will switch to the "thirder" position, then where's the dividing line?
I like my variant because it makes the question feel well defined. The original... I just don't know.
"But... I think if you were to imagine the experiment being re-run many times, the "thirder" position sounds true to me, because you'd expect the unlikely result to happen at least once, and then it becomes overwhelming."
I think the question "What is the expected percentage of awakenings in which the coin lands on heads?" (Answer=1/3) is a distinct question from the question being asked in Sleeping Beauty problem: "Upon awakening, what ought your credence be that the coin landed on heads?"
I think the correct answer to the SB question is the same regardless of whether it's the first time you're participating in the experiment as SB or the nth time. (I answer 1/2 regardless.)
I broadly agree with you that Linch is wrong / that SIA is wrong. I too am a halfer / "Anti".
I use the same alternate version of the SB problem as you, except I used the numbers 1/1,000 chance of a million awakenings (or clonings) rather than a 1/1,000,000 chance of a billion billion awakenings.
Your version is too extreme as a billion billion awakenings would take 31 billion years even if each awakening took only 1 second with no time in between.
I prefer clonings/copies to memory erasure because I think they're easier to think about and are functionally identical. What is memory erasure after all if not a copy/clone of a previous brain state? (I think they're the same.)
I decided to only include the thirder position in the main section a) because I personally believe it's correct, b) because it's the mainstream position, and c) because for a 101 class for babies, many things needed to be cut and I thought giving people the confidence to think about anthropic problems for themselves was more valuable than presenting the existing scholarly disputes.
I'm not sure I agree with your presentation on "observation selection not mattering" in the examples you mention, but will think about it!
Oh my gosh, I'm only now realizing that I got so distracted with Sleeping Beauty, I never actually finished reading your post!!!
In my defense, I was reading your post and working on my post in the downtime I had while vacationing in Japan, which was a really fun trip. But maybe suboptimal for my reading focus, haha
I'm not offended haha! You have the distinct honor of being my first critic who has written a full post as a reply, and I'm happy it's about an anthropics dispute rather than politics or animal welfare!
The original Sleeping Beauty Problem (this is an edited quote from Arnold Zuboff) went more like this: "a hypnotist puts SB into a hypnotic sleep that is to last for N days, where N>>1. The hypnotist will interrupt her sleep either daily, and thus N times, or else only once on a randomly selected day among the N. Which of these two numbers of awakenings there will be is determined by a toss of a fair coin whose outcome is kept secret from SB. At the end of any period of wakefulness, SB must be hypnotized into forgetting it completely and permanently before being made to sleep again. SB, who knows everything just explained, can score a win only if she correctly answers the question of which number of times she is being awakened."
Adam Elga's problem statement altered this to N=2, Heads meaning one waking (Zuboff didn't specify), and he asked the equivalent question about the coin landing Heads.
But his problem statement did not specify on which day the one waking would occur. That he added in his solution, so he could identify the possible days. I suggest that he did this to avoid having to resort to slightly more involved Mathematics that might confuse some, but what it actually accomplished was to start the entire controversy.
I'll go back to a randomly selected day, the Heads schedule, and the Heads question using the following method: At the same time the coin is flipped, roll an N-sided die. Each day, wake SB if the coin is showing Tails or the die is showing "1." Each day after SB is put back to sleep, either rotate the die down one number, or set it to N if it was already on 1.
On any day during the experiment, the "outside" or "prior" probability that the coin is showing Heads is 1/2. And the "outside" or "prior" probability that the die is showing 1 is 1/N. These are independent random variables, so the "outside" or "prior" probability of any combination is 1/(2N). And there are 2N combinations that can occur during the experiment.
But when she is awake, SB knows only that she is in one of the N combinations where the coin is showing Tails, or the one where the coin is showing Heads and the die is showing 1. So the probability that the coin is showing Heads is trivially 1/(N+1).
As you said yourself: The probability of Heads is 1/2.
This is true before and after SB wakes; it's true of her belief of the coin before or after she wakes.
What corresponds to ~1/N is how often she'll guess Heads in repeated experiments, and her belief in her own guesses, which is also unchanged by the act of sleeping or waking.
Any real solution can't have Sleeping Beauty changing her credences after learning no new information.
The *prior* probability of Heads is 1/2. That means that, on any day during the experiment it is equally likely to be Heads or Tails.
The *prior* probability of Monday is 1/2. That meas that, on any day during the experiment it is equally likely to be Monday or Tuesday.
These probabilities are independent; that means you can multiply them to get the *prior* probability of any combination
But once information is learned, you need to use conditional probability. Those probabilities do not take SB's awake/sleep state ON A SINGLE DAY into account. "Information" means any change to the possibilities that depend on the actual values of COIN and/or DAY.
Being awake, since it can't happen when COIN=HEADS and DAY=TUESDAY, is such information. Being awake can happen in three combinations only; that's prior probability 3/4. This is not the same as "I will be awake during the experiment," it applies individually to each combination.
Yes, so viewed one way the P(Heads + Mon) is 1/2 and the other two possibilities are 1/4. Or they're all 1/2, depending on how you define your terms.
If you'd like, you can show me a specific set of definitions and a calculation for arriving at the 1/N value, and I can either show you how that maps out to guesses rather than coinflips, or I can show you where there is an unwarranted assumption.
There is no valid definition where they are all 1/2.
Try this: Wake SB both days. In the morning, have her write down the probabilities for Pr(H), Pr(T), Pr(Mon), and Pr(Tue). They should all be 1/2. Then the probabilities for Pr(H&Mon), Pr(T&Mon), Pr(H&Tue), and Pr(T&Tue). They should all be 1/4.
At noon, feed her a lunch that includes the amnesia drug and, if it is H&Tue, add in the sleep drug. In the afternoon, if she is awake, give her the answers from the morning and ask her to update them based on the information that it can't be T&Tue.
Let C be the event [(H&Mon) or (T&Mon) or (T&Tue)]
In order to get 1/2, you need to define T&Mon and T&Tue to be the outcome to SB. Yes, one occurring does guarantee that the other will occur, but THEY DO NOT OCCUR AT THE SAME TIME, SO SB CANNOT CALL THEM THE SAME OUTCOME.
One way to demonstrate this, is to keep SB awake on the afternoon of H&Tue, but take her shopping at Saks Fifth Avenue instead of having her update the probabilities. If T&Mon and T&Tue are the same outcome, then so are H&Mon and H&Tue. But SB clearly can experience only one at a time.
Now, please show me your definitions that say otherwise, using this procedure.
"At noon, feed her a lunch that includes the amnesia drug and, if it is H&Tue, add in the sleep drug. In the afternoon, if she is awake, give her the answers from the morning and ask her to update them based on the information that it can't be T&Tue."
On Sunday Night, have SB write "Heads and Monday" on the back of a 3x5 card. Then "Tails & Monday" on a second, and "Tails & Tuesday" on a third. Turn them over, shuffle them, and have her right "A", "B", and "C" on them. So for each letter, there is a 1/3 chance that it says "Heads" on the opposite side, and a 2/3 chance that it says "Tails."
On each waking day of the experiment, put the appropriate card on the interview table with the letter side up. When SB sees the letter, regardless of what letter it is, she knows that there still is a 1/3 chance that it says "Heads" and 2/3 chance that it says "Tails." And so a 1/3 chance that the coin landed Heads, and a 2/3 chance that it landed Tails.
+++++
A "result," or "outcome" of experiment, from Beauty's viewpoint, is a single day in the experiment. Because, to her, the amnesia severs any connection between days. The sample space (an event space is a set of subsets of the sample space) of these days is {"H&Mon", "T&Mon", "H&Tue", "T&Tue"}. SB being conscious has nothing to do with whether a day exists. To see this, just add a fourth 3x5 card to the above, that will never be used.
The entire controversy surrounding the problem is about how to rationalize considering just awake days, when all four "days" (I'll use that word to mean the coin and the weekday) are valid outcomes. Each has a prior probability of 1/4; to see this, wake SB on H&Tue but take her on a shopping spree instead of interviewing her about probability. So as she wakes up, she sees a 1/4 probability of a shopping spree.
Viewed this way, the problem becomes a conditional probability problem where, by the fact that Beauty is awake and interviewed on the day in question, eliminates H&Tue. The conditional probabilities for the other three days update to 1/3.
There are three cards. One says "Heads," and two say "Tails."
The letters "A", "B", and "C" are randomly assigned to one card each.
So, how exactly do you get "he Heads letter is 50%"? Do I need to explain the many ways this produces a contradiction? For example, if "A" has a 50% chance, then so does "B" and also "C." They add up to 150%
Beauty won't know which letter is assigned Heads, so her best guess has to be 1/3 per letter. However, the letter that is presented has a 50% chance of Heads.
1/6 - Heads + A
1/6 - Heads + B
1/6 - Heads + C
1/6 - Tails + A
1/6 - Tails + B
1/6 - Tails + C
The only way to get an answer that isn't 50% corresponding to Heads or Tails is if you instead ask about the likelihood of Beauty's *guesses* about the coin. I know that sounds like it should mean the same thing, but it actually doesn't.
I plan on updating this post soon (today or tomorrow) to make the distinction more clear, I'm hoping that will help.
"Each must have the same probability of being assigned to Heads"
True, as is the case with the probabilities I showed.
"These probabilities must add up to 1. So each is 1/3."
Also true, as is the case with the probabilities I showed: Each letter has 1/6 + 1/6.
"Please explain, with Mathematics and not what you want to be true"
I did, with the probabilities I showed. What you seem to be neglecting is the Heads probability space is 1/6 + 1/6 + 1/6.
----
All that aside, I really don't think this is the best way to talk about the problem. I just updated my post. It might be a bit rough around the edges because I haven't had the time to edit yet, but I wanted to get it out ASAP. I think if you give it another read, things will hopefully click into place.
"However, the letter that is presented has a 50% chance of Heads." Why? You are asserting the answer that you want, and claiming it proves the answer you want.
AGAIN: There are THREE cards. ONE says Heads, and TWO say Tails. So setting down sample space where there are the same number of each is patently false.
You also seem to have no idea what a probability space is. Hint: you attempted to list a sample space, but you did so incorrectly. A sample space is the set of all distinct, fully described results, called outcomes, of the experiment in question. Here is the sample space for how the cards can be labeled:
A=H&Mon, B=T&Mon, C=T&Tue.
A=H&Mon, C=T&Mon, B=T&Tue.
B=H&Mon, A=T&Mon, C=T&Tue.
B=H&Mon, C=T&Mon, A=T&Tue.
C=H&Mon, A=T&Mon, B=T&Tue.
C=H&Mon, B=T&Mon, A=T&Tue.
The incorrect sample space you listed included some of these results twice. For example, A=Heads is the same outcome where B and C are tails.
A "probability space" is a different concept, but it does include a sample space. It's a bit more complicated, so I'll use a simpler version where we just have a probability distribution; that is, a probability for each outcome in the sample space (we can construct a formal probability space from it, but that is overkill here). They must add up to one. Since there are six outcomes, and they are indifferent, each has a 1/6 probability.
Now the probability that A is equivalent to Heads is easily seen to be 1/6+1/6=1/3.
>In this variant, Chloe should guess she’s the only one.
SIA says that she should bet on the many-Chloe alternative in this case, assuming she thinks every other Chloe in existence is getting the same offer. If we aren't assuming that, the SIA answer depends on how probable she thinks it is that her clones (if there are any) are getting the offer.
For the Chloe problem here's a way to motivate the alternate answer. Allow a betting, insurance or future's market and a sufficiently large number of repetitions.
Let's also say instead of creating a billion billion clones in the case of rolling 1 million only a million million are created, but each of them receive a million if they correctly guess (so a million million, million (10^18) dollars are paid out by the happy parents).
The bet will be as follows:
1) to accept the bet Chloe voluntarily shakes the bookie's hand and then does a speech act, the speech act for Chloe accepting the bet is guessing that 1 million was rolled but counter-intutively she will thus be betting that 1 million was not in fact rolled
2) if 1 million is not rolled Chloe wins the bet and receives 1 million dollars from the booky, but nothing from her parents as she did not guess correctly.
3) If the million is rolled Chloe loses the bet (but receives the million dollars from her parents) and pays out a single dollar to the booky.
Thus on the average 1 000 000 iterations (where 1 million is rolled once) the booky pays out $999 999 million but receives $ million million dollars and so the booky will be ahead $ 1 million dollars.
Neither Chloe nor the booky (I'm imaging the booky as a R0G3R type battle droid, or a million million in the case where there are a million-million identical bookies to let Choe take the bet) know what the die roll was. Yet given what they know both do something fairly reasonable (arguably Chloe is slightly shortchanged on the deal, but only by a millionth of a dollar and so we can chock it up to risk aversion and necessary friction to compensate the bookie).
This reflects the fact that from the point of view of the galaxy the expected utility to Chloe guessing the 1-in-a-million is much greater than the utility of her guessing not that. This also reflects the fact that for every Chloe-like point of view where a number other than a million was rolled there are a million Chloe-like points of view where it was (assuming enough iterations to average it out). Although this assumes the million-million Chloes would not exceed the carrying capacity of the Galaxy and throw the Galaxy into a hard scrabble struggle for survival. Or that the parents are not just basically printing money to give Chloe and so creating a massive inflation.
On Linch’s blog I advocated “utility side SIA”; basically use halfer probability but multiply the utility by the # of copies. More generally, I suggested you could take the view that “only PU matters”, and freely shuffle factors between P and U depending on what makes things easier to think about.
Whenever there’s a “canonical” final outcome (as in your examples), then it’s usually obvious what you should do and as you show sometimes this looks like the halfer move and sometimes line the thirder.
In the anthropic dilemmas I find interesting there is no canonical outcome. Eg for the presumptuous philosopher you just have to say if you care more about being responsive to the experimental evidence or about how many of the philosopher’s copies are right. I’m sympathetic to the latter because the SIA move is what you’d choose given a “view from nowhere”, but there are problems with this (eg as I raised with Lich-sometimes universes with many copies seem too “cheap” wrt standard priors). Feels like there should be a solution, but idk what it is.
> In the previous version, we determined that the chance is 1 in 1 million. This time, though, the long tail of wakings finally comes into play and the chance is 1 in a billion billion.
p(1 million and day 1) is 1 in a million billion billion (on halfer logic), p(day 1) ~ 1, so conditionalise and you get 1 in a million billion billion (to a close approximation).
I think you're right! I think? But why would it be approximate rather than exact?
To condition, you divide the joint probability by the marginal: P(1 million and day 1)/P(day 1). P(day 1) is approximately (1 million -1) / 1 million (because in only 1 case out of a million do you wake on any other day, and in that case you are almost almost guaranteed to wake up on another day). So you're dividing 1 in a million billion billion by a tiny bit less than 1.
Woops, I thought you were referring to my second version if this, but I just checked the quote, and for the first version P(Day 1) = 1 / 1 million, isn't it?
Also, not sure about P(1 million and day 1)/P(day 1), because that assumes independent variables? Or does it not? My stats 101 is admittedly rusty.
Oh no, I thought you were talking about the original problem with day labels....
The conditioning formula is valid regardless of dependence, however.
Ok I've read the day labelling procedure more carefully and I still think P(day 1) \approx 1. If the dice does not land on 1 million, then P(day 1) = 1. But that happens with probability (1 million - 1)/1 million, so P(day 1) has probability of at least that. Why do you think it is 1/1million?
*facepalm* I was confusing myself. Maybe I'll make an edit for clarity. And I'll fix the probability, thank you for that!
"The original question is ambiguous2 about which goal you should be maximizing for, but for my variant, the goal isn't “How do I prevent myself from being wrong the most number of times?”. The goal is, “What should I guess to maximize the likelihood that I’m not wrong right now?”"
The original SB asks what your credence upon awakening that the coin landed on heads *ought to be*. Does this depend on what goal you have? Perhaps, but if your goal is to be rational in your beliefs then perhaps that disambiguates.
There'a a famous LessWrong blog post by Yudkowsky about how the rational thing to do is to "make your beliefs pay rent in anticipated experiences."
Perhaps you ought to believe there is an invisible dragon in your garage if you value believing in dragons and don't care about truth. But if you want to be rational and don't actually anticipate that the bucket of flour you are holding will reveal the outline of a dragon when you dump the flour out into the garage, and actually anticipate that it will fall to the floor as if the garage is empty, then the rational thing to believe is that there is no invisible dragon in your garage.
So upon waking in Sleeping Beauty, is the rational thing to believe that the coin landed on heads with p=1/2 or p=1/3? Well that depends on what you anticipate experiencing. Do you anticipate that you are just as likely to learn that the coin landed on heads as you are to learne that the coin landed on tails? Or do you anticipate being twice as likely to learn that the coin landed on tails?
Extremizing it to p=0.001 (halfer view) and p=~0.999 (thirder view) by using a 1,000-sided die instead of a fair coin and having one million awakenings if and only if the die lands on 1,000 (and one awakening otherwise), the question is thus, upon awakening, do you anticipate that when you look over at the die to see what it landed on that you will almost definitely see that it landed on 1,000 (the SIA thirder view) or do you anticipate almost definitely seeing that it landed on some value from 1-999 (the SSA halfer view)?
I take the halfer view because I would anticipate seeing the die on some random value like 812. That is, I would almost definitely expect to see that it landed on 1-999.
The arguable ambiguity of the original phrasing of the question is something I skimmed over because (1) I didn't want to lose any readers who might disagree over it and (2) I cared more about arguing my case for the variant that makes most sense to me.
But... I think if you were to imagine the experiment being re-run many times, the "thirder" position sounds true to me, because you'd expect the unlikely result to happen at least once, and then it becomes overwhelming. Surely your "credence" must be impacted by the observation selection effect at that point.
What throws me, though, is the case where the experiment only gets run once. Does it make more sense to base your credence off the coin flip / dice roll, or to base it off the expected value of your observations? Maybe the "halver" position makes more sense here, but if that's true, and it's true that enough repeats will switch to the "thirder" position, then where's the dividing line?
I like my variant because it makes the question feel well defined. The original... I just don't know.
"But... I think if you were to imagine the experiment being re-run many times, the "thirder" position sounds true to me, because you'd expect the unlikely result to happen at least once, and then it becomes overwhelming."
I think the question "What is the expected percentage of awakenings in which the coin lands on heads?" (Answer=1/3) is a distinct question from the question being asked in Sleeping Beauty problem: "Upon awakening, what ought your credence be that the coin landed on heads?"
I think the correct answer to the SB question is the same regardless of whether it's the first time you're participating in the experiment as SB or the nth time. (I answer 1/2 regardless.)
I broadly agree with you that Linch is wrong / that SIA is wrong. I too am a halfer / "Anti".
I use the same alternate version of the SB problem as you, except I used the numbers 1/1,000 chance of a million awakenings (or clonings) rather than a 1/1,000,000 chance of a billion billion awakenings.
Your version is too extreme as a billion billion awakenings would take 31 billion years even if each awakening took only 1 second with no time in between.
I prefer clonings/copies to memory erasure because I think they're easier to think about and are functionally identical. What is memory erasure after all if not a copy/clone of a previous brain state? (I think they're the same.)
I agree the thirder position (aka SIA) is controversial in philosophy; eg Bostrom takes the halfer *(aka SSA) position. I covered the dispute under: https://linch.substack.com/i/169429562/further-reading
I decided to only include the thirder position in the main section a) because I personally believe it's correct, b) because it's the mainstream position, and c) because for a 101 class for babies, many things needed to be cut and I thought giving people the confidence to think about anthropic problems for themselves was more valuable than presenting the existing scholarly disputes.
I'm not sure I agree with your presentation on "observation selection not mattering" in the examples you mention, but will think about it!
Oh my gosh, I'm only now realizing that I got so distracted with Sleeping Beauty, I never actually finished reading your post!!!
In my defense, I was reading your post and working on my post in the downtime I had while vacationing in Japan, which was a really fun trip. But maybe suboptimal for my reading focus, haha
I'm not offended haha! You have the distinct honor of being my first critic who has written a full post as a reply, and I'm happy it's about an anthropics dispute rather than politics or animal welfare!
I'm sure if there's anything more difficult to talk about than the Sleeping Beauty Problem, it's politics haha.
I haven't really discussed animal welfare much online, is it just as bad??
well friends of the blog Bentham's Bulldog and Flo Bacus gets a lot of flack for their animal welfare takes.
Then I wish them luck in the fight!
The original Sleeping Beauty Problem (this is an edited quote from Arnold Zuboff) went more like this: "a hypnotist puts SB into a hypnotic sleep that is to last for N days, where N>>1. The hypnotist will interrupt her sleep either daily, and thus N times, or else only once on a randomly selected day among the N. Which of these two numbers of awakenings there will be is determined by a toss of a fair coin whose outcome is kept secret from SB. At the end of any period of wakefulness, SB must be hypnotized into forgetting it completely and permanently before being made to sleep again. SB, who knows everything just explained, can score a win only if she correctly answers the question of which number of times she is being awakened."
Adam Elga's problem statement altered this to N=2, Heads meaning one waking (Zuboff didn't specify), and he asked the equivalent question about the coin landing Heads.
But his problem statement did not specify on which day the one waking would occur. That he added in his solution, so he could identify the possible days. I suggest that he did this to avoid having to resort to slightly more involved Mathematics that might confuse some, but what it actually accomplished was to start the entire controversy.
I'll go back to a randomly selected day, the Heads schedule, and the Heads question using the following method: At the same time the coin is flipped, roll an N-sided die. Each day, wake SB if the coin is showing Tails or the die is showing "1." Each day after SB is put back to sleep, either rotate the die down one number, or set it to N if it was already on 1.
On any day during the experiment, the "outside" or "prior" probability that the coin is showing Heads is 1/2. And the "outside" or "prior" probability that the die is showing 1 is 1/N. These are independent random variables, so the "outside" or "prior" probability of any combination is 1/(2N). And there are 2N combinations that can occur during the experiment.
But when she is awake, SB knows only that she is in one of the N combinations where the coin is showing Tails, or the one where the coin is showing Heads and the die is showing 1. So the probability that the coin is showing Heads is trivially 1/(N+1).
As you said yourself: The probability of Heads is 1/2.
This is true before and after SB wakes; it's true of her belief of the coin before or after she wakes.
What corresponds to ~1/N is how often she'll guess Heads in repeated experiments, and her belief in her own guesses, which is also unchanged by the act of sleeping or waking.
Any real solution can't have Sleeping Beauty changing her credences after learning no new information.
The *prior* probability of Heads is 1/2. That means that, on any day during the experiment it is equally likely to be Heads or Tails.
The *prior* probability of Monday is 1/2. That meas that, on any day during the experiment it is equally likely to be Monday or Tuesday.
These probabilities are independent; that means you can multiply them to get the *prior* probability of any combination
But once information is learned, you need to use conditional probability. Those probabilities do not take SB's awake/sleep state ON A SINGLE DAY into account. "Information" means any change to the possibilities that depend on the actual values of COIN and/or DAY.
Being awake, since it can't happen when COIN=HEADS and DAY=TUESDAY, is such information. Being awake can happen in three combinations only; that's prior probability 3/4. This is not the same as "I will be awake during the experiment," it applies individually to each combination.
Yes, so viewed one way the P(Heads + Mon) is 1/2 and the other two possibilities are 1/4. Or they're all 1/2, depending on how you define your terms.
If you'd like, you can show me a specific set of definitions and a calculation for arriving at the 1/N value, and I can either show you how that maps out to guesses rather than coinflips, or I can show you where there is an unwarranted assumption.
There is no valid definition where they are all 1/2.
Try this: Wake SB both days. In the morning, have her write down the probabilities for Pr(H), Pr(T), Pr(Mon), and Pr(Tue). They should all be 1/2. Then the probabilities for Pr(H&Mon), Pr(T&Mon), Pr(H&Tue), and Pr(T&Tue). They should all be 1/4.
At noon, feed her a lunch that includes the amnesia drug and, if it is H&Tue, add in the sleep drug. In the afternoon, if she is awake, give her the answers from the morning and ask her to update them based on the information that it can't be T&Tue.
Let C be the event [(H&Mon) or (T&Mon) or (T&Tue)]
Pr(C) = Pr(H&Mon) + Pr(T&Mon) + Pr(T&Tue) = (1/4)+(1/4)+(1/4) = 3/4
BY THE DEFINITION OF CONDITIONAL PROBABILITY:
Pr(H|C) = Pr(H & C)/Pr(C) = Pr(H&Mon)/Pr(C) = (1/4)/(3/4) = 1/3.
Pr(T|C) = Pr(T & C)/Pr(C) = Pr(T&Mon or T&Tue)/Pr(C) = 2/3.
In order to get 1/2, you need to define T&Mon and T&Tue to be the outcome to SB. Yes, one occurring does guarantee that the other will occur, but THEY DO NOT OCCUR AT THE SAME TIME, SO SB CANNOT CALL THEM THE SAME OUTCOME.
One way to demonstrate this, is to keep SB awake on the afternoon of H&Tue, but take her shopping at Saks Fifth Avenue instead of having her update the probabilities. If T&Mon and T&Tue are the same outcome, then so are H&Mon and H&Tue. But SB clearly can experience only one at a time.
Now, please show me your definitions that say otherwise, using this procedure.
"At noon, feed her a lunch that includes the amnesia drug and, if it is H&Tue, add in the sleep drug. In the afternoon, if she is awake, give her the answers from the morning and ask her to update them based on the information that it can't be T&Tue."
I'm sorry, I'm not understanding this setup?
On Sunday Night, have SB write "Heads and Monday" on the back of a 3x5 card. Then "Tails & Monday" on a second, and "Tails & Tuesday" on a third. Turn them over, shuffle them, and have her right "A", "B", and "C" on them. So for each letter, there is a 1/3 chance that it says "Heads" on the opposite side, and a 2/3 chance that it says "Tails."
On each waking day of the experiment, put the appropriate card on the interview table with the letter side up. When SB sees the letter, regardless of what letter it is, she knows that there still is a 1/3 chance that it says "Heads" and 2/3 chance that it says "Tails." And so a 1/3 chance that the coin landed Heads, and a 2/3 chance that it landed Tails.
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A "result," or "outcome" of experiment, from Beauty's viewpoint, is a single day in the experiment. Because, to her, the amnesia severs any connection between days. The sample space (an event space is a set of subsets of the sample space) of these days is {"H&Mon", "T&Mon", "H&Tue", "T&Tue"}. SB being conscious has nothing to do with whether a day exists. To see this, just add a fourth 3x5 card to the above, that will never be used.
The entire controversy surrounding the problem is about how to rationalize considering just awake days, when all four "days" (I'll use that word to mean the coin and the weekday) are valid outcomes. Each has a prior probability of 1/4; to see this, wake SB on H&Tue but take her on a shopping spree instead of interviewing her about probability. So as she wakes up, she sees a 1/4 probability of a shopping spree.
Viewed this way, the problem becomes a conditional probability problem where, by the fact that Beauty is awake and interviewed on the day in question, eliminates H&Tue. The conditional probabilities for the other three days update to 1/3.
Thanks for the comment.
The Heads letter is 50%, however.
There are three cards. One says "Heads," and two say "Tails."
The letters "A", "B", and "C" are randomly assigned to one card each.
So, how exactly do you get "he Heads letter is 50%"? Do I need to explain the many ways this produces a contradiction? For example, if "A" has a 50% chance, then so does "B" and also "C." They add up to 150%
Beauty won't know which letter is assigned Heads, so her best guess has to be 1/3 per letter. However, the letter that is presented has a 50% chance of Heads.
1/6 - Heads + A
1/6 - Heads + B
1/6 - Heads + C
1/6 - Tails + A
1/6 - Tails + B
1/6 - Tails + C
The only way to get an answer that isn't 50% corresponding to Heads or Tails is if you instead ask about the likelihood of Beauty's *guesses* about the coin. I know that sounds like it should mean the same thing, but it actually doesn't.
I plan on updating this post soon (today or tomorrow) to make the distinction more clear, I'm hoping that will help.
There are three letters. Each must have the same probability of being assigned to Heads. These probabilities must add up to 1. So each is 1/3.
There. That's the mathematical reasoning. Please explain, with Mathematics and not what you want to be true, how to get 1/2.
"Each must have the same probability of being assigned to Heads"
True, as is the case with the probabilities I showed.
"These probabilities must add up to 1. So each is 1/3."
Also true, as is the case with the probabilities I showed: Each letter has 1/6 + 1/6.
"Please explain, with Mathematics and not what you want to be true"
I did, with the probabilities I showed. What you seem to be neglecting is the Heads probability space is 1/6 + 1/6 + 1/6.
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All that aside, I really don't think this is the best way to talk about the problem. I just updated my post. It might be a bit rough around the edges because I haven't had the time to edit yet, but I wanted to get it out ASAP. I think if you give it another read, things will hopefully click into place.
"However, the letter that is presented has a 50% chance of Heads." Why? You are asserting the answer that you want, and claiming it proves the answer you want.
AGAIN: There are THREE cards. ONE says Heads, and TWO say Tails. So setting down sample space where there are the same number of each is patently false.
You also seem to have no idea what a probability space is. Hint: you attempted to list a sample space, but you did so incorrectly. A sample space is the set of all distinct, fully described results, called outcomes, of the experiment in question. Here is the sample space for how the cards can be labeled:
A=H&Mon, B=T&Mon, C=T&Tue.
A=H&Mon, C=T&Mon, B=T&Tue.
B=H&Mon, A=T&Mon, C=T&Tue.
B=H&Mon, C=T&Mon, A=T&Tue.
C=H&Mon, A=T&Mon, B=T&Tue.
C=H&Mon, B=T&Mon, A=T&Tue.
The incorrect sample space you listed included some of these results twice. For example, A=Heads is the same outcome where B and C are tails.
A "probability space" is a different concept, but it does include a sample space. It's a bit more complicated, so I'll use a simpler version where we just have a probability distribution; that is, a probability for each outcome in the sample space (we can construct a formal probability space from it, but that is overkill here). They must add up to one. Since there are six outcomes, and they are indifferent, each has a 1/6 probability.
Now the probability that A is equivalent to Heads is easily seen to be 1/6+1/6=1/3.
>In this variant, Chloe should guess she’s the only one.
SIA says that she should bet on the many-Chloe alternative in this case, assuming she thinks every other Chloe in existence is getting the same offer. If we aren't assuming that, the SIA answer depends on how probable she thinks it is that her clones (if there are any) are getting the offer.