One cannot extract a meaningful notion of probabilities — sime

Why not?

I know that one envelope contains twice as much as the other. I pick one at random. What is the probability that I picked the smaller envelope? It seems perfectly correct to say $1\over2$.

The same supposed paradox occurs even if we know the possible values.

Assume that one envelope contains £10 and the other envelope contains £20.
Let y be the value of my chosen envelope and z be the value of the other envelope.

$P(y = 10) = P(z = 20) = P(z = 2y) = {1\over2}\\P(y = 20) = P(z = 10) = P(z = {y\over2}) = {1\over2}$

The expected value of z is allegedly:

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y$

This commits a mathematical fallacy, conflating three different values of $y$, resulting in a third value.

$\begin{aligned}E(z) &= {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12\end{aligned}$

But my envelope doesn't contain £12, given that one envelope contains £10 and the other envelope contains £20.

This fallacy is committed even when we only know the value of one envelope and even when we don't know the value of any envelope.

The paradox has nothing to do with probability at all. It's just an improper use of algebraic variables when calculating the expected value.

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? — fdrake

It's explained in the OP:

Let x be the amount in one envelope and 2x be the amount in the other envelope.
Let y be the amount in my envelope and z be the amount in the other envelope.

So to bring together my various posts:

$A: P(y = x) = P(z = 2x) = P(z = 2y) = {1\over2}\\B: P(y = 2x) = P(z = x) = P(z = {y\over2}) = {1\over2}$

The expected value of z is allegedly:

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y$

However, if we assume that the amount in the other envelope (z) is £10 then:

$A: P(y = x | z = 10) = P(z = 2x | z = 10) = P(z = 2y | z = 10) = {1\over2}\\B: P(y = 2x | z = 10) = P(z = x | z = 10) = P(z = {y\over2} | z = 10) = {1\over2}$

Notice that in A, $y = {z\over2} = 5$ and in B, $y = 2z = 20$. That makes the addition performed in E(z) above a mathematical fallacy. You cannot add $y$ to $y\over4$, where each y has a different value, to get ${5\over4}y$.

E(z) is properly represented as:

$\begin{aligned}y_a &= {z\over2}\\\\y_b &= 2z\\\\E(z) &= {1\over2}2y_a + {1\over2}({y_b\over2})\\\\&={1\over2}({2z\over2}) + {1\over2}({2z\over2})\\\\&=z=2y_a= {y_b\over2}\end{aligned}$

Which of course tells us nothing we didn't already know; that the other envelope contains either twice as much or half as much as my envelope.

And so there's no reason to switch, as expected.

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.

But what I believe I have shown is that the formula used to calculate the expected value conflates two different values of y:

$\begin{aligned}y_1 &= {z\over2}\\\\y_2 &= 2z\\\\E(z) &= {1\over2}2y_1 + {1\over2}({y_2\over2})\\\\&={1\over2}({2z\over2}) + {1\over2}({2z\over2})\\\\&=z=2y_1= {y_2\over2}\end{aligned}$

And as such it is not more rational to switch.

It is a mathematical fallacy that leads to the conclusion that $E(z) = {5\over4}y$, much like the fallacy that covertly divides by zero to prove that $1 = 2$.

The trouble is on the RHS, the probability that I am being interviewed given that my coin was heads. — Srap Tasmaner

Which is 1. I know that I will be interviewed if the coin lands heads.

Consider a simpler version of the experiment. If heads then I will be interviewed once. If tails then I will be interviewed once. Bayes' theorem is unproblematically:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

We don't say that because half of all questions are when tails then P(Questioned | Heads) = 0.5. That would give us this very clearly wrong calculation:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over2}*{1\over2}}\over1}\\&={1\over4}\end{aligned}$

What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner

Yes, that's the left hand side of the theorem that we're trying to solve: P(Heads|Questioned).

We use the known values on the right hand side to determine it.

And how do you expect to apply Bayes's rule without any base rate information? SB can reason as I have described to determine what those base rates would be were the experiment repeated a number of times, and set her subjective probabilities accordingly. — Srap Tasmaner

We have the base rate information.

The probability of a coin landing heads is 0.5. The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. So:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Compare with Bayes' theorem as applied to my variation:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

It should be clear that the reasoning is sound in the second case, and so too is it sound in the first case.

I don't think your numbers are accurate there.

But in this case we're not asking about an outsider's analysis of frequencies over many experiments, but Sleeping Beauty's when it is known that just a single experiment is being run.

The absence of that thing is informative, it amounts to "it was tails or this is your first interview," and this is true as well for stock SB. Being asked is itself information you can condition on. — Srap Tasmaner

How do you condition on such a thing? What values do you place into Bayes' theorem?

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\end{aligned}$

I switched back shortly after that post; it's right there in the thread. The argument that convinced me was this: consider a variation, "Informative SB", in which Beauty is told she will be awakened twice either way, but if it was heads she will be told at the second interview that it was heads and that this is her second interview; at none of the others will she be given such information. — Srap Tasmaner

I don't understand how this is different to the original. In the original she's woken up on Wednesday and told the result of the coin flip (whether heads or tails). So there are two interviews if heads, three interviews if tails.

All the average return on the envelope tells us is that if we keep playing over the long term, its going to average a return greater than one. — Philosophim

That's the supposed paradox. Switching doesn't increase our expected return, but the reasoning given suggests that it does. So we need to make sense of this contradiction.

Possible answer: because the material conditions for US citizens are conducive to him getting a platform. If the USA didn't have so many poor, didn't have so many people one healthcare invoice away from being poor, then nobody would take Trump seriously. — Benkei

If economic issues were the concern then they'd be voting for Democrats.

It's clearly social issues (the "culture war") that elicit support for Trump and the Republicans.

Given that the coin flip just determines if a Tuesday interview will happen, Beauty is being asked for her credence that a Tuesday interview will happen.

That she's interviewed twice as often if a Tuesday interview will happen isn't that a Tuesday interview will happen twice as often.

That she's interviewed twice as often if it's tails isn't that it's tails twice as often.

If it helps, consider that the coin toss happens after the Monday interview.

No, I'm in the double halfer camp now. The post right above explains my current thinking.

((This is, I don't know, maybe the third time I've argued with Michael about something and then concluded he was right all along.)) — Srap Tasmaner

So you've switched back to being a thirder?

This is straightforwardly true, but from the perspective of an observer of the experiment. But to answer the problem you must adopt SB's perspective. That makes all the difference. — hypericin

Why does it make a difference?

But SB is asked on every wakening, and is woken twice as much on tails. This must influence the odds — hypericin

Why would it influence the odds?

Let's say that I wanted to bet on a coin toss. I bet £100 that it will be tails. To increase the odds that it's tails, I ask you to put me to sleep, wake me up, put me back to sleep, wake me up, put me back to sleep, wake me up, and so on. Does that make any sense?

Influences the gambling odds, even though the coin toss is fair in both cases — hypericin

It doesn't influence the odds. You just get to bet twice, hence twice the payout.

Consider a coin toss.

P(Heads) = 0.5
P(Tails) = 0.5

I bet £1 on a coin toss. If I bet correctly I will win £1. What is my expected return if I bet on heads?

There's a 0.5 chance of losing £1 and a 0.5 chance of winning £1, so the expected return is:

(0.5 * -1) + (0.5 * 1) = 0

So there's no point in betting.

But if I can win £1.50 for betting correctly then the expected return is:

(0.5 * -1) + (0.5 * 1.5) = 0.25

So it's rational to bet.

In the case of the two envelopes paradox, the claim is that there's a 0.5 chance of doubling my winnings and a 0.5 chance of halving my winnings, and so the expected return is:

(0.5 * 2y) + (0.5 * y/2) = (5/4)y

I'm not sure how you're getting the equation that you are, not only in what you've done, but by what the equation stands for. — Philosophim

Then you should read this, this, and this.

So there's no possible world in which the other envelope may with equal probability contain either two fifty or a tenner, which is the only way to make a profit by switching. — Baden

The issue isn't with probability. Given that there's £10 in my envelope, the probability that the other envelope contains £5 is equal to the probability that the other envelope contains £20; that probability being 1/2.

As an analogy, if I flip a coin and hide it from you, it is correct for you to say that the probability that it landed heads is equal to the probability that it landed tails; that probability being 1/2.

The error made in the paradox is with it's calculation of the expected value of switching. It claims that the expected value is 0.5 * 2y + 0.5 * y/2, which is technically correct, but what it fails to explain is that it's 0.5 * 2y where y = x/2 + 0.5 * y/2 where y = 2x.

So a more accurate formulation is:

$E(x) = {1\over2}({2x\over2}) + {1\over2}({2x\over2}) = {x\over2} + {x\over2} = x$

Here's a Venn diagram to show how the probabilities interlink (where "Monday" should be read as "she will be woken on Monday" and "Tuesday” as "she will be woken on Tuesday"):



The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.

As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.

A variation of my variation.

Let's say that there are two beauties; Michael and Jane. They are put to sleep and assigned a random number from {1, 2}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

If we use Bayes' theorem then:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over2}*{1\over2}}\over{1\over2}}\\&={1\over2}\end{aligned}$

The number assignment and the coin toss do not affect each other, and so they are independent events.

But there is something different about my example. If we use Bayes' theorem then:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

Given that P(Heads|Awake) ≠ P(Heads), waking does in fact provide new relevant evidence. And the above shows that the correct answer is 1/3.

Applying this same formula to the original problem gives P(Heads|Awake) = P(Heads), and so is consistent with Lewis' reasoning and his conclusion that the correct answer is 1/2.

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

David Lewis' answer, summarised here, states that:

Sleeping Beauty receives no new non-self-locating information throughout the experiment because she is told the details of the experiment. Since her credence before the experiment is P(Heads) = 1/2, she ought to continue to have a credence of P(Heads) = 1/2 since she gains no new relevant evidence when she wakes up during the experiment.

Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment.

However, given that if woken he is 1 iff the coin landed heads, he ought to have a credence of either P(Heads) = 1/3 or P(1) = 1/2.

Do you think that when "they let you do it", it is assault? — NOS4A2

If they don't consent then yes. Letting someone do something and consenting to them doing something are different things. The former only implies that they don't object and/or resist.

For example, you let the government tax you, but you've made it clear in the past that you don't consent to it.

And even if Trump meant it in the sense of "consent", that he says that they do isn't that they do, only that he thinks that they do. If he just assumes consent and so "just starts kissing them ... [without] even wait[ing]" then it's assault.

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). — fdrake

Given my previous post that shows that under S5 we cannot assume that ◇∃x□Px is true for any logically consistent Px, the third alternative is that “possibly necessary” under S5 means something different to what it means under other systems (or natural English).

You can call NOS4A2 a moronic rape apologist.

So if anything it's a reductio ad absurdum against the assumption that ◇∃x□Fx is true for every logically consistent Fx. — Michael

On this point, consider this:

1. ∃xFx → ∃x∀y(Fy ↔ (x = y))
2. ◇∃x□(Fx ∧ Ax) ∴ ∃x□(Fx ∧ Ax)
3. ◇∃x□(Fx ∧ ¬Ax) ∴ ∃x□(Fx ∧ ¬Ax)

The first premise asserts that if there is an x such that Fx then there is exactly one x such that Fx. The second asserts that there is an x such that it is necessary that both Fx and Ax. The third asserts that there is an x such that it is necessary that both Fx and ¬Ax. Obviously this is a contradiction.

If we take Fx to mean something like "x is the sole creator of the world" then 1 is true, and as both 2 and 3 are valid under S5 it must be that one or both antecedents are false, and so one or both of these is true:

4. ¬◇∃x□(Fx ∧ Ax)
5. ¬◇∃x□(Fx ∧ ¬Ax)

Therefore we cannot assume that ◇∃x□Px is true for any logically consistent Px, and so cannot assume that it is possible that something necessarily created the world.

Or we have to reject S5, but if we reject S5 then the modal ontological argument is invalid as “possibly necessary” wouldn’t entail “necessary”.

Gets found liable for defamation, proceeds to defame on TV the following day.

He just invites his own problems. Either a second lawsuit or more evidence for the second trial if he gets an appeal.

The jury disagreed.

What evidence? There was no evidence of either rape or battery. But they went with one and not the other, for whatever reason. — NOS4A2

I wasn't at the trial so I don't know. You'd have to ask the jury.

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. — fdrake

I did offer a more substantial notion of God here. The argument attempts to show that there exists something which necessarily created the world.

But you are right that we can posit any entity. So if anything it's a reductio ad absurdum against the assumption that ◇∃x□Fx is true for every logically consistent Fx.

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? — fdrake

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.

They disagreed with her rape accusation. So it’s clear they thought she was lying. — NOS4A2

Clearly they thought that either a) the evidence of what Trump did didn't satisfy the legal definition of rape or b) the evidence of rape wasn't a "preponderance" of evidence.

I’m quite aware he lost. But the fact remains there is no evidence of his supposed crimes. — NOS4A2

Clearly the jury disagreed. There was more evidence that he was guilty than there was evidence that Carroll and the others were lying.

So, Trump didn't object to an anonymous jury, didn't testify, missed the deadline for providing DNA evidence, and didn't offer his own defence witnesses.

There's no grand unjust conspiracy here @NOS4A2. He just lost.

Incidentally, https://s3.documentcloud.org/documents/23721313/kaplan-ruling-in-carroll-v-trump-jury.pdf

The trial of this case will begin on April 25, 2023. On March 11, 2023, the Court
directed the parties to file any objections to trying the case before an anonymous jury. Neither objected.

Yes, the sordid fears of a New York judge take precedence over an individual’s right to an impartial jury. — NOS4A2

Anonymous doesn't mean not impartial.

That seems to be why he lost, not because E. Jean Carrol established anything beyond a reasonable doubt. — NOS4A2

In a civil trial the requirement is a preponderance of evidence, not proof beyond reasonable doubt.

Because it was Donald Trump. — NOS4A2

Yes, the jury were especially at risk because it was Donald Trump who was being tried. If he wasn't such a twat and if his supporters weren't such psychos then the jury wouldn't have been kept anonymous.

https://www.reuters.com/legal/trump-face-anonymous-jury-high-profile-new-york-defamation-trial-2023-03-23/

Kaplan said the need for juror anonymity reflected the "unprecedented circumstances in which this trial will take place, including the extensive pretrial publicity and a very strong risk that jurors will fear harassment, unwanted invasions of their privacy, and retaliation."

...

In his decision, Kaplan cited Trump's March 18 call for protest if he were indicted in a Manhattan's district attorney case for covering up a hush money payment to porn star Stormy Daniels before the 2016 election.

Kaplan said Trump's reaction "has been perceived by some as an incitement to violence," and said some people charged over the Jan. 6, 2021, attack on the U.S. Capitol "rightly or wrongly" attributed their actions to incitement by Trump.

The judge also said Trump has "repeatedly" attacked courts, judges, law enforcement and even individual jurors.

These, the judge said, included the forepersons of the grand jury looking into whether Trump tried to sway the 2020 election results in Georgia, and the jury at longtime Trump adviser Roger Stone's 2019 obstruction trial.

"If jurors' identities were disclosed, there would be a strong likelihood of unwanted media attention to the jurors, influence attempts, and/or of harassment or worse of jurors by supporters of Mr. Trump," Kaplan wrote.

Trump only has himself to blame.

So then why was it not allowed in this case? — NOS4A2

Because it doesn't apply to States, although most States choose to follow it to some extent. In this case the judge ruled that the danger posed to the jury warranted them remaining anonymous.

Although actually, I'm not sure this response is even warranted. Even if our "ordinary" concept of God is one that isn't to be interpreted as being modally necessary, the modal ontological argument does posit such a God, and their argument should be addressed on that premise.

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

Then why would they have voir dire in civil cases? — NOS4A2

Because of the 7th Amendment?

Edit: Although it doesn't apply to States, but many States choose to follow it anyway.