EDIT: I just realized that quoting you out of context like that makes it look a bit like you’re defending incels. Sorry about that :grin: — Jamal

fdrake incel stan confirmed.

What disturbed me were the intrusive alien thoughts that I disagreed with, the misogyny in embryo. — Jamal

Know the feel bro. "Political is personal" is also in your mind maaan.

Maybe you’re saying that that’s just how male frustration manifests itself (in this society etc.), and that this in itself is not indicative of incel tendencies, though it’s probably a necessary condition. — Jamal

Yes. Necessary but not sufficient. I think "the embryo", as you put it, is a similar mechanism to violent or transgressive intrusive thoughts. Like suddenly wanting to whack the person in front of you in the queue for taking too long. I get the feeling misogynist-lite intrusive thoughts are a bit easier to apply to reality; more seductive; but the violent rage at the queue fucker is definitely very visceral. Maybe such thoughts turn to misogyny when the intrusive thoughts become egosyntonic. When anger becomes justice.

There are two reasonable ways to assign variables to the set from which you will select your envelope: {x, 2x} and {x, x/2}. Either works so long as you stick with it, but if you backtrack over your variable assignment, you have to completely switch to the other assignment scheme. — Srap Tasmaner

I agree with that. I think I wrote something similar, but with more words, in my reply here. I think that's the right generating mechanism for what's random in the scenario. The agent doesn't know which of (x,2x) they have. But they know (x,2x) are the amounts in the envelopes. Rather than the agent doesn't know which of (x/2, 2x) is in the other envelope - since (x/2, 2x) isn't a pair of envelopes, it's a representation of the agent's epistemic uncertainty that doesn't reflect the uncertainty of allocating amounts to the envelope pair. The variable they ascribe to the other envelope doesn't describe the allocation of amounts to envelopes at all; so their expectation, based on that variable, can't be expected to reflect the true expectation of switching.

But if in reality, someone told them they had 10 in their envelope, flipped a coin to choose between 5 and 20, shoved the resultant amount into another envelope, and presented the choice to switch to the agent - in that scenario they should switch.

Do we have the same take?

It doesn’t hurt that you have a rich set of radically progressive beliefs to keep you away from the dark side. When I was going through it I had already been calling myself a communist for a couple of years, so I wonder if this helped prevent my descent into misogyny. I don’t think that was the main thing though. — Jamal

I dunno. It would be a nice thought.

To be completely honest, I went through it again around 2016-2017. I went a bit mad with resentment and, the flipside, an unhealthy infatuation (which never led me to do anything abusive or creepy, I should add. Well, maybe mildly creepy.). Come to think of it, over the course of my life I seem to have oscillated between periods of quiet incel resentment that I was saved from at the last minute by the women in my life. — Jamal

Makes sense. I think it's so commonplace it's kinda silly to call it "being an incel" or whatever. It's just getting frustrated at being romantically/sexually unfulfilled and alone. You can see the same kinda attitude in a few "pretty privilege" criticising TikToks from women. It looks like: men are shit because they don't pay romantic attention to me... But if they did it would be objectification, fuck men.

Anyway, it’s probably better to target one’s ressentiment at the abstract woman than actual women. I don’t know if the former leads to the latter in a smooth progression or if something just breaks at some point based on individual psychology or circumstances. — Jamal

Also yes. It isn't "most of me" that thinks this. I'm very used to having intrusive thoughts of various flavours. My Inner Incel is just trying to stop me from acknowledging my own loneliness and romantic shame!

I can imagine the former leading to the latter, though. If I were to start seeing women under the light of that abstracted object of desire/resentment, it starts to look a lot like misogyny. Whereas the loneliness which underpins those errant thoughts isn't anything of the sort.

Edit: so I guess you can say I'm wary of pathologising it - getting angry about other emotions and hallucinating problems to solve is what blokes grow up doing. But indulging in misogyny is an obvious no-no. Can't a guy just be frustrated at being alone?

I have personal experience of it. In my youth I was unsuccessful with women, and I noticed I was feeling a rising resentment about it. I knew this was wrong so I didn't let it develop too far. At the time I did not have many friends, let alone female friends. All of the toxic feelings disappeared once my sex life became as astonishingly rich and exciting as it remains today. — Jamal

Same. I'm going through a period like myself actually. Well, I have quite a lot of friends but no romance going on. I get these errant thoughts of resentment. They've all got this character where I take the "isolation/fragmentation of modern life" phenomenon, turn it into a personal failing, then externalise that judged personal failing onto an abstract representation of my "object of desire" - a largely "arbitrary" woman. Ressentiment through and through. Though socially inculcated.

So if Bob is mad enough to reason with subjective probability distributions (which IMO should never be used in science, and which can be avoided even when discussing credences by using imprecise probabilities — sime

As an aside, AFAIK Bayesian applied statistics uses priors not based on previously collected data all the time. I don't know if researchers care about the distinction between subjective and objective priors nowadays.

@Wayfarer @180 Proof - Metzinger's lectures on "Minimal Phenomenal Selfhood" would make a great thread. It'd troll everyone!

Luckily for us tho, I think women are more inclined to entertain our emotions and feelings on the subject and to support us in being equals rather then becoming as toxically matriarchal as patriarchy has been towards women in the past. We have shit to learn from women. We should listen up. — Benj96

I think this is a moral imperative, rather than an anthropological observation. It will tell us how we ought to behave, not why inceldom is becoming more popular (if indeed it's becoming popular). In the interest of systemic critique, there should be structural reasons for why inceldom is becoming more popular, rather than individual moral failings. Things that are true now of society that weren't true "then", whenever then is.

The underlying reasons I've speculated about are as follows. Can group them into two categories. The first is what I'd call "exposure" related, and tries to show the increase in inceldom is overstated. The second is what I'd call "generative conditions" - things that generate inceldom and make it possible.

Exposure related:
1) There aren't more incels now than there were, we're just more exposed to extreme ideologies more often.
1a) A bloke being frustrated at dating being a horrifying, alienating experience for them can sound a lot like being an incel.
1b) Talking about differences between men's and women's (het) experiences dating can also make you sound like an incel. EG the statement "women have all the power of selection" might make you sound like a misogynist to uncharitable ears. To charitable ears, as far as online dating goes, it's a question of realised match given desired match being more frequent for women.

Generative conditions:
2) Means of finding a romantic partner have now been disrupted by the current configuration of capitalism and patriarchy (or social gender norms, whatevs).
2a) Dating media's success piggybacks on the current configuration of capitalism and current gender norms.
2b) Current gender norms leave men, often, without any intimate friends or support networks. Especially networks that allow deep emotional disclosure. That role, traditionally, was held only by intimate partners. If you lose your partner and cannot find a new one long term, in those conditions you easily become isolated and alone; I call this the "old man pub principle". This style of relationship is now seen as "toxic", and regardless is antithetical to current dating culture.
2c) Current gender norms enable women to have more casual emotional disclosure, that creates deeper bonds in extant social networks and forms new ones.
2d) The latter two points create an asymmetry of emotional dependence. Prosaically, men have demand for women to gain greater emotional intimacy, women have supply but don't wish to "produce" the good of "toxic" asymmetric relationships.
2e) The degradation of traditional gender norms makes those "toxic" relationships undesirable - or alternatively, reveals them as having been undesirable/stifling all along.
2f) Dating media takes the demand for emotional intimacy, from men, and commodifies it. Their algorithms control exposure of men to (what is often seen as) their only means of intimate expression.
2g) Dating media have come to have a monopoly on obtaining intimate partnerships.
2h) Dating media's monopoly on obtaining intimate partnerships derives from a fragmentation of social life; the nuclear family is dying, birth rates are going down, the desire to "home make" and "settle down" is rarer and rarer. Simultaneously, people work longer hours, social activities are becoming more expensive in real terms (median real wages have been stagnant or declining since the 70s). That blocks access to social spaces and diminishes social time.
2i) People are more likely to find intimate and satisfying relationships when they have access to suitable social spaces and have time to do so.

tl;dr - capitalism, death of social norms regarding family and courtship, the norms dying "slower" for men, fragmentation of social life.

This doesn't require an additional coin flip. She either is in the (5, 10) case or the (10, 20) case, with equal prior (and equal posterior) probabilities in this scenario. However, this is just one hypothetical situation. — Pierre-Normand

I agree that it doesn't require an additional coin flip when you've interpreted it like you have. It's still a different interpretation than assigning a random deviate to the unobserved envelope after you've seen your envelope has 10 in it. The former is a distribution on pairs that you condition on the observation that your envelope is 10, then down to a univariate distribution. The latter takes the subject's epistemic state upon seeing their envelope and treats the unobserved envelope as a univariate coinflip. For the stated reasons, these aren't the same scenario as the random variables in them are different.

I do agree that if you fix (5,10) and (10,20) as the possible pairs, and you're in the bivariate case, it's rational to switch. The wikipedia article, instead, treats "A" as a random quantity, which is univariate. A prior on A is thus a prior on one variable. And it's not necessarily bounded - as you've pointed out previously, I believe.

In contrast, the priors in the bivariate case are all bounded - because they're on the probabilities of amounts being in envelopes. Probabilities get bounded priors since they lay in [0,1].

Why I'm drawing such a distinction between the Wikipedia case, and the one discussed in this thread, is because the Wikipedia case treats A as a random deviate, which already isn't a bivariate random variable. In that case, the remarks about unbounded support on A (the amount in the other envelope) make sense.

Taking into account all three possibilities regarding the contents of her initially chosen envelope, her EV for switching is the weighted sum of the updated (i.e. conditioned) EVs for each case, where the weights are the prior probabilities for the three potential contents of the initial envelope. Regardless of the initial bivariate distribution, this calculation invariably results in an overall EV of zero for switching. — Pierre-Normand

I agree in the bivariate set up. The (5,10) and (10,5) cases have opposite gains from switching, so do the (10,20) and (20,10) cases. They're all equally likely. So the expected gain of switching is 0. That's the expectation of switching on the joint, rather than conditional, distribution.

This approach also underlines the flaw the popular argument that, if sound, would generate the paradox. If we consider an initial bivariate distribution where the potential contents of the larger envelope range from $2 to $(2^m) (with m being very large) and are evenly distributed, it appears that the Expected Value (EV) of switching, conditioned on the content of the envelope being n, is positive in all cases except for the special case where n=2^m. — Pierre-Normand

This makes sense. If we're in the case on the wiki, where A is a random variable, A's set of possible values are in [0, inf). A "uniform" prior there gives you an infinite expectation (or put another way, no expectation). At that point you get kind of a "wu" answer, since expected loss doesn't make any sense with that "distribution".

If you assume both envelopes are unobserved in the Wiki case, and you want to force an expectation to exist, there's a few analyses you can do:

Give the support of a distribution on A an upper bound, [0,n], Making it a genuine uniform distribution. The expectation of A is n/2. The other envelope is stipulated to contain (2n) or (n/4) based on a coin flip. If this really reflects the unobserved envelope's sampling mechanism, the expectation of the other envelope is n(1+1/8), which gives you switching being optimal. Switching will always be optimal. What stops you from "reasoning the other way" in this scenario is that the order of operations is important, if my envelope has values (2n), (n/4) with expectation (1+1/8)n it can't be identical with the other envelope's random variable which has expectation (n/2). One is a coinflip, one is a guess about the value of A. This might be an engineer's proof, though. "Why can't you equate the scenarios?" "You know they have different expectations, because the order you assign deviates to the envelopes matters from the set up. "Your" envelope and "the other" envelope aren't symmetric".

Now consider the case where the values A and 2A are assigned to the envelopes prior to opening. This gives the pair (A, 2A) are the possibilities. The envelope is unopened. The agent gets allocated one of the two (A, or 2A) based on the result of a coinflip. If they know the possible amounts in their envelope are A and 2A, and they just don't know whether A is in their envelope and 2A is in the other, or vice versa, then the expectation of their envelope is 1.5A. So is the expectation of the other envelope. Switching is then of expected value 0.

The fallacy in the 'always switch' strategy is somewhat akin to the flaw in Martingale roulette strategies. — Pierre-Normand

I think that if the "always switch" strategy is wrong, it's only wrong when the sampling mechanism which proves it doesn't apply to the problem. It just isn't contestable that if you flip a coin, get 20 half the time and 5 half the time, the expectation is 12.5. The devil is in how this doesn't apply to the scenario.

In my view that comes down to misspecification of the sampling mechanism (like the infinite EV one), or losing track of how the random variables are defined. Like missing the asymmetry between your envelope and the other envelope's deviates, or equating the deviate of drawing a given pair (or the value A) with the subject's assignment of a deviate to the other envelope.

Maybe my response here helps.

Another way of looking at it @Pierre-Normand. If you wanna update on what you see, it's gotta be the realisation of a random variable. I set it up like this:

Initial Speciication

1. Someone flips a coin determining if the envelope pair is Heads=(5,10), Tails=(10,20)
2. They give you the resultant envelope pair.
3. You open your envelope and see 10.

3. doesn't tell you anything about the result of the coin flip in one. What's the expectation of "the other envelope"? That means assigning a random variable to it. How do you do that? I think that scenario plays out like this:

Two Flips Specifications

1. Someone flips a coin determining if the envelope pair is Heads_1=(5,10), Tails_1=(10,20). Call this random variable Flip_1.
2. They give you the resultant envelope pair.
3. You open your envelope and see 10.
4. You flip a coin, Heads_2 means you're in case (5,10) and the other envelope is 5. Tails_2 means you're in the case (10,20) and the other envelope is 20. Call this random variable Flip_2.

I'm abbreviating Heads to H, Tails to T and Flip to F.

The paradox is arising due to stipulated relationships between Flip_1 and Flip_2.

The following reflects the intuition that the coin flip in 1 totally determines the realisation of the envelope in 4. It equates the two random variables. Let's say that we can equate the random variable Flip_1 and Flip_2, that means they have the same expectation. The expectation of Flip_1 would be... how do you average a pair? You don't. The expectation of Flip_2 is transparently 12.5. That means you can't equate the random variables, since they have differing expectations. In essence, the first random variable chooses between apples, and the latter chooses between oranges.

Let's say that instead of there being an equation of Flip_1 with Flip_2, there's the relationship:

P( (H_1=H_2) = 1 )
P( (T_1 = T_2) = 1 )

But that would mean H_1=(5,10) is equal to H_2 = 5, and they aren't equal. Similarly with T_1 and T_2. So the relationship can't be that.

Then there's a conditional specification:

P(F_2 = H_2 | F_1=H_1)=1
P(F_2 = H_2 | F_1=T_1)=0
P(F_2 = T_2 | F_1=H_1)=0
P(F_2 = T_2 | F_1=T_1)=1

Which legitimately makes sense. Then we can ask the question: what does "You open your envelope and see 10" relate to? It can't be a result of F_1, since those are pairs. It can't be a result of F_2, since that's the other envelope. It seems I can't plug it in anywhere in that specification. It's been used in the definition of F_2, to "remove" 10 from the appropriate space of values and render that a coin flip.

In fact, "opening my envelope and seeing 10" and updating on that is only interpretable as another modification of the set up:

Bivariate Distribution Specification

1) You generate a deviate from the following distribution (S,R), where S is the other envelope's value and R is your envelope's value. The possible values for S are (5,10,20) and the possible value for R are the same. S and R have a bivariate distribution (S,R) with the constraint (S=2R or R=2S). It has the following probabilities defining it:

P(S=5, R=10) = 0.25
P(S=10, R=20) =0.25
P(S=20, R=10) = 0.25
P(S=10, R=5) =0.25

A Bayesian solution modifies this distribution. I've just assumed all envelope pairs are equally likely.

2) You then observe R=10.
3) That gives you a conditional distribution P(S=5)=0.5, P(S=20)=0.5
4) Its expectation is 12.5 as desired.
5) You should switch, as S|R=10 has higher expectation than 10.

A different loss function modifies step 4.

And with that set up, you switch. With that set up, if you don't open your envelope, you don't switch.

There's the question of whether the "Bivariate Distribution Specification" reflects the envelope problem. It doesn't reflect the one on Wiki. The reason being the one on the wiki generates the deviate (A,A/2) OR (A,2A) exclusively when allocating the envelope, which isn't reflected in the agent's state of uncertainty surrounding the "other envelope" being (A/2, 2A).

It only resembles the one on the Wiki if you introduce the following extra deviate, another "flip" coinciding to the subject's state of uncertainty when pondering "the other envelope":

1) You generate a deviate from the following distribution (S,R), where S is the other envelope's value and R is your envelope's value. The possible values for S are (5,10,20) and the possible value for R are the same. S and R have a bivariate distribution (S,R) with the constraint (S=2R or R=2S). It has the following probabilities defining it:

P(S=5, R=10) = 0.25
P(S=10, R=20) =0.25
P(S=20, R=10) = 0.25
P(S=10, R=5) =0.25

2) You get the envelope 10, meaning the other envelope is 5 or 20 with equal probability. Call the random variable associated with these values F_1.
3) That gives you a conditional distribution P(F_1=5)=0.5, P(F_1=20)=0.5
4) Its expectation is 12.5 as desired.
5) You should switch, as F_1 produces a gain of 2.5 over not switching.

In that model, there's no relationship between the bivariate distribution (S,R) and F_1. Trying to force one gets us back to the Two Flips Specification, and its resultant equivocations. The presentation above is not supposed to make sense as a calculation, it's supposed to highlight where the account would not make sense.

Since you're an R user, you might find it interesting to define a model in RStan, using different choices for the prior for the smallest amount S put into a envelope. Provided the chosen prior P(S) is proper, a sample from the posterior distribution P( S | X) , where X is the observed quantity of one of the envelopes, will not be uniform, resulting in consistent and intuitive conditional expectations for E [ Y | X] (where Y refers to the quantity in the other envelope) — sime

It would!

However, it seems that you've misunderstood my use of Bayesian updating. I am not arguing that observing the $10 allows me to switch between cases. Rather, I'm saying that, given an observation of $10, I can update my beliefs about the probability of being in the (5,10) case or the (10,20) case. — Pierre-Normand

I think I see what you mean. Would you agree that which envelope pair you're in is conditionally independent of the observation that your envelope is 10? Assuming that equation (my envelope = 10) lets that "random variable which equals 10" remain random, anyway.

Seeing that there is $10 in one envelope still leaves it open that there might be $5 or $20 dollars in the other one. — Pierre-Normand

I think this depends how you've set up your random variables to model the problem. If you can assign a random variable to the other envelope and "flip its coin", so to speak, you end up in case C where there's a positive EV of switching.

I can see that you've assigned probabilities to the pairs (5,10) and (10,20), I'm still unclear on what random variables you've assigned probabilities to, though. Where are you imagining the randomness comes in your set up?

What does the cut correspond to? — Srap Tasmaner

Answering that gives you the origin of the paradox, right?

I don’t think your simulation is relevant to our disagreement given that I don’t believe that it is rational to switch. — Michael

I don't believe it's rational to switch either. I'd believe it's rational to switch if we were in Case C.

You seem to suggest I just arbitrarily whipped up some code and said "hey guys, code! problem solved!" — hypericin

Not what I intended, apologies for any offence caused. I whipped up some code myself to illustrate different scenarios in another thread.

Therefore, she should retain her $10, as her prior for Joe having included $20 is sufficiently low. Regardless, before she inspects the second envelope, both outcomes ($5 or $20) remain possible. — Pierre-Normand

You can conclude either strategy is optimal if you can vary the odds (Bayes or nonconstant probability) or the loss function (not expected value). Like if you don't care about amounts under 20 pounds, the optimal strategy is switching. Thus, I'm only really interested in the version where "all results are equally likely", since that seems essential to the ambiguity to me.

Your assertion that 'only two values are possible' for the contents of the envelopes in the two-envelope paradox deserves further exploration. If we consider that the potential amounts are $(5, 10, 20), we might postulate some prior probabilities as follows: — Pierre-Normand

As I wrote, the prior probabilities wouldn't be assigned to the numbers (5,10,20), they'd be assigned to the pairs (5,10) and (10,20). If your prior probability that the gameshow host would award someone a tiny amount like 5 is much lower than the gigantic amount 20, you'd switch if you observed 10. But if there's no difference in prior probabilities between (5,10) and (10,20), you gain nothing from seeing the event ("my envelope is 10"), because that's equivalent to the disjunctive event ( the pair is (5,10) or (10,20) ) and each constituent event is equally likely

Edit: then you've got to calculate the expectation of switching within the case (5,10) or (10,20). If you specify your envelope is 10 within case... that makes the other envelope nonrandom. If you specify it as 10 here and think that specification impacts which case you're in - (informing whether you're in (5,10) or (10,20), that's close to a category error. Specifically, that error tells you the other envelope could have been assigned 5 or 20, even though you're conditioning upon 10 within an already fixed sub-case; (5,10) or (10,20).

The conflation in the edit, I believe, is where the paradox arises from. Natural language phrasing doesn't distinguish between conditioning "at the start" (your conditioning influencing the assignment of the pair (5,10) or (10,20) - no influence) or "at the end" (your conditioning influencing which of (5,10) you have, or which of (10,20) you have, which is totally deterministic given you've determined the case you're in).

@Michael

Simulation in R to demonstrate the different sampling mechanisms.


Onto something that sounds like Case_A, and something you can to do it to make it superficially resemble Case_C.


You notice that when you condition on "having your envelope be 10" in that set up, you're subsetting to cases where the gain is -5 or 10, and those occur with equal probability since the cases (5,10) and (10,20) occur with equal probability. If you were to condition on your envelope being 10 in the loop (case index=1), you end up with exactly the same gain numerically but it represents a subjectively different belief state for an agent. Why? The subsetting done at the end lets you look at the whole ensemble of cases where the first envelope was 10, where the assignment of 10 was random. Fixing the envelope as 10 within case makes the assignment of 10 nonrandom.


The thing we're butting heads on, in my view, is the Case_A inner loop line 1, which is where the randomness comes in through the allocation of pairs. If at any point, in the loop, the agent *knows* what case they're in, their gain is deterministic. When you grant that knowledge hypothetically, you either enter subcase A 1 (the first gain line) or subcase A 2 (the second gain line). Once you've done those hypothetical calculations, you reintroduce the randomness of allocating envelopes in the next line to choose the received gain.

If you're in subcase 1 and you have £10 and you switch then you lose 5, if you're in subcase 2 and you have £10 then you gain £10. Each of these is equally likely, hence E(z) suggesting you should switch. — Michael

Eh, we're just asserting the same thing over and over again at this point. I stop.

If you're in subcase 1 then you lose £5 by switching. If you're in subcase 2 then you gain £10 by switching. There is a 50% chance that you're in subcase 1 and a 50% chance that you're in subcase 2. So according to the E(z) calculation, it is rational to switch. — Michael

You don't know whether you're in Case A subcase 1 or case A subcase 2. Each of those has probability half. If you're in case A subcase 1, if you switch you gain 5 or lose 5. If you're in case A subcase 2, if you switch you gain 10 or lose 10. Each of those has 0 gain.

The relevant question is; let's say you condition on having 10, does that give you any information on what pair you're in (IE, whether you're in subcase A 1 or subcase A 2), and the answer is no. If you knew what case you're in, knowing that you have 10 makes the choice deterministic. The conflation between case C and case A turns precisely on this point.

If you know you have 10, it becomes very tempting to say that you know the other envelope has 5 or 20. Whereas you don't know that, you just know that either (your envelope is 10 and the other is 20) or (your envelope is 10 and the other envelope is 5), and that tells you nothing about whether you're in case A subcase 1 or case A subcase 2.

When you're stipulating that you "have 10", you're using that to resolve uncertainty within subcases. Whereas the point you're shown it, you could only gain information about which subcase you're in - and it provides no information there.

What's the probability of your envelope being 10? Why can you condition on it?

You both seem to be mixing up the participant's subjective assessment and some God's eye view objective assessment. — Michael

Not true, a different subjective assessment. Your uncertainty concerns the values (5,20), which is case C. Case A's uncertainty concerns the pairs (5,10) and (10,20). When you switch, you don't know if you're in case A subcase 1 or case A subcase 2. So you average the gain of switching over each of those. Which is 0.

Yes, and I think it is perfectly correct to say, in case A, that the probability that the other envelope contains £20 is 50%. — Michael

It isn't though. "The other envelope contains 20" in case A subcase 1 has probability 0. It just isn't an event in that case. The only events in that case are "my envelope contains 5 and the other envelope contains 10" and "my envelope contains 10 and the other envelope contains 5". In case A subcase 2, "the other envelope contains 20" has probability half, the events in that one are "my envelope contains 10 and the other envelope contains 20" and "my envelope contains 20 and the other envelope contains 10".

Nowhere in that set up is there ever a way of considering that the other envelope contains 5 or 20! That's only an available move in case C.

They are aware. They are told before the experiment starts that one envelope contains twice as much as the other. They open their envelope to find £10. They know that the possible values in the other envelope are £5 or £20. — Michael

Their awareness corresponds to a different fact. I don't think we're making any progress here. Do you understand the difference between case A and case C?

If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael

To break it down, this comes with the following steps:
1) The possible ball types are red, blue, white.
2) Two distinct types of ball are taken.
3) They're put in two distinct envelopes.
4) I choose an envelope.
5 ) I know that the only balls I could have chosen are red, blue, white
6 ) I observe white.
7 ) I know that the only balls remaining are red, blue
8 ) Each occurs with probability half

5 ) There corresponds to telling the agent, in the envelope case, that the possible values in the other envelope are 5 or 20. This makes your framing case C.

Case A, however, does not have the agent aware that the possible values in the other envelope are 5 or 20. Precisely because they are epistemically indifferent to the cases where the pair is (5,10) or (5,20). The randomness there is epistemic. But it's equivalent to randomly assigning red or blue to the other package at that time by a coin flip. They represent exactly the same set up. But no coin flip is done at that stage; IE, nothing actually happens to the agent in the envelope set up in case A. But they would in case C or your ball example.

Exactly. You asked me to pick one, then treated that like drawing a ball from the bag.

↪fdrake If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael

That's case C! Drawing a new ball is the same as assigning a value to the other envelope. It's just not assigned at that step in the envelope set up. Hence, the distinction between (A and B) and C. It seems to me you're having difficulty seeing the distinction between A and C?

If there is a 50% chance that I am in (5, 10) and a 50% chance that I am in (10, 20), and if I have £10, then there is a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. According to — Michael

That changes the sample space. The subcase in A5's sample space is (5,10) - IE an envelope contains 5 or 10, the subject just doesn't know that. The subcase in A6's sample space is (10,20), the subject just doesn't know that.

When you assign (5,20) to the other envelope, that's not an event in the sample space in A5 or A7. It's only an event in the sample space in case C.

The framing is the paradox. I pick one of two envelopes at random. One is twice the value of the other. Given that the probability that I picked the more valuable envelope, it is rational to switch. But also to then switch back. — Michael

And the distinction between Case C and Case A is my solution to it. Ambiguous phrasing suggests we're in Case C, whereas we're actually in case A or B.

I'm not sure what you mean. Perhaps you could answer the questions I posed earlier? — Michael

It's a "wu" thing, the presumptions in your questions already define away how to dissolve the problem IMO. So I don't think it's wise of me to answer them. I do agree, given what you just said and your framing, that the calculation of gain is correct. What's wrong is the framing, not the calculation.

What I'm saying: case C makes the assignment of (5,20) to envelopes random. Even though (5,20) could never be an assignment of envelopes. Case C has a perfectly cromulent way of calculating expectations, it's just not the random assignment mechanism to the envelopes.

If I gave you an envelope containing 10 pounds, and I told you I had an envelope containing (5 or 20), you should switch. Just... That's not what situation we're in. We're in a situation where we don't know whether the pairs of envelopes are (5,10) or (10,20).

I don't think the problem with this formula is with its probability assignments. If I know that one envelope contains more money than the other, and if I pick one at random, then whether I open it or not the probability that I picked the envelope with the more money is — Michael

Probability assignments are done with respect to a space of events. The event spaces in case A and C are different.

I tell you that heads is worth £30. What is the probability that it landed on the more valuable side? 1/2. You knowing that it landed on heads and that heads is worth £30 makes no difference as you don't know which of heads and tails is the more valuable or how valuable the more valuable side is. — Michael

In general I think it's one of these cases where describing exactly what is random, what isn't, and what is conditioned upon is completely necessary. The debate in thread consists in different ways of setting up the randomness in the problem.

I'm of the opinion that knowing the value of your envelope tells you nothing about what you should do, so long as you can assume the probabilities of each value being in each envelope are equal. Here is what I think describes the randomness in the situation.

1 ) There are two pairs of envelopes. One of them contains (5,10), one of them contains (10,20).
2 ) Someone flips a coin and assigns you one of these pairs of envelopes.

Then depending upon the formulation, either you open it or don't.
Case A, you open it:
A3 ) If you open it and see 10, you don't know if your 10 is in the (5,10) pair or the (10,20) pair.
A4 ) Each of those is equally likely.
A5 ) Assume you're in the (5,10) pair, switching has 0 gain there under equal probability and expected loss.
A6 ) Assume you're in the (10,20) pair, switching has 0 gain under the same assumptions.
A7 ) The expected value of switching is 0.

Edit: A5 is a conditioning step (let your envelope pair be 5,10), A6 is a conditioning step (let your envelope pair by 10,20), A7 is a calculation using the law of total expectation

Case B, you don't open it:
B3 ) If you don't open it, you don't know if your pair is the (5,10) pair or the (10,20) pair.
The reasoning is exactly the same.

Open or don't open, with that framing, there's no gain from switching

If you want to assume the unknown envelope contains (5,20) given that you observe 10 in your envelope, the randomness mechanism looks like:

C1) Someone assigns 10 to your envelope and you open it.
C2 ) Someone assigns 5 or 20 to the other envelope with equal probability
C3 ) If you switch, you either lose 5 or gain 10, with equal probability
C4 ) You'd gain 2.5 by switching.

The major contrast is between the randomness coming in C2 with the randomness coming in the steps 1) and A3 or B3. Those are totally distinct sampling mechanisms (A3 and B3 are the same).

Edit: there actually isn't a conditioning step in this presentation, since the assignment of 10 to the envelope you see is nonrandom. You simply stipulate you open an envelope with 10 in it. If you want to frame this as conditioning, then the probability distribution is just assigning probability 1 to the value 10, for the value in your envelope. (And strictly speaking that isn't a probability distribution which comes with its own subdiscussion... And its support isn't clear, which comes with its own subdiscussion)

There is nothing there that I disagree with. But I don't think the paradox arises if the values of the two envelopes are stipulated in advance ($10 and $20, say). The paradox arises when we stipulate that the probability of the second envelope containing the largest amount is 1/2 and is not conditionally dependent on the value of the initially chosen envelope. In the example that you give, the probability of the second envelope containing $20 conditionally on the first envelope containing $10 is 1 (and vice versa). — Pierre-Normand

Ultimately the disagreement you're having with @Michael there is about the representation of the agent's belief state. If the agent knew that one envelope contained 10, and one envelope contained 20, and they saw 10, the other envelope has 20 in it with probability 1. Notably, this requires the agent knowing the values in both envelopes. In effect, there is a secondary conditioning step in your analysis; you're also conditioning on the set of values {n=10 or n=20} being the values in the envelope, and aggregating a (probably?) uniform prior on probability (the chance of each amount being in each envelope) into the n=10 and n=20 case.

The ability to condition on {n=10 or n=20} isn't something the agent can do while representing their state of evidence, if they don't know that n=10 or n=20 at the start.

Nevertheless, if they observe n=10 in the first envelope, I still think there's a problem with assigning a probability distribution on the values (5, 20) in the other envelope. This is because that stipulates there being three possible values in the envelopes combined; (5, 10, 20); whereas the agent knows only two are possible.

So IMO the issue with "conditioning" on (n=10) when opening the envelope isn't with the conditioning operation itself, it's with the background specification of the sample space's events. Two envelopes (pairs), where one must be half the other. On the possible values (5,10,20), this just (5,10) and (10,20).

A further illustration of why this is weird; if the agent really thinks that the possible values in the envelopes are (5,10,20), and they observe (10) in theirs, then assign equal probability mass to (5,20) based on that observation, that means initially there will have been a nonzero probability assigned to each (5,10,20). A choice between (5,10,20) needs a sampling mechanism associated with it (whence the randomisation), and there isn't one which produces three values anywhere in the problem.

In other words, given that your envelope is 10, making "the other envelope" (5,20) doesn't give you an event which is possible given the (alleged) randomisation that puts the pairs of values in the envelopes in the first place.

Edited: changed 1/3 to nonzero.

As a general rule, simulating probabilities doesn't resolve disputes about which computations are appropriate. Every randomness has a generating mechanism. Addressing these paradoxes means addressing the generating mechanism.

Demonstrative example:

1) The moon is made of cheese or eggs
2) both are equally likely
3) the probability the moon is made of cheese given by the following R script:

So the moon is made of cheese with 50% probability.

It's just right, look at the code!

All coding something up does is let you check a calculation for it being correct, not whether it's the appropriate calculation to do.

Do real anarchists have meetings? — frank

Endless meetings, meetings that continue on communication media afterwards, that get revisited the next social, then talked about at the next meeting.

One angle the author's maybe missed is that the broader conditions of a society which allows virtual child molestation are distinguished from societies which allow virtual murder. I read the article as focussing on the expression of individual desire, and an individual's engagement with such a simulation (molestation or murder).

Perhaps an uncovered counter argument is that it's impossible for a society to avoid the simulation of murder - in fiction, fantasy and so on - but possible for a society to avoid the simulation of child molestation. One reason for that might be that violent conflict can be construed as a cultural universal, whereas child molestation - despite being commonplace - is not typically a codified component of human experience.

This interfaces with the argument in the paper in the following way; for something to be morally impermissible, it has to be possible not to do. It is impossible for a society not to virtually murder, therefore virtually murdering cannot be impermissible. Whereas we know (perhaps) it is possible for a society to virtually murder and not virtually molest.

Effectively, this would entail virtual murder on a societal level is unavoidable, whereas virtual molestation on a societal level is avoidable. So only the second can present a moral problem for societies.

I stand by the claim that the probability that the other envelope contains twice as much as my envelope is equal to the probability that the other envelope contains half as much as my envelope, that probability being 1/2. — Michael

I think this is true regardless of framing.

P(the other envelope = my envelope/2)

You can get that by counting the ways this happens. In the first case (X,X/2), the probability is 0.5. In the second case (2X, X), the probability is 0.5. The probability that we're in the first case is 0.5, which means the probability that we're in the first case and the other envelope contains half mine is 0.25 total. You get another 0.25 from the same reasoning in the other case. Law of total probability gives you 0.5 total for that probability.

No values are conflated in this case, as there's a hierarchy of random variables.

The first random variable is a coinflip, it tells you whether your envelope values are (X, X/2) or (2X, X). Within each case, you have a random variable that has probability half for each event. The only "conflation" which arises is when forgetting the distinction between the realisation of a random variable (it attaining a particular fixed value) and the event of a random variable taking a particular value (which may or may not have happened).

It isn't clear what your random variable z means in your calculation, since its sample space isn't defined. z is a particular fixed value in the first envelope, no? So it's nonrandom, as soon as you open it. So it doesn't have a probability to BE anything else.

We talked about this 5 years ago! Probability calculations are based on models and intuitions about situations. In this case, disagreements are not about calculations, it's about what is an appropriate description of the sampling mechanism of the envelopes.

One way of framing it: You see two envelopes on the table. You open one. It contains X. You know that the remaining envelope contains 2X or X/2. Assuming both are equally likely, half the time you switch and get an extra X, and half the time you switch and lose an X/2, so you stand to gain 0.5X - 0.25X=0.25X by switching.

Another way of framing it: Two envelopes are on the table. You know one is twice the value of the other. You open your envelope and see it contains an amount. That fixes the other envelope to contain the other amount. You open your envelope and see you have the amount X. Do you know whether the other envelope has 2X or X/2 in it? Nope. What situations can this arise in?

Either you were given X and the other envelope contains X/2, xor you were given X and the other envelope contains 2X. Those are separate cases. If you were given X and the other envelope contains X/2, you'd lose X/2. If you were given X/2 and the other envelope contains X, you'd gain X/2. In the second case, if you were given X and the other envelope contains 2X, you'd gain X, if you were given 2X and the other envelope contains X, you'd lose X. If you're in the first case, the expected gain of switching is 0, if you're in the second case, the expected gain of switching is 0, both cases are equally likely, so the expected gain of switching is 0.

The relevant question which determines the accuracy of each of these is whether you model the value of the unobserved card as being "twice or half" for a given observed value of X (2X, 0.5X), or whether you model the value of the unobserved card as being drawn from the two scenarios (X, 0.5X) or (2X, X) and switching a move in two distinct ones.

The first framing construes "the sampling mechanism", which is where randomness arises in this scenario, as being the conditioning step - observing the value in the first envelope. The second framing construes "the sampling mechanism" as the allocation of pairs of values to the envelopes. In the first case, observing the value in the first envelope randomly assigns 0.5X or 2X to the unseen envelope, in the second case, observing the value in the first envelope tells you nothing about whether the unseen envelope and your envelope have possible values (X, X/2) or (X, 2X).

After 5 years I remain of the opinion that the second framing is appropriate, since you really do gain no information about what's in the second envelope given what's in the first.

But, nevertheless, if you relax equal probability assumptions of all cases or use a different loss function than expected loss, it can still be more rational to switch (or not switch) depending upon your problem set up. Like if you really needed $200, but didn't care about $100 at all, you'd switch if you got $100.

Stop. I am the law.

Aye! Nothing tells you the meaning of the two modal operators other than the context they're applied to.

Gentlemen, don't fight here, this is the war room.

The argument does depend on S5 where the accessibility relation is universal. From my reading there are good reasons to accept S5 so it would be shortsighted to deny it simply to dismiss the modal ontological argument, and special pleading to deny it only for the modal ontological argument. — Michael

Eh, possibly necessary => necessary is reasonably easy to argue against. I don't like it for the above stated reasons. That lets you conjure up the luminiferous aether, assuming the "true logic of metaphysics" lets you do possibly necessary implies necessary. In that regard, either we'd have to rejected that the luminiferous aether isn't possibly physically necessary, or the law of logic which leads to the inference. I'm inclined to reject the latter, since I intuit that things like physical laws are "physically necessary" (whatever that means).

I wouldn't want to deny that S5 has applications, just that demonstrating it as the "true logic of metaphysics" is a project unto itself. That a logic applies to a domain isn't something that can be taken for granted, I think. The above argument regarding physical necessity is a reason to reject the application of any logic which allows the inference pattern (possibly necessary => necessary) to metaphysics in general.

Does it show that a modally necessary God exists? Is ◊◻∃xGx true and does ◊◻∃xGx entail ◻∃xGx? — Michael

First quibble: it isn't demonstrated to be any particular god, just an entity which satisfies G. The only thing which makes it god-ish is that G is associated with god. It would need to be argued that any given god has the property G, which is established independently (and is a theological thing, right). Moreover, it would need to be argued that an entity could, in principle, have that property.

Second quibble: possibly there exists x such that Gx is unsupported. Modal logics do lots of different things. You can say that 1 is possible for 2 under the accessibility relation "less than or equal to" in the integers. Whether the relevant sense of modality in the logic models an appropriate notion of metaphysical necessity is still something that you can quibble with. Why would you need something like an equivalence accessibility relation between worlds?

An example of that quibble: it was possibly physically necessary that the luminiferous aether existed, therefore it was physically necessary that the luminiferous aether existed, therefore the luminiferous aether existed. A sense of metaphysical necessity which lets you do this conjuring trick is... well, it needs a good argument to support.

Third quibble: you can always deny that it's possible that any particular god exists. And in that case the entity in question would not exist in any world.

Nevertheless, it might be the case that the underlying metaphysics that facilitates the argument is the correct one. It just still would have relatively little to do with a god. Or, as with other ontological arguments, you can perform the same conjuring trick where you posit an entity with G and then it suddenly exists. Like the aether example.