Was a bit worried about the pump-against-thirders there, but I think that objection makes sense. Somehow two of my favourite problems (Sleeping Beauty, and Newcomb's) have gotten all tangled up. For reference, playing Prisoner's Dilemma against your clone is basically Newcomb's problem. And so this pump-against-thirders works only against CDT thirders, I think. EDT and FDT would refuse to cancel their bet, saying "if I cancel, then I'd cancel in all 3 worlds, which is bad". That being said, I saw a money pump (-ish) where CDT beats EDT, so thirder+FDT is looking like the best combo here. (My thirder argument: modify the problem so that Tuesday+Heads instead involves waking up but being told "and the experiment is over, you don't need to answer anything". So upon waking up, you have 25% on all 4 worlds, then when you don't get told the experiment's over, you eliminate the Tuesday+Heads world, leaving 2/3 chance of Tails)
> That being said, I saw a money pump (-ish) where CDT beats EDT, so thirder+FDT is looking like the best combo here.
Why wouldn't the exact same money pump work for FDT thirdism as well?
> My thirder argument
There are only two possible outcomes of the experiment: Heads and Tails. In both of them "Not told once" happens, so this event doesn't update you in any way.
Your mistake is in assuming that
P(Not told once) + P(Told once) = 1,
while, in fact, on Heads both of these events happen.
I don't agree that this doesn't update you. Do you agree that before the information (i.e. just after waking up; suppose we knew that "experiment over" info happens 10 minutes after we wake up, if it happens) we should be equally split between the four cases, e.g. {monday+heads, ...}? That is, we have a prior of 25% on these 4 cases. Then, if we wait and no-one comes to tell us the experiment is over, we eliminate the Tuesday+Heads case. So we have P(Tails | (not tuesday+heads)) = P((not tuesday+heads) | tails) * P(tails) / P(not tuesday+heads) = 1 * 0.5 / (3/4) = 2/3. Which step here do you disagree with?
> In probability theory, the sample space (also called sample description space,[1] possibility space,[2] or outcome space[3]) of an experiment or random trial is the set of all possible outcomes or results of that experiment.
> The outcomes must be mutually exclusive, i.e. if [one] occurs, then no other will take place
In the same iteration of this experiment both Monday and Tuesday awakenings happen. Therefore, individual awakenings are not mutually exclusive. Therefore they can't form the sample space for this experiment.
Meanwhile, states of the coin are mutually exclusive. In every iteration where the coin is Tails it's not Heads and vice versa. Therefore they can form sample space for this experiment.
The correct model is:
P(Heads) = P(Tails) = 1/2
P(Monday) = P(Tuesday) = 1
Where "Monday" means "Moday awakening happens in this iteration of the experiment", and likewise for Tuesday.
When you are told that experiments is over it means that you learn that Tuesday awakening has happened and that the coin is Heads. This naturally makes you certain that the coin is Heads:
P(Heads|Heads&Tuesday) = P(Heads|Heads) = 1
When you are not told that experiment is over you learn that Monday awakening has happened and the coin is either Heads or Tails or that Tuesday awakening has happened and the coin is Tails, this correspond to the whole sample space so you do not update:
P(Heads|(Heads or Tails)&Monday or Tails&Tuesday) = P(Heads|Heads or Tails or Tails) = P(Heads|Heads or Tails) = P(Heads) = 1/2
Maybe there's some philosophical disagreement here? The reason I say that they're mutually exclusive is that P(It's monday right now and it's tuesday right now)=0. Is there some reason this shouldn't be allowed?
Hold on though, P(Monday happens during the experiment) is not the same as P(Monday right now), which is what I'm using. P(heads and monday right now) and so on ARE mutually exclusive, and should sum to 1. Under this definition, what would you say is P(monday right now | experiment-not-over) ?
> P(Monday happens during the experiment) is not the same as P(Monday right now), which is what I'm using.
"Monday happens during this trial" is a well-defined event, which in every iteration of the experiment has a stable truth value. And as it's True in 100% iteration of experiment, probability of this event is 1.
While "Monday right now" is an ill-defined event, which doesn't have a stable truth value in any iteration of this experiment. In every trial it is supposed to be simultaneously True (on Monday) and False (on Tuesday). Therefore it doesn't have coherent probability measure.
Now, we can come up with some different experiment where these events are well-defined. That should be an experiment where Monday and Tuesday awakenings do not happen in the same iteration - are mutually exclusive. Such as:
You will be awakened either on Monday or on Tuesday. And a coin is tossed. If it's Tuesday Heads you are told that. Otherwise you are told nothing. You are awakened and told nothing what is the probability that the coin is Heads?
As you can see "Monday right now" is well-defined here. In every iteration of the experiment it is either True or False, because there is not a single iteration where both awakenings happen.
Is it ill-defined? Maybe that's where our disagreement is. If I ask you "what is the time right now", I consider this a well-defined and objective question, even though the (nominally) same question can lead to different answers if asked repeatedly. Perhaps it's me making some Bayesian assumption, but I see it as as normal to put a probability on "monday right now" representing your uncertainty -- and yes, this can change over time. I imagine that I already effectively do this: I'm at about 95% currently, in a few hours it might decrease as I might've forgotten the distinctness of the memory of recently having checked the day, in even more hours it will decrease more as I wonder whether we've passed midnight, and then if I check and it has then it'll go very low. Does your framework not accept this sort of thing? I agree that I'm not being super rigorous, but I still think there's something clear underneath. I could maybe try to formalise what I'm imagining here if it would help
It makes me somewhat emotional to say that I've seen Bentham's Bulldog typing a post, in the flesh! (And also heard him talk, and was surprised to realize that he talks in a way that's very similar to the way he writes) Anyway, the handful of minutes I got to spend with BB seems to confirm my impression that he's not only an effective altruist, but also a cool guy all around.
If you accept that your credences are/should be strongly tied to your betting odds, and that it's a bad thing when your betting odds predictably lead to losing all your money in some situations, it shouldn't matter how complicated those situations are. (Not that they even are all that complicated in this case.) It's an argument against the rationality of your credences either way.
I'm very confused about the double halving money pump.
>But she also thinks that there’s a 1/8 chance that it’s currently day two and the coin that will be flipped at the end of today will come up heads.
Why would it be 1/8? The whole point of being a double-halver is that you have the same experience no matter which side the coin landed on, so your experience doesn't give you any evidence. So no matter what a double-halver experiences they're going to assume that the first coin flip had 1/2 chance of being heads or tails, and whether the first coin comes up heads or tails has no effect on whether the next coin flip will be heads or tails.
>Offer Beauty a bet before she learns what day it is that pays out 46 cents if the coin that will be flipped at the end of today comes up heads, but costs 54 cents if it comes up tails. Naturally, Beauty thinks there’s a 5/8 chance that the coin will come up heads, so she takes the bet.
Why would she think there's a 5/8 chance? What day it is won't effect the probability of a coin being flipped at the end of that day.
The 1/8 number isn't very important. There's a 1/2 chance the first coin came up heads--which entails that it's currently day 1 and the coin will come up heads. There is also, quite certainly, some chance that it's currently day 2 and the coin came up tails. The precise number doesn't much matter.
The reason that it's 1/8 is that three things have to happen for it to be day 2 and it being the case that the coin will come up heads. The first coin has to come up tails. It has to presently be the second day (of two). And the second coin has to come up heads. Each has 1/8 probability.
She thinks there's a 5/8 chance because there are two scenarios in which it's true that "the coin that will be flipped at the end of today will come up heads." One of them has 1/2 odds. The other has 1/8 odds.
>There's a 1/2 chance the first coin came up heads--which entails that it's currently day 1 and the coin will come up heads. There is also, quite certainly, some chance that it's currently day 2 and the coin came up tails. The precise number doesn't much matter.
Actually, it matters a lot and after five days of chewing over this (and thanks to Ape in the Coat linking to the Boy or Girl Paradox) I've finally figured out what's wrong with your calculation!
You've got it laid out like this:
Odds that it is Monday and the coin will come up heads: 1/2
Odds that it is Monday and the coin will come up tails: 1/2
Odds that it is Tuesday and the coin will come up heads: 1/8
Odds that it is Tuesday and the coin will come up tails: 1/8
Therefore the odds that the coin will come up heads is 5/8
Which of course doesn't work (all the probabilities add up to more than 1)
But that math is incorrect. If you are a halfer then the probabilities come out like this:
Odds that it is Monday and the coin will come up heads: 2/6
Odds that it is Monday and the coin will come up tails: 2/6
Odds that it is Tuesday and the coin will come up heads: 1/6
Odds that it is Tuesday and the Con will come up heads: 1/6
Therefore the odds that the coin will come up heads is 3/6, or 1/2
The reason for this is straightforward:
The odds that Beauty will be awake on a Monday=1/1 (No matter what, she's awake on Monday)
The odds that Beauty will be awake on a Tuesday=1/2 (She's only awake on Tuesday if the coin came up tails on Monday, which as a 1/2 chance)
Therefore, it is twice as likely that Beauty is awake on a Monday than that she is awake on a Tuesday. In other words:
Probability that it is Monday: 2/3
Probability that it is Tuesday: 1/3
With that in mind, what is the probability that today is Monday and the coin will come up heads? It's the probability that it is Monday (2/3) times the probability that a coin will come up heads (1/2), which equals 2/6. Same probability for it being Monday and coming up tails. And what is the probability that it's Tuesday and the coin will come up heads? It's the probability that it is Tuesday (1/3) times the probability of a coin coming up heads (1/2) which equals 1/6
In other words, on waking the halfer should think there's a 2/6 chance it is Monday and the coin will come up heads, and a 1/6 chance that it is Tuesday and the coin will come up heads, for an overall odds of 3/6 that the coin will come up heads. On learning that today is Monday the odds change to a 1/2 chance that it is Monday and the coin will come up heads (1*1/2), and since 3/6 is equal to 1/2 the betting odds remain the same.
That's not what I'd say that double halfers credences are. What they are:
P(it's day 1 and the coin will come up heads)=1/2
P(it's day 1 and the coin will come up tails=1/4 (given tails there are two equiprobable days--the odds of tails are 1/2 and the odds it would be day 1 given tails are 1/2).
P(it's day 2 and the coin will come up tails)=1/8
P(it's day 2 and the coin will come up heads)=1/8
Your math emphatically negates halferism. It implies that in the version where there's a coinflip on the second day, you should think it's twice as likely the coin flipped on day 1 came up tails as heads so that there are two awakenings,
There are two different questions being asked: what day is it, and what are the odds that a coin flipped on that day will come up heads. If not told the day, Beauty should assume that it is twice as likely to be Monday as it is to be Tuesday, so 2/3rds chance it is Monday. And because Beauty is a double-halfer she knows that the odds of the coin flip remain at 1/2 no matter what day it is. So the odds that it is Monday and the coin will come up heads is 2/6, and the odds that it is Tuesday and the coin will come up heads is 1/6, so altogether the odds of the coin coming up heads today is still 1/2.
Your version of the double halfer credences only works if the double-halfer thinks it is both 1/2 likely that the coin will come up heads and 1/2 likely that today is Monday. But that's not the double halfer position, the double halfer position is that even if Beauty knows it is Monday it shouldn't change her credence about the coin odds. And it's clear why: if we don't know what day it is the odds are 2/6+1/6 that the coin will come up heads, if we know that it's Monday then the 1/6 chance disappears and it become 3/6, so the update in information does not change the odds for the coin flip.
No dude, by halfer logic she should think it's 3x likelier to be Monday as Tuesday. There's a 1/2 chance that there's only one awakening. That's the core claim of the halfer position. Conditional on there being two awakenings, odds are 1/2 that the present day would be a Monday awakening. So overall, for it to be tuesday, two things with odds of 1/2 have to happen--overall it has 1/4 odds, making the other one twice as likely.
The double halfer position--by virtue of being a halfer position--think that before knowing her birth rank, there's a 1/2 chance that the coin flipped at the end of day 1 comes up heads. After all, the double halfer thinks that when the coin that determines whether beauty wakes up a second time is flipped on Sunday--changing when in time the coin is flipped shouldn't change her credence in its outcomes. But if the coin comes up heads on day 1, that straightforwardly entails that the present day is day 1 and the coin will come up heads. Therefore, she has to think that has probability 1/2 and therefore the rest of the probabilities are as I desccribed.
You're absolutely right: it is 3x times likelier to be Monday as Tuesday, not 2x. But the math still checks out!
Odds of it being Monday: 3/4
Odds of it being Tuesday: 1/4
Odds that it is Monday and the first coin is heads is 3/4*1/2=3/8
Odds that it is Monday and the first coin is tails is 3/4*1/2=3/8
Odds that it is Tuesday and the first coin is heads and the second coin is heads is 0/4*1/2=0
Odds that it is Tuesday and the first coin is heads and the second coin is tails is 0/4*1/2=0
Odds that it is Tuesday and the first coin is tails and the second coin is heads is 1/4*1/2=1/8
Odds that it is Tuesday and first coin is tails and the second coin is heads is 1/4*1/2=1/8
Which means that if I don't know the day then the odds that today's coin will be heads is 3/8+1/8=4/8=1/2. Then when I learn that it is Monday it becomes 1*1/2=1/2. No money pump.
>She thinks there's a 5/8 chance because there are two scenarios in which it's true that "the coin that will be flipped at the end of today will come up heads." One of them has 1/2 odds. The other has 1/8 odds.
I think this is the crux of my confusion: no matter what day it is, the odds of a coin being flipped at the end of that day remains 1/2, doesn’t it? Both scenarios have 1/2 odds when it comes to a coin flip at the end of that day. How does it matter whether it’s day 1 or day 2?
It's because "today's coin" is a different coin depending on when today is, and what day today is is entangled with what values of "today's coin" have shown up in the experiment so far.
I'll be honest, I don't think a halfer is bound to respect BB's calculation, since at least some halfers take their position on the basis that shenanigans involving what "today" is are the root of misunderstanding in the SB case to begin with.
Basically, the total set of possibilities is:
1. Coin 1 heads, today is Monday
2. Coin 1 tails, today is Monday, coin 2 heads
3. Coin 1 tails, today is Monday, coin 2 tails
4. Coin 1 tails, today is Tuesday, coin 2 heads
5. Coin 1 tails, today is Tuesday, coin 2 tails
No "coin 1 heads, today is Tuesday" appear because there is no Tuesday awakening if coin 1 is heads, and no coin 2 is listed for coin 1 heads for the same reason: there is no coin 2 if coin 1 is heads.
If you take the view that p(coin 1 heads) = p(coin 1 tails) =1/2, and similarly for coin 2, and that conditional on coin 1 being tails, each of Monday or Tuesday is equally likely, then all probabilities are 1 except p(today is Monday|coin 1 heads) which equals 1.
So the probability of each outcome above is p=1/2 for outcome 1, and p=1/8 for the others, which sums up to 1.
Finally, notice that for situations 1,2,3 "today's coin" is coin 1, whereas for 4,5 it's coin 2. So the outcomes where "today's coin lands heads" are 1 and 4 only, with probabilities 1/2 and 1/8 respectively, hence overall probability that "today's coin is heads" is 1/2 + 1/8 = 5/8.
One thing I notice in my working this through is, it didn't actually require double-halfing? I think it only used the fact that p(coin 1 is heads)=1/2, i.e. it just relied on SB being a halfer?
I appreciate you trying to explain this to me, but I'm still confused. Why in the world would what day it is or what the first coin result was affect the probability of what a coin flip at the end of the day will be? No matter what the first coin flip was I get woken up and asked to bet on whether a coin flipped at the end of the day will be heads, right? Since the first coin flip doesn't affect the odds of getting a "heads" on future coin flips, what the first coin did is immaterial to the question. And since coins don't flip with different odds on Tuesdays than they do on Mondays, what day it is is also irrelevant to the question.
So I assume I'm confused on what the actual question we're trying to answer is here.
Fwiw I agree with you, or at least, I don't think I quite buy BB's argument. I agree that what you say sounds correct and intuitive, and I'm not suggesting that BB's reasoning _overrides_ yours; I'm just saying BB's reasoning also has a certain logic to it, and yet gives a different answer.
"Since the first coin flip doesn't affect the odds of getting a "heads" on future coin flips, what the first coin did is immaterial to the question."
It's not that it affects the odds of heads on future coin flips, it affects the odds of there _being_ future coin flips, and the probability you can assign to the different values of what "today" is.
I think your argument is something like, if X is "today's coin flip lands heads", and A is "today is Monday", then p(X|A)=p(X|~A)=1/2, i.e. regardless of the day, the probability of getting heads on today's coin is 1/2, and since p(A)+p(~A)=1 (it's either Monday or Tuesday and those are exhaustive), we should be able to calculate
p(X)=p(X|A)p(A)+p(X|~A)p(~A)
= 1/2[p(A)+p(~A)]=1/2
where in the last step I use that both conditional probabilities are 1/2 so I can factor that common factor out.
But as I noted elsewhere p(X|A) is "today's coin is heads, given that today is Monday", whereas p(X|~A) is "today's coin is heads, given that today is Tuesday"--so "today's coin" refers to two different coins in these cases. It's not really clear that we can just use an identity over the variable X when X is isn't referring to a fixed coin, it takes different values at different terms in the equation.
So this formalized version of your reasoning is a bit suspect. That doesn't mean you can't tighten it up, but it should at least make you a little more open to an analysis that lays out the different possibilities by day and coin, and tries to count over those, and doing that you might plausibly get BB's answer.
I really do appreciate you helping me figure this out. Is it really the case that p(X|A) and p(X|~A) are referring to two different coins? For all we know they use the same coin no matter what day it is, and even if they don't any fair coin is going to have the same probability of coming up heads. If it is a problem, then X should really be "A fair coin flipped at the end of today will land on heads".
Just to say a little more, I think this whole line of reasoning is something a halfer is likely to reject as this kind of "listing out the possibilities and crossing them off" is what you have to reject to be a halfer in the first place. After all, a simple thirder argument is, there are four possibilities:
1. Heads and it's Monday
2. Heads and it's Tuesday
3. Tails and it's Monday
4. Tails and it's Tuesday,
with option 2 being impossible, resulting in 3 possibilities, two of which have the coin being tails, hence p(T)=2/3.
So it's not clear that a halfer would actually buy the calculation BB is doing.
>Suppose you have two actions that are correlated, such that either action individually guarantees that you’ll take the other action. In this case, I think each action should only be given credit for half of the value that you get from the sequence of actions. If there are two buttons where if you press them both you’ll get 4 dollars, even if the button-pressings are correlated, they should each only count for two dollars.
It doesn't really make sense to ascribe dollar amounts to individual components of your choices/strategy, instead of to the totality of your choices/strategy (unless the consequences of that component are in some sense independent of those of all other components). The component choice "pressing the left button" underspecifies an outcome (or even probability distribution over outcomes!) if both buttons are necessary to get the four dollars. Hence it doesn't have a dollar amount attached expected value-wise.
Both CDT and EDT agree that EV caclulations only make sense when you hold fixed certain things that are sufficient to determine relevant probability distributions over outcomes (though they may disagree as to which things those are). "Pressing the left button and *possibly* the right button" in your example doesn't do that on either CDT or EDT.
"If any view holds thirding is right, it must think that the existence of more candidates for being your present self makes the existence of your present self likelier."
Is this true? Being a thirder requires holding SIA-ish views? What's the argument for that?
This is good but I have some disagreements with your "if there are two actions that together get you one payout, then you should only treat each individual action as responsible for half the payout" principle.
1) The principle is not neutral wrt personal identity. What I mean by this is that it suddenly matters a lot to your decision-making if some action was made by YOU, or someone else. If it's someone else, you treat it normally by assigning credences to what that person does, and doing expected utility. If it's you, you instead do this division of responsibility thing. So rational action is going to depend heavily on getting into the weeds of personal identity in some possible cases. A lot of decision theorists, e.g. Brian Hedden, really want personal-identity-neutrality, so will not like this principle.
2) In Sleeping Beauty cases, Beauty's action in accepting/rejecting the bet, or pressing/not-pressing 'cancel', does not affect the number of awakenings. This is not a necessity. In Absent-Minded-Driver-like cases, the agent's decision at some node affects how many future awakenings or person-moments there will be. It is hard to see how to generalise your principle to such cases, since the number of individual actions that you're dividing responsibility between is dependent on what the agent does.
Fwiw, I think the way to go is an epistemic principle. Note that in the Korzukhin case, the CDT-thirder is permitted both to accept the Dutch Book or to reject it. Oesterheld and Conitzer (2024) - well, really it's Piccione and Rubinstein (1997) - prove that in a very wide class of cases there is always a non-Dutch Book policy compatible with CDT-thirding, although a Dutch Book may also be permissible. If you supplement CDT-thirding with the epistemic principle that Beauty should have a prior of 1 that she will choose according to the non-Dutch Book policy, this combined view will be immune to Dutch Books in the cases covered by their theorem, which includes all the Sleeping Beauty and Absent Minded Driver cases.
If she has a prior of 1 that other instances of herself will choose according to the non-Dutch Book policy, she will always be incentivised to choose according to that policy too. So this principle guarantees that all the Beauties coordinate around that policy. Ironically, it is possible that they would ALL rather choose a Dutch Book policy if they could choose for all the other instances though! So they can all be unhappy that they are coordinating on the ex-ante optimal policy, since ex-interim it doesn't seem optimal.
Since you recognize that it's plausible that most lives in this world may not be worth living (insects), how do you reconcile this with your extremely strange claim that a perfect being created this world ?
The simplest betting argument against thirdism is this:
The Beauty is told that the coin is Tails. And then proposed to bet that there is not a single Monday awakening during this experiment at slightly better than 3:1 odds.
Why would she agree to such a ridiculous bet? Well, according to thirdism:
P(Monday) = 1/2
And there are two awakenings, which she is supposed to treat as independent trials, due to amnesia, therefore
P(Monday at least once) = 1-(1-P(Monday))(1-P(Monday)) = 3/4
And so she believes that there is 1/4 chance that no Monday awakening happens.
A more complicated argument comes from Unbounded Sleeping Beauty, where the coin is tossed multiple times untill the first Heads outcome and then the Beauty receives 2 to the power of n awakenings, where n is number of Tails outcomes in the row.
The Beauty is proposed to wage a million dollars vs 1 cent that at least a thousand Tails in the row has happened. Then the same bet for two thousands Tails, four thousand Tails and so on till she runs out of all her money.
I can't tell what you are saying. In the first bet, she'd bet at odds of 1 that there's an awakening on Monday in the experiment because no matter whether the coin came up heads or tails, there's a Monday awakening.
Not if she strictly followed thirder (or single halfer) model.
> no matter whether the coin came up heads or tails, there's a Monday awakening.
This is true, but thirder and single halfer model do not capture this fact in any way.
If we grant their premise that "What is the current day?" is a coherent question in Tails Sleeping Beauty, and on every awakening the answer to it works the same way, isomorphic to a coin toss - two possible outcomes with probabilities 1/2 each, - then we arrive right into this conundrum. Consider:
A coin is tossed two times. What is the probability that the first toss is Heads?
P(Heads1) = 1/2
What is the probability that the second toss is Heads?
P(Heads2) = 1/2
What is the probability that at least one toss is Heads?
P(Heads1 or Heads2) = 1 - P(Tails1)P(Tails2) = 3/4
Likewise
You have two awakenings, they can happen either on Monday or Tuesday. What is the probability that your current awakening is on Monday?
P(Monday1) = 1/2
What is the probability that your other awakening is on Monday?
P(Monday2) = 1/2
What is the probability that at least one of the awakenings happens on Monday?
P(Monday1 or Monday2) = 1 - P(Tuesday1)P(Tuesday2) = 3/4
This is a level of confusion that verges on the infinite. The Monday awakenings are not like the first coin coming up heads because they happen whether or not the first coin comes up heads. No matter which outcome is true today, she is certain that there is at least one Monday wakening.
Monday and Tuesday awakening can not be correctly modelled as two independent random events. But that's what thirders and single halfers are doing. Therefore thirders and single halfers are wrong
To correctly represent SB problem we need to acknowledge that awakenings are casually connected and therefore are part of a single outcome. Which is - reason as a double halfer.
I don’t think the first money pump has anything to do with single-halfers, we can do the same with SIA. Take a sleeping beauty variant where with heads you get two awakenings (Monday, Tuesday), with tails you get three awakenings (Monday, Tuesday, Wednesday). Then SIA tells you that upon waking up, you should have probability 2/5 in heads, 3/5 in tails. You are offered the bet you mention before the experiment, with you paying 49 cents and getting 51 cents if heads. On Monday night and Tuesday-tail night, you are going to be told that you are either in Monday or Tuesday-tails.
By conditionalization once given that information, you will be willing to give 15 cents to receive your 49 cents back, so you have a guaranteed loss. In fact, you can do this money pump for any reasonable anthropic theory c, as you can do it even in the boring sleeping beauty variant where with both heads and tails you get two awakening in Monday and Tuesday, and c tells the agent to be indifferent between the four awakenings.
> First of all, I think that thirders have a way out of this money pump
Well sure, if you re-adjust your utilities to compensate for the shifts in probabilities then you will not be money pumped. But this itself is a blow against thirdism as it requires you to postulate an extra principle and now your utilities are unstable, which leads to all kind of complications. Double Halfism that adjust neither probabilities nor utilities, is, therefore, more parsimonious.
> After waking up in a red room, you should obviously think at 2/3 odds that the coin came up heads.
Not on Double Halfism, you don't. P(Red) = 1 as you observe red room in every iteration of the experiment, therefore P(Heads|Red) = 1/2 and therefore no money pump.
Your really need to refactor this post. In its current stance it's nearly incomprehensible. Some exmaples:
> Suppose that Beauty is put to sleep on Monday. Then a coin is flipped. If it comes up tails, she is woken up again on Tuesday with no memories. If it comes up tails, she’s woken up again
> Cancelling the Sunday bet at a cost no longer pays!
The fact that you switch from talking about Sleeping Beauty to talking about clones in the Double Halfing Section is also very confusing.
Anyway the correct (Double Halfer) model for SB is this:
"At least one awakening has happened in this trial": P(Awake) = 1
"Monday awakening has happened in this trial": P(Monday) = 1
"Tuesday awakening has happened in this trial": P(Tuesday)=1/2
"The coin is heads in this trial": P(Heads) = 1/2
"The coin is Tails in this trial": P(Tails)=1/2
Events talking about "Today" are ill-defined due to amnesia, unless they can be reformulated as talking about "Every day".
There are no money pumps that this model is vulnerable to. First money pump doesn't work because this statement:
> But she also thinks that there’s a 1/8 chance that it’s currently day two and the coin that will be flipped at the end of today will come up heads.
Is wrong. This event is ill-defined.
P(Heads) = P(Heads|Monday) = P(Heads|Awake) = 1/2 for both coins
If you reformulate your second money pump in terms of Sleeping Beauty problem, I'd be happy to show that it doesn't work as well.
Thanks for the comment: you are 100% correct, that paper gets the math wrong.
Because you don't know what day's coin you're looking at the possibilities are not TT 0 HT 1/3 TH 1/3 HH 1/3 (assuming the coin you can see is heads), it's TT 0 HT 1/2 TH 1/2 HH 1. So after seeing a coin the odds of both coins being the same stays at 1/2. Then if they tell you what day it is (lets say day 2) the odds become TT 0 HT 0 TH 1 HH 1, so it's still 1/2.
Also, for "clones", I don't think your reasoning makes sense, either.
Suppose on Tails when lights go up two people find themselves in rooms of different colors: one is in red room and the other is in blue room. While on Heads both are either in red rooms or blue rooms.
Before lights go up P(Heads|Darkness) = P(Tails|Darkness) = 1/2
After lights go up
P(Heads|Blue) = P(Tails|Blue) = 1/2
P(Heads|Red) = P(Tails|Red) = 1/2
So no update happens. The existence of another person doesn't add anything to the problem.
Real quick, am i just being dense or is there a typo here: "Suppose that Beauty is put to sleep on Monday. Then a coin is flipped. If it comes up tails, she is woken up again on Tuesday with no memories. If it comes up tails, she’s woken up again"? Is it supposed to be "tails" on both scenarios?
Was a bit worried about the pump-against-thirders there, but I think that objection makes sense. Somehow two of my favourite problems (Sleeping Beauty, and Newcomb's) have gotten all tangled up. For reference, playing Prisoner's Dilemma against your clone is basically Newcomb's problem. And so this pump-against-thirders works only against CDT thirders, I think. EDT and FDT would refuse to cancel their bet, saying "if I cancel, then I'd cancel in all 3 worlds, which is bad". That being said, I saw a money pump (-ish) where CDT beats EDT, so thirder+FDT is looking like the best combo here. (My thirder argument: modify the problem so that Tuesday+Heads instead involves waking up but being told "and the experiment is over, you don't need to answer anything". So upon waking up, you have 25% on all 4 worlds, then when you don't get told the experiment's over, you eliminate the Tuesday+Heads world, leaving 2/3 chance of Tails)
> That being said, I saw a money pump (-ish) where CDT beats EDT, so thirder+FDT is looking like the best combo here.
Why wouldn't the exact same money pump work for FDT thirdism as well?
> My thirder argument
There are only two possible outcomes of the experiment: Heads and Tails. In both of them "Not told once" happens, so this event doesn't update you in any way.
Your mistake is in assuming that
P(Not told once) + P(Told once) = 1,
while, in fact, on Heads both of these events happen.
I don't agree that this doesn't update you. Do you agree that before the information (i.e. just after waking up; suppose we knew that "experiment over" info happens 10 minutes after we wake up, if it happens) we should be equally split between the four cases, e.g. {monday+heads, ...}? That is, we have a prior of 25% on these 4 cases. Then, if we wait and no-one comes to tell us the experiment is over, we eliminate the Tuesday+Heads case. So we have P(Tails | (not tuesday+heads)) = P((not tuesday+heads) | tails) * P(tails) / P(not tuesday+heads) = 1 * 0.5 / (3/4) = 2/3. Which step here do you disagree with?
> Do you agree that ... we should be equally split between the four cases
No I don't. We should be equally splitted between two cases.
Our core disagreement is this.
You believe that sample space consits of four possible outcomes, corresponding to each individual awakening:
{ Heads&Monday, Heads&Tuesday, Tails&Monday, Tails&Tuesday }
While I believe that it consists of two possible outcomes, correspinding to two different states of the coin:
{Heads&Monday&Tuesday, Tails&Monday&Tuesday}
This disagreement is easy to solve. You just need to check this page:
https://en.wikipedia.org/wiki/Sample_space
relevant passages are:
> In probability theory, the sample space (also called sample description space,[1] possibility space,[2] or outcome space[3]) of an experiment or random trial is the set of all possible outcomes or results of that experiment.
> The outcomes must be mutually exclusive, i.e. if [one] occurs, then no other will take place
In the same iteration of this experiment both Monday and Tuesday awakenings happen. Therefore, individual awakenings are not mutually exclusive. Therefore they can't form the sample space for this experiment.
Meanwhile, states of the coin are mutually exclusive. In every iteration where the coin is Tails it's not Heads and vice versa. Therefore they can form sample space for this experiment.
The correct model is:
P(Heads) = P(Tails) = 1/2
P(Monday) = P(Tuesday) = 1
Where "Monday" means "Moday awakening happens in this iteration of the experiment", and likewise for Tuesday.
When you are told that experiments is over it means that you learn that Tuesday awakening has happened and that the coin is Heads. This naturally makes you certain that the coin is Heads:
P(Heads|Heads&Tuesday) = P(Heads|Heads) = 1
When you are not told that experiment is over you learn that Monday awakening has happened and the coin is either Heads or Tails or that Tuesday awakening has happened and the coin is Tails, this correspond to the whole sample space so you do not update:
P(Heads|(Heads or Tails)&Monday or Tails&Tuesday) = P(Heads|Heads or Tails or Tails) = P(Heads|Heads or Tails) = P(Heads) = 1/2
Maybe there's some philosophical disagreement here? The reason I say that they're mutually exclusive is that P(It's monday right now and it's tuesday right now)=0. Is there some reason this shouldn't be allowed?
Hold on though, P(Monday happens during the experiment) is not the same as P(Monday right now), which is what I'm using. P(heads and monday right now) and so on ARE mutually exclusive, and should sum to 1. Under this definition, what would you say is P(monday right now | experiment-not-over) ?
> P(Monday happens during the experiment) is not the same as P(Monday right now), which is what I'm using.
"Monday happens during this trial" is a well-defined event, which in every iteration of the experiment has a stable truth value. And as it's True in 100% iteration of experiment, probability of this event is 1.
While "Monday right now" is an ill-defined event, which doesn't have a stable truth value in any iteration of this experiment. In every trial it is supposed to be simultaneously True (on Monday) and False (on Tuesday). Therefore it doesn't have coherent probability measure.
See this: https://en.wikipedia.org/wiki/Well-defined_expression
Now, we can come up with some different experiment where these events are well-defined. That should be an experiment where Monday and Tuesday awakenings do not happen in the same iteration - are mutually exclusive. Such as:
You will be awakened either on Monday or on Tuesday. And a coin is tossed. If it's Tuesday Heads you are told that. Otherwise you are told nothing. You are awakened and told nothing what is the probability that the coin is Heads?
As you can see "Monday right now" is well-defined here. In every iteration of the experiment it is either True or False, because there is not a single iteration where both awakenings happen.
Is it ill-defined? Maybe that's where our disagreement is. If I ask you "what is the time right now", I consider this a well-defined and objective question, even though the (nominally) same question can lead to different answers if asked repeatedly. Perhaps it's me making some Bayesian assumption, but I see it as as normal to put a probability on "monday right now" representing your uncertainty -- and yes, this can change over time. I imagine that I already effectively do this: I'm at about 95% currently, in a few hours it might decrease as I might've forgotten the distinctness of the memory of recently having checked the day, in even more hours it will decrease more as I wonder whether we've passed midnight, and then if I check and it has then it'll go very low. Does your framework not accept this sort of thing? I agree that I'm not being super rigorous, but I still think there's something clear underneath. I could maybe try to formalise what I'm imagining here if it would help
It makes me somewhat emotional to say that I've seen Bentham's Bulldog typing a post, in the flesh! (And also heard him talk, and was surprised to realize that he talks in a way that's very similar to the way he writes) Anyway, the handful of minutes I got to spend with BB seems to confirm my impression that he's not only an effective altruist, but also a cool guy all around.
Many people are saying this!
These are the sorts of articles that i absolutely love from you, buddy!
everything here seems right to me!
My kingdom to the actual published philosopher who creates positions and arguments without resorting to rube goldberg machines
Huh?
These guys trying to disprove SIA are resorting to wacky, complicated gambling scenarios instead of just questioning the assumptions. Poor form
If you accept that your credences are/should be strongly tied to your betting odds, and that it's a bad thing when your betting odds predictably lead to losing all your money in some situations, it shouldn't matter how complicated those situations are. (Not that they even are all that complicated in this case.) It's an argument against the rationality of your credences either way.
I'm very confused about the double halving money pump.
>But she also thinks that there’s a 1/8 chance that it’s currently day two and the coin that will be flipped at the end of today will come up heads.
Why would it be 1/8? The whole point of being a double-halver is that you have the same experience no matter which side the coin landed on, so your experience doesn't give you any evidence. So no matter what a double-halver experiences they're going to assume that the first coin flip had 1/2 chance of being heads or tails, and whether the first coin comes up heads or tails has no effect on whether the next coin flip will be heads or tails.
>Offer Beauty a bet before she learns what day it is that pays out 46 cents if the coin that will be flipped at the end of today comes up heads, but costs 54 cents if it comes up tails. Naturally, Beauty thinks there’s a 5/8 chance that the coin will come up heads, so she takes the bet.
Why would she think there's a 5/8 chance? What day it is won't effect the probability of a coin being flipped at the end of that day.
The 1/8 number isn't very important. There's a 1/2 chance the first coin came up heads--which entails that it's currently day 1 and the coin will come up heads. There is also, quite certainly, some chance that it's currently day 2 and the coin came up tails. The precise number doesn't much matter.
The reason that it's 1/8 is that three things have to happen for it to be day 2 and it being the case that the coin will come up heads. The first coin has to come up tails. It has to presently be the second day (of two). And the second coin has to come up heads. Each has 1/8 probability.
She thinks there's a 5/8 chance because there are two scenarios in which it's true that "the coin that will be flipped at the end of today will come up heads." One of them has 1/2 odds. The other has 1/8 odds.
>There's a 1/2 chance the first coin came up heads--which entails that it's currently day 1 and the coin will come up heads. There is also, quite certainly, some chance that it's currently day 2 and the coin came up tails. The precise number doesn't much matter.
Actually, it matters a lot and after five days of chewing over this (and thanks to Ape in the Coat linking to the Boy or Girl Paradox) I've finally figured out what's wrong with your calculation!
You've got it laid out like this:
Odds that it is Monday and the coin will come up heads: 1/2
Odds that it is Monday and the coin will come up tails: 1/2
Odds that it is Tuesday and the coin will come up heads: 1/8
Odds that it is Tuesday and the coin will come up tails: 1/8
Therefore the odds that the coin will come up heads is 5/8
Which of course doesn't work (all the probabilities add up to more than 1)
But that math is incorrect. If you are a halfer then the probabilities come out like this:
Odds that it is Monday and the coin will come up heads: 2/6
Odds that it is Monday and the coin will come up tails: 2/6
Odds that it is Tuesday and the coin will come up heads: 1/6
Odds that it is Tuesday and the Con will come up heads: 1/6
Therefore the odds that the coin will come up heads is 3/6, or 1/2
The reason for this is straightforward:
The odds that Beauty will be awake on a Monday=1/1 (No matter what, she's awake on Monday)
The odds that Beauty will be awake on a Tuesday=1/2 (She's only awake on Tuesday if the coin came up tails on Monday, which as a 1/2 chance)
Therefore, it is twice as likely that Beauty is awake on a Monday than that she is awake on a Tuesday. In other words:
Probability that it is Monday: 2/3
Probability that it is Tuesday: 1/3
With that in mind, what is the probability that today is Monday and the coin will come up heads? It's the probability that it is Monday (2/3) times the probability that a coin will come up heads (1/2), which equals 2/6. Same probability for it being Monday and coming up tails. And what is the probability that it's Tuesday and the coin will come up heads? It's the probability that it is Tuesday (1/3) times the probability of a coin coming up heads (1/2) which equals 1/6
In other words, on waking the halfer should think there's a 2/6 chance it is Monday and the coin will come up heads, and a 1/6 chance that it is Tuesday and the coin will come up heads, for an overall odds of 3/6 that the coin will come up heads. On learning that today is Monday the odds change to a 1/2 chance that it is Monday and the coin will come up heads (1*1/2), and since 3/6 is equal to 1/2 the betting odds remain the same.
That's not what I'd say that double halfers credences are. What they are:
P(it's day 1 and the coin will come up heads)=1/2
P(it's day 1 and the coin will come up tails=1/4 (given tails there are two equiprobable days--the odds of tails are 1/2 and the odds it would be day 1 given tails are 1/2).
P(it's day 2 and the coin will come up tails)=1/8
P(it's day 2 and the coin will come up heads)=1/8
Your math emphatically negates halferism. It implies that in the version where there's a coinflip on the second day, you should think it's twice as likely the coin flipped on day 1 came up tails as heads so that there are two awakenings,
There are two different questions being asked: what day is it, and what are the odds that a coin flipped on that day will come up heads. If not told the day, Beauty should assume that it is twice as likely to be Monday as it is to be Tuesday, so 2/3rds chance it is Monday. And because Beauty is a double-halfer she knows that the odds of the coin flip remain at 1/2 no matter what day it is. So the odds that it is Monday and the coin will come up heads is 2/6, and the odds that it is Tuesday and the coin will come up heads is 1/6, so altogether the odds of the coin coming up heads today is still 1/2.
Your version of the double halfer credences only works if the double-halfer thinks it is both 1/2 likely that the coin will come up heads and 1/2 likely that today is Monday. But that's not the double halfer position, the double halfer position is that even if Beauty knows it is Monday it shouldn't change her credence about the coin odds. And it's clear why: if we don't know what day it is the odds are 2/6+1/6 that the coin will come up heads, if we know that it's Monday then the 1/6 chance disappears and it become 3/6, so the update in information does not change the odds for the coin flip.
No dude, by halfer logic she should think it's 3x likelier to be Monday as Tuesday. There's a 1/2 chance that there's only one awakening. That's the core claim of the halfer position. Conditional on there being two awakenings, odds are 1/2 that the present day would be a Monday awakening. So overall, for it to be tuesday, two things with odds of 1/2 have to happen--overall it has 1/4 odds, making the other one twice as likely.
The double halfer position--by virtue of being a halfer position--think that before knowing her birth rank, there's a 1/2 chance that the coin flipped at the end of day 1 comes up heads. After all, the double halfer thinks that when the coin that determines whether beauty wakes up a second time is flipped on Sunday--changing when in time the coin is flipped shouldn't change her credence in its outcomes. But if the coin comes up heads on day 1, that straightforwardly entails that the present day is day 1 and the coin will come up heads. Therefore, she has to think that has probability 1/2 and therefore the rest of the probabilities are as I desccribed.
You're absolutely right: it is 3x times likelier to be Monday as Tuesday, not 2x. But the math still checks out!
Odds of it being Monday: 3/4
Odds of it being Tuesday: 1/4
Odds that it is Monday and the first coin is heads is 3/4*1/2=3/8
Odds that it is Monday and the first coin is tails is 3/4*1/2=3/8
Odds that it is Tuesday and the first coin is heads and the second coin is heads is 0/4*1/2=0
Odds that it is Tuesday and the first coin is heads and the second coin is tails is 0/4*1/2=0
Odds that it is Tuesday and the first coin is tails and the second coin is heads is 1/4*1/2=1/8
Odds that it is Tuesday and first coin is tails and the second coin is heads is 1/4*1/2=1/8
Which means that if I don't know the day then the odds that today's coin will be heads is 3/8+1/8=4/8=1/2. Then when I learn that it is Monday it becomes 1*1/2=1/2. No money pump.
>She thinks there's a 5/8 chance because there are two scenarios in which it's true that "the coin that will be flipped at the end of today will come up heads." One of them has 1/2 odds. The other has 1/8 odds.
I think this is the crux of my confusion: no matter what day it is, the odds of a coin being flipped at the end of that day remains 1/2, doesn’t it? Both scenarios have 1/2 odds when it comes to a coin flip at the end of that day. How does it matter whether it’s day 1 or day 2?
It's because "today's coin" is a different coin depending on when today is, and what day today is is entangled with what values of "today's coin" have shown up in the experiment so far.
I'll be honest, I don't think a halfer is bound to respect BB's calculation, since at least some halfers take their position on the basis that shenanigans involving what "today" is are the root of misunderstanding in the SB case to begin with.
Basically, the total set of possibilities is:
1. Coin 1 heads, today is Monday
2. Coin 1 tails, today is Monday, coin 2 heads
3. Coin 1 tails, today is Monday, coin 2 tails
4. Coin 1 tails, today is Tuesday, coin 2 heads
5. Coin 1 tails, today is Tuesday, coin 2 tails
No "coin 1 heads, today is Tuesday" appear because there is no Tuesday awakening if coin 1 is heads, and no coin 2 is listed for coin 1 heads for the same reason: there is no coin 2 if coin 1 is heads.
If you take the view that p(coin 1 heads) = p(coin 1 tails) =1/2, and similarly for coin 2, and that conditional on coin 1 being tails, each of Monday or Tuesday is equally likely, then all probabilities are 1 except p(today is Monday|coin 1 heads) which equals 1.
So the probability of each outcome above is p=1/2 for outcome 1, and p=1/8 for the others, which sums up to 1.
Finally, notice that for situations 1,2,3 "today's coin" is coin 1, whereas for 4,5 it's coin 2. So the outcomes where "today's coin lands heads" are 1 and 4 only, with probabilities 1/2 and 1/8 respectively, hence overall probability that "today's coin is heads" is 1/2 + 1/8 = 5/8.
One thing I notice in my working this through is, it didn't actually require double-halfing? I think it only used the fact that p(coin 1 is heads)=1/2, i.e. it just relied on SB being a halfer?
I appreciate you trying to explain this to me, but I'm still confused. Why in the world would what day it is or what the first coin result was affect the probability of what a coin flip at the end of the day will be? No matter what the first coin flip was I get woken up and asked to bet on whether a coin flipped at the end of the day will be heads, right? Since the first coin flip doesn't affect the odds of getting a "heads" on future coin flips, what the first coin did is immaterial to the question. And since coins don't flip with different odds on Tuesdays than they do on Mondays, what day it is is also irrelevant to the question.
So I assume I'm confused on what the actual question we're trying to answer is here.
Fwiw I agree with you, or at least, I don't think I quite buy BB's argument. I agree that what you say sounds correct and intuitive, and I'm not suggesting that BB's reasoning _overrides_ yours; I'm just saying BB's reasoning also has a certain logic to it, and yet gives a different answer.
"Since the first coin flip doesn't affect the odds of getting a "heads" on future coin flips, what the first coin did is immaterial to the question."
It's not that it affects the odds of heads on future coin flips, it affects the odds of there _being_ future coin flips, and the probability you can assign to the different values of what "today" is.
I think your argument is something like, if X is "today's coin flip lands heads", and A is "today is Monday", then p(X|A)=p(X|~A)=1/2, i.e. regardless of the day, the probability of getting heads on today's coin is 1/2, and since p(A)+p(~A)=1 (it's either Monday or Tuesday and those are exhaustive), we should be able to calculate
p(X)=p(X|A)p(A)+p(X|~A)p(~A)
= 1/2[p(A)+p(~A)]=1/2
where in the last step I use that both conditional probabilities are 1/2 so I can factor that common factor out.
But as I noted elsewhere p(X|A) is "today's coin is heads, given that today is Monday", whereas p(X|~A) is "today's coin is heads, given that today is Tuesday"--so "today's coin" refers to two different coins in these cases. It's not really clear that we can just use an identity over the variable X when X is isn't referring to a fixed coin, it takes different values at different terms in the equation.
So this formalized version of your reasoning is a bit suspect. That doesn't mean you can't tighten it up, but it should at least make you a little more open to an analysis that lays out the different possibilities by day and coin, and tries to count over those, and doing that you might plausibly get BB's answer.
I really do appreciate you helping me figure this out. Is it really the case that p(X|A) and p(X|~A) are referring to two different coins? For all we know they use the same coin no matter what day it is, and even if they don't any fair coin is going to have the same probability of coming up heads. If it is a problem, then X should really be "A fair coin flipped at the end of today will land on heads".
Just to say a little more, I think this whole line of reasoning is something a halfer is likely to reject as this kind of "listing out the possibilities and crossing them off" is what you have to reject to be a halfer in the first place. After all, a simple thirder argument is, there are four possibilities:
1. Heads and it's Monday
2. Heads and it's Tuesday
3. Tails and it's Monday
4. Tails and it's Tuesday,
with option 2 being impossible, resulting in 3 possibilities, two of which have the coin being tails, hence p(T)=2/3.
So it's not clear that a halfer would actually buy the calculation BB is doing.
I don't think I understand the proposal here.
>Suppose you have two actions that are correlated, such that either action individually guarantees that you’ll take the other action. In this case, I think each action should only be given credit for half of the value that you get from the sequence of actions. If there are two buttons where if you press them both you’ll get 4 dollars, even if the button-pressings are correlated, they should each only count for two dollars.
It doesn't really make sense to ascribe dollar amounts to individual components of your choices/strategy, instead of to the totality of your choices/strategy (unless the consequences of that component are in some sense independent of those of all other components). The component choice "pressing the left button" underspecifies an outcome (or even probability distribution over outcomes!) if both buttons are necessary to get the four dollars. Hence it doesn't have a dollar amount attached expected value-wise.
But each time you take the bet you are taking an action not a strategy. CDTers generally think actions have expected values, not just plans.
Both CDT and EDT agree that EV caclulations only make sense when you hold fixed certain things that are sufficient to determine relevant probability distributions over outcomes (though they may disagree as to which things those are). "Pressing the left button and *possibly* the right button" in your example doesn't do that on either CDT or EDT.
"If any view holds thirding is right, it must think that the existence of more candidates for being your present self makes the existence of your present self likelier."
Is this true? Being a thirder requires holding SIA-ish views? What's the argument for that?
This is good but I have some disagreements with your "if there are two actions that together get you one payout, then you should only treat each individual action as responsible for half the payout" principle.
1) The principle is not neutral wrt personal identity. What I mean by this is that it suddenly matters a lot to your decision-making if some action was made by YOU, or someone else. If it's someone else, you treat it normally by assigning credences to what that person does, and doing expected utility. If it's you, you instead do this division of responsibility thing. So rational action is going to depend heavily on getting into the weeds of personal identity in some possible cases. A lot of decision theorists, e.g. Brian Hedden, really want personal-identity-neutrality, so will not like this principle.
2) In Sleeping Beauty cases, Beauty's action in accepting/rejecting the bet, or pressing/not-pressing 'cancel', does not affect the number of awakenings. This is not a necessity. In Absent-Minded-Driver-like cases, the agent's decision at some node affects how many future awakenings or person-moments there will be. It is hard to see how to generalise your principle to such cases, since the number of individual actions that you're dividing responsibility between is dependent on what the agent does.
Fwiw, I think the way to go is an epistemic principle. Note that in the Korzukhin case, the CDT-thirder is permitted both to accept the Dutch Book or to reject it. Oesterheld and Conitzer (2024) - well, really it's Piccione and Rubinstein (1997) - prove that in a very wide class of cases there is always a non-Dutch Book policy compatible with CDT-thirding, although a Dutch Book may also be permissible. If you supplement CDT-thirding with the epistemic principle that Beauty should have a prior of 1 that she will choose according to the non-Dutch Book policy, this combined view will be immune to Dutch Books in the cases covered by their theorem, which includes all the Sleeping Beauty and Absent Minded Driver cases.
If she has a prior of 1 that other instances of herself will choose according to the non-Dutch Book policy, she will always be incentivised to choose according to that policy too. So this principle guarantees that all the Beauties coordinate around that policy. Ironically, it is possible that they would ALL rather choose a Dutch Book policy if they could choose for all the other instances though! So they can all be unhappy that they are coordinating on the ex-ante optimal policy, since ex-interim it doesn't seem optimal.
Since you recognize that it's plausible that most lives in this world may not be worth living (insects), how do you reconcile this with your extremely strange claim that a perfect being created this world ?
The simplest betting argument against thirdism is this:
The Beauty is told that the coin is Tails. And then proposed to bet that there is not a single Monday awakening during this experiment at slightly better than 3:1 odds.
Why would she agree to such a ridiculous bet? Well, according to thirdism:
P(Monday) = 1/2
And there are two awakenings, which she is supposed to treat as independent trials, due to amnesia, therefore
P(Monday at least once) = 1-(1-P(Monday))(1-P(Monday)) = 3/4
And so she believes that there is 1/4 chance that no Monday awakening happens.
A more complicated argument comes from Unbounded Sleeping Beauty, where the coin is tossed multiple times untill the first Heads outcome and then the Beauty receives 2 to the power of n awakenings, where n is number of Tails outcomes in the row.
The Beauty is proposed to wage a million dollars vs 1 cent that at least a thousand Tails in the row has happened. Then the same bet for two thousands Tails, four thousand Tails and so on till she runs out of all her money.
I can't tell what you are saying. In the first bet, she'd bet at odds of 1 that there's an awakening on Monday in the experiment because no matter whether the coin came up heads or tails, there's a Monday awakening.
> In the first bet, she'd bet at odds of 1
Not if she strictly followed thirder (or single halfer) model.
> no matter whether the coin came up heads or tails, there's a Monday awakening.
This is true, but thirder and single halfer model do not capture this fact in any way.
If we grant their premise that "What is the current day?" is a coherent question in Tails Sleeping Beauty, and on every awakening the answer to it works the same way, isomorphic to a coin toss - two possible outcomes with probabilities 1/2 each, - then we arrive right into this conundrum. Consider:
A coin is tossed two times. What is the probability that the first toss is Heads?
P(Heads1) = 1/2
What is the probability that the second toss is Heads?
P(Heads2) = 1/2
What is the probability that at least one toss is Heads?
P(Heads1 or Heads2) = 1 - P(Tails1)P(Tails2) = 3/4
Likewise
You have two awakenings, they can happen either on Monday or Tuesday. What is the probability that your current awakening is on Monday?
P(Monday1) = 1/2
What is the probability that your other awakening is on Monday?
P(Monday2) = 1/2
What is the probability that at least one of the awakenings happens on Monday?
P(Monday1 or Monday2) = 1 - P(Tuesday1)P(Tuesday2) = 3/4
This is a level of confusion that verges on the infinite. The Monday awakenings are not like the first coin coming up heads because they happen whether or not the first coin comes up heads. No matter which outcome is true today, she is certain that there is at least one Monday wakening.
My point exactly!
Monday and Tuesday awakening can not be correctly modelled as two independent random events. But that's what thirders and single halfers are doing. Therefore thirders and single halfers are wrong
To correctly represent SB problem we need to acknowledge that awakenings are casually connected and therefore are part of a single outcome. Which is - reason as a double halfer.
Thirders don’t treat them as independent obviously.
I don’t think the first money pump has anything to do with single-halfers, we can do the same with SIA. Take a sleeping beauty variant where with heads you get two awakenings (Monday, Tuesday), with tails you get three awakenings (Monday, Tuesday, Wednesday). Then SIA tells you that upon waking up, you should have probability 2/5 in heads, 3/5 in tails. You are offered the bet you mention before the experiment, with you paying 49 cents and getting 51 cents if heads. On Monday night and Tuesday-tail night, you are going to be told that you are either in Monday or Tuesday-tails.
By conditionalization once given that information, you will be willing to give 15 cents to receive your 49 cents back, so you have a guaranteed loss. In fact, you can do this money pump for any reasonable anthropic theory c, as you can do it even in the boring sleeping beauty variant where with both heads and tails you get two awakening in Monday and Tuesday, and c tells the agent to be indifferent between the four awakenings.
You can't give your money back twice on both Monday night and Tuesday tail night!
Upon being told it is Monday or Tuesday tails, you should update to think tails is twice as likely as heads. But there's no money pump here.
> First of all, I think that thirders have a way out of this money pump
Well sure, if you re-adjust your utilities to compensate for the shifts in probabilities then you will not be money pumped. But this itself is a blow against thirdism as it requires you to postulate an extra principle and now your utilities are unstable, which leads to all kind of complications. Double Halfism that adjust neither probabilities nor utilities, is, therefore, more parsimonious.
> After waking up in a red room, you should obviously think at 2/3 odds that the coin came up heads.
Not on Double Halfism, you don't. P(Red) = 1 as you observe red room in every iteration of the experiment, therefore P(Heads|Red) = 1/2 and therefore no money pump.
Your really need to refactor this post. In its current stance it's nearly incomprehensible. Some exmaples:
> Suppose that Beauty is put to sleep on Monday. Then a coin is flipped. If it comes up tails, she is woken up again on Tuesday with no memories. If it comes up tails, she’s woken up again
> Cancelling the Sunday bet at a cost no longer pays!
The fact that you switch from talking about Sleeping Beauty to talking about clones in the Double Halfing Section is also very confusing.
Anyway the correct (Double Halfer) model for SB is this:
"At least one awakening has happened in this trial": P(Awake) = 1
"Monday awakening has happened in this trial": P(Monday) = 1
"Tuesday awakening has happened in this trial": P(Tuesday)=1/2
"The coin is heads in this trial": P(Heads) = 1/2
"The coin is Tails in this trial": P(Tails)=1/2
Events talking about "Today" are ill-defined due to amnesia, unless they can be reformulated as talking about "Every day".
There are no money pumps that this model is vulnerable to. First money pump doesn't work because this statement:
> But she also thinks that there’s a 1/8 chance that it’s currently day two and the coin that will be flipped at the end of today will come up heads.
Is wrong. This event is ill-defined.
P(Heads) = P(Heads|Monday) = P(Heads|Awake) = 1/2 for both coins
If you reformulate your second money pump in terms of Sleeping Beauty problem, I'd be happy to show that it doesn't work as well.
See also this case https://www.cs.cmu.edu/~conitzer/devastatingPHILSTUD.pdf
The author of this paper simply reinvented boy or girl paradox, and is being confused at it:
https://en.wikipedia.org/wiki/Boy_or_girl_paradox
Thanks for the comment: you are 100% correct, that paper gets the math wrong.
Because you don't know what day's coin you're looking at the possibilities are not TT 0 HT 1/3 TH 1/3 HH 1/3 (assuming the coin you can see is heads), it's TT 0 HT 1/2 TH 1/2 HH 1. So after seeing a coin the odds of both coins being the same stays at 1/2. Then if they tell you what day it is (lets say day 2) the odds become TT 0 HT 0 TH 1 HH 1, so it's still 1/2.
That's not true. See the second money pump in section 3.
Also, for "clones", I don't think your reasoning makes sense, either.
Suppose on Tails when lights go up two people find themselves in rooms of different colors: one is in red room and the other is in blue room. While on Heads both are either in red rooms or blue rooms.
Before lights go up P(Heads|Darkness) = P(Tails|Darkness) = 1/2
After lights go up
P(Heads|Blue) = P(Tails|Blue) = 1/2
P(Heads|Red) = P(Tails|Red) = 1/2
So no update happens. The existence of another person doesn't add anything to the problem.
As I've already said:
> If you reformulate your second money pump in terms of Sleeping Beauty problem, I'd be happy to show that it doesn't work as well.
Until then it's not relevant to Sleeping Beauty.
Real quick, am i just being dense or is there a typo here: "Suppose that Beauty is put to sleep on Monday. Then a coin is flipped. If it comes up tails, she is woken up again on Tuesday with no memories. If it comes up tails, she’s woken up again"? Is it supposed to be "tails" on both scenarios?
Oops first is supposed to be heads